In this article we will talk about universal trigonometric substitution. It involves expressing the sine, cosine, tangent and cotangent of any angle through the tangent of a half angle. Moreover, such a replacement is carried out rationally, that is, without roots.
First, we will write down formulas expressing sine, cosine, tangent and cotangent in terms of the tangent of a half angle. Next we will show the derivation of these formulas. In conclusion, let's look at a few examples of using the universal trigonometric substitution.
Page navigation.
Sine, cosine, tangent and cotangent through the tangent of a half angle
First, let's write down four formulas expressing sine, cosine, tangent and cotangent of an angle through the tangent of a half angle.
The indicated formulas are valid for all angles at which the tangents and cotangents included in them are defined:
Deriving formulas
Let us analyze the derivation of formulas expressing sine, cosine, tangent and cotangent of an angle through the tangent of a half angle. Let's start with the formulas for sine and cosine.
Let's represent sine and cosine using the double angle formulas as 
And 
respectively. Now the expressions 
And 
we write it in the form of fractions with a denominator of 1 as 
And 
. Next, based on the main trigonometric identity, we replace the units in the denominator with the sum of the squares of sine and cosine, after which we get 
And 
. Finally, we divide the numerator and denominator of the resulting fractions by (its value is different from zero provided 
). As a result, the entire chain of actions looks like this: 
And 
This completes the derivation of formulas expressing sine and cosine through the tangent of a half angle.
It remains to derive formulas for tangent and cotangent. Now, taking into account the formulas obtained above, both formulas and 
, we immediately obtain formulas expressing the tangent and cotangent through the tangent of the half angle: 
So, we have derived all the formulas for the universal trigonometric substitution.
Examples of using universal trigonometric substitution
First, let's look at an example of using universal trigonometric substitution when transforming expressions.
Example.
Give an expression 
to an expression containing only one trigonometric function.
Solution.
Answer:
.
Bibliography.
- Algebra: Textbook for 9th grade. avg. school/Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova; Ed. S. A. Telyakovsky.- M.: Education, 1990.- 272 p.: ill.- isbn 5-09-002727-7
- Bashmakov M. I. Algebra and the beginnings of analysis: Textbook. for 10-11 grades. avg. school - 3rd ed. - M.: Education, 1993. - 351 p.: ill. - ISBN 5-09-004617-4.
- Algebra and the beginning of analysis: Proc. for 10-11 grades. general education institutions / A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn and others; Ed. A. N. Kolmogorov. - 14th ed. - M.: Education, 2004. - 384 pp.: ill. - ISBN 5-09-013651-3.
- Gusev V. A., Mordkovich A. G. Mathematics (a manual for those entering technical schools): Proc. allowance.- M.; Higher school, 1984.-351 p., ill.
Cosine of the sum and difference of two angles
In this section the following two formulas will be proved:
cos (α + β) = cos α cos β - sin α sin β, (1)
cos (α - β) = cos α cos β + sin α sin β. (2)
The cosine of the sum (difference) of two angles is equal to the product of the cosines of these angles minus (plus) the product of the sines of these angles.
It will be more convenient for us to start with the proof of formula (2). For simplicity of presentation, let us first assume that the angles α And β satisfy the following conditions:
1) each of these angles is non-negative and less 2π:
0 < α <2π, 0< β < 2π;
2) α > β .
Let the positive part of the 0x axis be the common starting side of the angles α And β .
We denote the end sides of these angles by 0A and 0B, respectively. Obviously the angle α - β can be considered as the angle by which beam 0B needs to be rotated around point 0 counterclockwise so that its direction coincides with the direction of beam 0A.
On rays 0A and 0B we mark points M and N, located at a distance of 1 from the origin of coordinates 0, so that 0M = 0N = 1.
In the x0y coordinate system, point M has coordinates ( cos α, sin α), and point N is the coordinates ( cos β, sin β). Therefore, the square of the distance between them is:
d 1 2 = (cos α - cos β) 2 + (sin α - sin β) 2 = cos 2 α - 2 cos α cos β +
+ cos 2 β + sin 2 α - 2sin α sin β + sin 2 β = .
In our calculations we used the identity
sin 2 φ + cos 2 φ = 1.
Now consider another coordinate system B0C, which is obtained by rotating the 0x and 0y axes around point 0 counterclockwise by an angle β .
In this coordinate system, point M has coordinates (cos ( α - β ), sin ( α - β )), and the point N is coordinates (1,0). Therefore, the square of the distance between them is:
d 2 2 = 2 + 2 = cos 2 (α - β) - 2 cos (α - β) + 1 +
+ sin 2 (α - β) = 2 .
