Home / Science, Engineering & Maths / Maths / Advanced Precalculus: Geometry, Trigonometry and Exponentials / It’s your turn on trigonometric identities
It’s your turn on trigonometric identities
solutions to the exercises
Hello. Welcome to the section “It’s Your Turn on Trigonometric Identities”. Let us start with the first point. Then we have to compute the sine of pi over 4 plus pi over 3. OK. This is equal to the sine of the first angle times the cosine of the second angle plus the cosine of the first angle times the sine of the second angle.
OK. And therefore, we get the sine of pi over 4, which is the square root of 2 over 2