But the distance between points M and N does not depend on which coordinate system we are considering these points in relation to. That's why
d 1 2 = d 2 2
2 (1 - cos α cos β - sin α sin β) = 2 .
This is where formula (2) follows.
Now we should remember those two restrictions that we imposed for simplicity of presentation on the angles α And β .
The requirement that each of the corners α And β was non-negative, not really significant. After all, to any of these angles you can add an angle that is a multiple of 2, which will not affect the validity of formula (2). In the same way, from each of these angles you can subtract an angle that is a multiple of 2π. Therefore we can assume that 0 < α < 2π, 0 < β < 2π.
The condition also turns out to be insignificant α > β . Indeed, if α < β , That β >α ; therefore, given the parity of the function cos X , we get:
cos (α - β) = cos (β - α) = cos β cos α + sin β sin α,
which essentially coincides with formula (2). So the formula
cos (α - β) = cos α cos β + sin α sin β
true for all angles α And β . In particular, replacing in it β on - β and given that the function cosX is even, and the function sinX odd, we get:
cos (α + β) = cos [α - (- β)] = cos α cos (-β) + sin α sin (-β) =
= cos α cos β - sin α sin β,
which proves formula (1).
So, formulas (1) and (2) are proven.
Examples.
1) cos 75° = cos (30° + 45°) = cos 30° cos 45°-sin 30°-sin 45° =
2) cos 15° = cos (45° - 30°) = cos 45° cos 30° + sin 45° sin 30° =
Exercises
1 . Calculate without using trigonometric tables:
a) cos 17° cos 43° - sin 17° sin 43°;
b) sin 3° sin 42° - cos 39° cos 42°;
c) cos 29° cos 74° + sin 29° sin 74°;
d) sin 97° sin 37° + cos 37° cos 97°;
e) cos 3π / 8 cos π / 8 + sin 3π / 8 sin π / 8 ;
e) sin 3π / 5 sin 7π / 5 - cos 3π / 5 cos 7π / 5 .
2.Simplify expressions:
a). cos( α + π/3 ) + cos(π/3 - α ) .
b). cos (36° + α ) cos (24° - α ) + sin (36° + α ) sin ( α - 24°).
V). sin(π/4 - α ) sin (π / 4 + α ) - cos (π / 4 + α ) cos (π / 4 - α )
d) cos 2 α + tg α sin 2 α .
3 . Calculate :
a) cos(α - β), If
cos α = - 2 / 5 , sin β = - 5 / 13 ;
90°< α < 180°, 180° < β < 270°;
b) cos ( α + π / 6), if cos α = 0,6;
3π/2< α < 2π.
4 . Find cos(α + β) and cos (α - β) ,if it is known that sin α = 7 / 25, cos β = - 5 / 13 and both angles ( α And β ) end in the same quarter.
5 .Calculate:
A). cos [ arcsin 1 / 3 + arccos 2 / 3 ]
b). cos [ arcsin 1 / 3 - arccos (- 2 / 3)] .
V). cos [ arctan 1 / 2 + arccos (- 2) ]
The concepts of sine (), cosine (), tangent (), cotangent () are inextricably linked with the concept of angle. To understand well these, at first glance, complex concepts(which cause a state of horror in many schoolchildren), and to make sure that “the devil is not as scary as he is painted,” let’s start from the very beginning and understand the concept of an angle.
Angle concept: radian, degree
Let's look at the picture. The vector has “turned” relative to the point by a certain amount. So the measure of this rotation relative to the initial position will be corner.
What else do you need to know about the concept of angle? Well, of course, angle units!
Angle, in both geometry and trigonometry, can be measured in degrees and radians.
An angle of (one degree) is called central angle in a circle, based on a circular arc equal to part of the circle. Thus, the entire circle consists of “pieces” of circular arcs, or the angle described by the circle is equal.
That is, the figure above shows an angle equal to, that is, this angle rests on a circular arc the size of the circumference.
An angle in radians is the central angle in a circle subtended by a circular arc whose length is equal to the radius of the circle. Well, did you figure it out? If not, then let's figure it out from the drawing.
So, the figure shows an angle equal to a radian, that is, this angle rests on a circular arc, the length of which is equal to the radius of the circle (the length is equal to the length or the radius is equal to the length of the arc). Thus, the arc length is calculated by the formula:
Where is the central angle in radians.
Well, knowing this, can you answer how many radians are contained in the angle described by the circle? Yes, for this you need to remember the formula for circumference. Here she is:
Well, now let’s correlate these two formulas and find that the angle described by the circle is equal. That is, by correlating the value in degrees and radians, we get that. Respectively, . As you can see, unlike "degrees", the word "radian" is omitted, since the unit of measurement is usually clear from the context.
How many radians are there? That's right!
Got it? Then go ahead and fix it:
Having difficulties? Then look answers:
Right triangle: sine, cosine, tangent, cotangent of angle
So, we figured out the concept of an angle. But what is sine, cosine, tangent, and cotangent of an angle? Let's figure it out. For this it will help us right triangle.
What are the sides of a right triangle called? That's right, hypotenuse and legs: the hypotenuse is the side that lies opposite the right angle (in our example this is the side); the legs are the two remaining sides and (those adjacent to the right angle), and if we consider the legs relative to the angle, then the leg is the adjacent leg, and the leg is the opposite. So, now let’s answer the question: what are sine, cosine, tangent and cotangent of an angle?
Sine of angle- this is the ratio of the opposite (distant) leg to the hypotenuse.
In our triangle.
Cosine of angle- this is the ratio of the adjacent (close) leg to the hypotenuse.
In our triangle.
Tangent of the angle- this is the ratio of the opposite (distant) side to the adjacent (close).
In our triangle.
Cotangent of angle- this is the ratio of the adjacent (close) leg to the opposite (far).
In our triangle.
These definitions are necessary remember! To make it easier to remember which leg to divide into what, you need to clearly understand that in tangent And cotangent only the legs sit, and the hypotenuse appears only in sinus And cosine. And then you can come up with a chain of associations. For example, this one:
Cosine→touch→touch→adjacent;
Cotangent→touch→touch→adjacent.
First of all, you need to remember that sine, cosine, tangent and cotangent as the ratios of the sides of a triangle do not depend on the lengths of these sides (at the same angle). Do not believe? Then make sure by looking at the picture:
Consider, for example, the cosine of an angle. By definition, from a triangle: , but we can calculate the cosine of an angle from a triangle: . You see, the lengths of the sides are different, but the value of the cosine of one angle is the same. Thus, the values of sine, cosine, tangent and cotangent depend solely on the magnitude of the angle.
If you understand the definitions, then go ahead and consolidate them!
For the triangle shown in the figure below, we find.
Well, did you get it? Then try it yourself: calculate the same for the angle.
Unit (trigonometric) circle
Understanding the concepts of degrees and radians, we considered a circle with a radius equal to. Such a circle is called single. It will be very useful when studying trigonometry. Therefore, let's look at it in a little more detail.
As you can see, this circle is constructed in the Cartesian coordinate system. The radius of the circle is equal to one, while the center of the circle lies at the origin of coordinates, the initial position of the radius vector is fixed along the positive direction of the axis (in our example, this is the radius).
Each point on the circle corresponds to two numbers: the axis coordinate and the axis coordinate. What are these coordinate numbers? And in general, what do they have to do with the topic at hand? To do this, we need to remember about the considered right triangle. In the figure above, you can see two whole right triangles. Consider a triangle. It is rectangular because it is perpendicular to the axis.
What is the triangle equal to? That's right. In addition, we know that is the radius of the unit circle, which means . Let's substitute this value into our formula for cosine. Here's what happens:
What is the triangle equal to? Well, of course, ! Substitute the radius value into this formula and get:
So, can you tell what coordinates a point belonging to a circle has? Well, no way? What if you realize that and are just numbers? Which coordinate does it correspond to? Well, of course, the coordinates! And what coordinate does it correspond to? That's right, coordinates! Thus, period.
What then are and equal to? That's right, let's use the corresponding definitions of tangent and cotangent and get that, a.
What if the angle is larger? For example, like in this picture:
What has changed in this example? Let's figure it out. To do this, let's turn again to a right triangle. Consider a right triangle: angle (as adjacent to an angle). What are the values of sine, cosine, tangent and cotangent for an angle? That's right, we adhere to the corresponding definitions of trigonometric functions:
Well, as you can see, the value of the sine of the angle still corresponds to the coordinate; the value of the cosine of the angle - the coordinate; and the values of tangent and cotangent to the corresponding ratios. Thus, these relations apply to any rotation of the radius vector.
It has already been mentioned that the initial position of the radius vector is along the positive direction of the axis. So far we have rotated this vector counterclockwise, but what happens if we rotate it clockwise? Nothing extraordinary, you will also get an angle of a certain value, but only it will be negative. Thus, when rotating the radius vector counterclockwise, we get positive angles, and when rotating clockwise - negative.
So, we know that a whole revolution of the radius vector around a circle is or. Is it possible to rotate the radius vector to or to? Well, of course you can! In the first case, therefore, the radius vector will make one full revolution and stop at position or.
In the second case, that is, the radius vector will make three full revolutions and stop at position or.
Thus, from the above examples we can conclude that angles that differ by or (where is any integer) correspond to the same position of the radius vector.
The figure below shows an angle. The same image corresponds to the corner, etc. This list can be continued indefinitely. All these angles can be written by the general formula or (where is any integer)
Now, knowing the definitions of the basic trigonometric functions and using the unit circle, try to answer what the values are:
Here's a unit circle to help you:
Having difficulties? Then let's figure it out. So we know that:
From here, we determine the coordinates of the points corresponding to certain angle measures. Well, let's start in order: the angle at corresponds to a point with coordinates, therefore:
Does not exist;
Further, adhering to the same logic, we find out that the corners in correspond to points with coordinates, respectively. Knowing this, it is easy to determine the values of trigonometric functions at the corresponding points. Try it yourself first, and then check the answers.
Answers:
Does not exist
Does not exist
Does not exist
Does not exist
Thus, we can make the following table:
There is no need to remember all these values. It is enough to remember the correspondence between the coordinates of points on the unit circle and the values of trigonometric functions:
But the values of the trigonometric functions of angles in and, given in the table below, must be remembered:
Don't be scared, now we'll show you one example quite simple to remember the corresponding values:
To use this method, it is vital to remember the values of the sine for all three measures of angle (), as well as the value of the tangent of the angle. Knowing these values, it is quite simple to restore the entire table - the cosine values are transferred in accordance with the arrows, that is:
Knowing this, you can restore the values for. The numerator " " will match and the denominator " " will match. Cotangent values are transferred in accordance with the arrows indicated in the figure. If you understand this and remember the diagram with the arrows, then it will be enough to remember all the values from the table.
Coordinates of a point on a circle
Is it possible to find a point (its coordinates) on a circle, knowing the coordinates of the center of the circle, its radius and angle of rotation?
Well, of course you can! Let's get it out general formula to find the coordinates of a point.
For example, here is a circle in front of us:
We are given that the point is the center of the circle. The radius of the circle is equal. It is necessary to find the coordinates of a point obtained by rotating the point by degrees.
As can be seen from the figure, the coordinate of the point corresponds to the length of the segment. The length of the segment corresponds to the coordinate of the center of the circle, that is, it is equal. The length of a segment can be expressed using the definition of cosine:
Then we have that for the point coordinate.
Using the same logic, we find the y coordinate value for the point. Thus,
So, in general view coordinates of points are determined by the formulas:
Coordinates of the center of the circle,
Circle radius,
The rotation angle of the vector radius.
As you can see, for the unit circle we are considering, these formulas are significantly reduced, since the coordinates of the center are equal to zero and the radius is equal to one:
Well, let's try out these formulas by practicing finding points on a circle?
1. Find the coordinates of a point on the unit circle obtained by rotating the point on.
2. Find the coordinates of a point on the unit circle obtained by rotating the point on.
3. Find the coordinates of a point on the unit circle obtained by rotating the point on.
4. The point is the center of the circle. The radius of the circle is equal. It is necessary to find the coordinates of the point obtained by rotating the initial radius vector by.
5. The point is the center of the circle. The radius of the circle is equal. It is necessary to find the coordinates of the point obtained by rotating the initial radius vector by.
Having trouble finding the coordinates of a point on a circle?
Solve these five examples (or get good at solving them) and you will learn to find them!
1.
You can notice that. But we know what corresponds to a full revolution of the starting point. Thus, the desired point will be in the same position as when turning to. Knowing this, we find the required coordinates of the point:
2. The unit circle is centered at a point, which means we can use simplified formulas:
You can notice that. We know what corresponds to two full revolutions of the starting point. Thus, the desired point will be in the same position as when turning to. Knowing this, we find the required coordinates of the point:
Sine and cosine are table values. We recall their meanings and get:
Thus, the desired point has coordinates.
3. The unit circle is centered at a point, which means we can use simplified formulas:
You can notice that. Let's depict the example in question in the figure:
The radius makes angles equal to and with the axis. Knowing that the table values of cosine and sine are equal, and having determined that the cosine here takes a negative value and the sine takes a positive value, we have:
Such examples are discussed in more detail when studying the formulas for reducing trigonometric functions in the topic.
Thus, the desired point has coordinates.
4.
Angle of rotation of the radius of the vector (by condition)
To determine the corresponding signs of sine and cosine, we construct a unit circle and angle:
As you can see, the value, that is, is positive, and the value, that is, is negative. Knowing the tabular values of the corresponding trigonometric functions, we obtain that:
Let's substitute the obtained values into our formula and find the coordinates:
Thus, the desired point has coordinates.
5. To solve this problem, we use formulas in general form, where
Coordinates of the center of the circle (in our example,
Circle radius (by condition)
Angle of rotation of the radius of the vector (by condition).
Let's substitute all the values into the formula and get:
and - table values. Let’s remember and substitute them into the formula:
Thus, the desired point has coordinates.
SUMMARY AND BASIC FORMULAS
The sine of an angle is the ratio of the opposite (far) leg to the hypotenuse.
The cosine of an angle is the ratio of the adjacent (close) leg to the hypotenuse.
The tangent of an angle is the ratio of the opposite (far) side to the adjacent (close) side.
The cotangent of an angle is the ratio of the adjacent (close) side to the opposite (far) side.
The relationships between the basic trigonometric functions - sine, cosine, tangent and cotangent - are given trigonometric formulas. And since there are quite a lot of connections between trigonometric functions, this explains the abundance of trigonometric formulas. Some formulas connect trigonometric functions of the same angle, others - functions of a multiple angle, others - allow you to reduce the degree, fourth - express all functions through the tangent of a half angle, etc.
In this article we will list in order all the main trigonometric formulas, which are sufficient to solve the vast majority of trigonometry problems. For ease of memorization and use, we will group them by purpose and enter them into tables.
Page navigation.
Basic trigonometric identities
Basic trigonometric identities define the relationship between sine, cosine, tangent and cotangent of one angle. They follow from the definition of sine, cosine, tangent and cotangent, as well as the concept of the unit circle. They allow you to express one trigonometric function in terms of any other.
For a detailed description of these trigonometry formulas, their derivation and examples of application, see the article.
Reduction formulas
Reduction formulas follow from the properties of sine, cosine, tangent and cotangent, that is, they reflect the property of periodicity of trigonometric functions, the property of symmetry, as well as the property of shift by a given angle. These trigonometric formulas allow you to move from working with arbitrary angles to working with angles ranging from zero to 90 degrees.
The rationale for these formulas, a mnemonic rule for memorizing them and examples of their application can be studied in the article.
Addition formulas
Trigonometric addition formulas show how trigonometric functions of the sum or difference of two angles are expressed in terms of trigonometric functions of those angles. These formulas serve as the basis for deriving the following trigonometric formulas.
Formulas for double, triple, etc. angle
Formulas for double, triple, etc. angle (they are also called multiple angle formulas) show how trigonometric functions of double, triple, etc. angles () are expressed in terms of trigonometric functions of a single angle. Their derivation is based on addition formulas.
More detailed information is collected in the article formulas for double, triple, etc. angle
Half angle formulas
Half angle formulas show how trigonometric functions of a half angle are expressed in terms of the cosine of a whole angle. These trigonometric formulas follow from the double angle formulas.
Their conclusion and examples of application can be found in the article.
Degree reduction formulas
Trigonometric formulas for reducing degrees are designed to facilitate the transition from natural powers of trigonometric functions to sines and cosines in the first degree, but multiple angles. In other words, they allow you to reduce the powers of trigonometric functions to the first.
Formulas for the sum and difference of trigonometric functions
The main purpose formulas for the sum and difference of trigonometric functions is to go to the product of functions, which is very useful when simplifying trigonometric expressions. These formulas are also widely used in solving trigonometric equations, as they allow you to factor the sum and difference of sines and cosines.
Formulas for the product of sines, cosines and sine by cosine
The transition from the product of trigonometric functions to a sum or difference is carried out using the formulas for the product of sines, cosines and sine by cosine.
Copyright by cleverstudents
All rights reserved.
Protected by copyright law. No part of the www.site, including internal materials and external design, may not be reproduced in any form or used without the prior written permission of the copyright holder.
Trigonometric identities- these are equalities that establish a relationship between sine, cosine, tangent and cotangent of one angle, which allows you to find any of these functions, provided that any other is known.
tg \alpha = \frac(\sin \alpha)(\cos \alpha), \enspace ctg \alpha = \frac(\cos \alpha)(\sin \alpha)
tg \alpha \cdot ctg \alpha = 1
This identity says that the sum of the square of the sine of one angle and the square of the cosine of one angle is equal to one, which in practice makes it possible to calculate the sine of one angle when its cosine is known and vice versa.
When converting trigonometric expressions, this identity is very often used, which allows you to replace the sum of the squares of the cosine and sine of one angle with one and also perform the replacement operation in the reverse order.
Finding tangent and cotangent using sine and cosine
tg \alpha = \frac(\sin \alpha)(\cos \alpha),\enspace
These identities are formed from the definitions of sine, cosine, tangent and cotangent. After all, if you look at it, then by definition the ordinate y is a sine, and the abscissa x is a cosine. Then the tangent will be equal to the ratio \frac(y)(x)=\frac(\sin \alpha)(\cos \alpha), and the ratio \frac(x)(y)=\frac(\cos \alpha)(\sin \alpha)- will be a cotangent.
Let us add that only for such angles \alpha at which the trigonometric functions included in them make sense, the identities will hold, ctg \alpha=\frac(\cos \alpha)(\sin \alpha).
For example: tg \alpha = \frac(\sin \alpha)(\cos \alpha) is valid for angles \alpha that are different from \frac(\pi)(2)+\pi z, A ctg \alpha=\frac(\cos \alpha)(\sin \alpha)- for an angle \alpha other than \pi z, z is an integer.
Relationship between tangent and cotangent
tg \alpha \cdot ctg \alpha=1
This identity is valid only for angles \alpha that are different from \frac(\pi)(2) z. Otherwise, either cotangent or tangent will not be determined.
Based on the above points, we obtain that tg \alpha = \frac(y)(x), A ctg \alpha=\frac(x)(y). It follows that tg \alpha \cdot ctg \alpha = \frac(y)(x) \cdot \frac(x)(y)=1. Thus, the tangent and cotangent of the same angle at which they make sense are mutually inverse numbers.
Relationships between tangent and cosine, cotangent and sine
tg^(2) \alpha + 1=\frac(1)(\cos^(2) \alpha)- the sum of the square of the tangent of the angle \alpha and 1 is equal to the inverse square of the cosine of this angle. This identity is valid for all \alpha other than \frac(\pi)(2)+ \pi z.
1+ctg^(2) \alpha=\frac(1)(\sin^(2)\alpha)- the sum of 1 and the square of the cotangent of the angle \alpha is equal to the inverse square of the sine of the given angle. This identity is valid for any \alpha different from \pi z.
Examples with solutions to problems using trigonometric identities
Example 1
Find \sin \alpha and tg \alpha if \cos \alpha=-\frac12 And \frac(\pi)(2)< \alpha < \pi ;
Show solution
Solution
The functions \sin \alpha and \cos \alpha are related by the formula \sin^(2)\alpha + \cos^(2) \alpha = 1. Substituting into this formula \cos \alpha = -\frac12, we get:
\sin^(2)\alpha + \left (-\frac12 \right)^2 = 1
This equation has 2 solutions:
\sin \alpha = \pm \sqrt(1-\frac14) = \pm \frac(\sqrt 3)(2)
By condition \frac(\pi)(2)< \alpha < \pi . In the second quarter the sine is positive, so \sin \alpha = \frac(\sqrt 3)(2).
In order to find tan \alpha, we use the formula tg \alpha = \frac(\sin \alpha)(\cos \alpha)
tg \alpha = \frac(\sqrt 3)(2) : \frac12 = \sqrt 3
Example 2
Find \cos \alpha and ctg \alpha if and \frac(\pi)(2)< \alpha < \pi .
Show solution
Solution
Substituting into the formula \sin^(2)\alpha + \cos^(2) \alpha = 1 given number \sin \alpha=\frac(\sqrt3)(2), we get \left (\frac(\sqrt3)(2)\right)^(2) + \cos^(2) \alpha = 1. This equation has two solutions \cos \alpha = \pm \sqrt(1-\frac34)=\pm\sqrt\frac14.
By condition \frac(\pi)(2)< \alpha < \pi . In the second quarter the cosine is negative, so \cos \alpha = -\sqrt\frac14=-\frac12.
In order to find ctg \alpha , we use the formula ctg \alpha = \frac(\cos \alpha)(\sin \alpha). We know the corresponding values.
ctg \alpha = -\frac12: \frac(\sqrt3)(2) = -\frac(1)(\sqrt 3).
Loading...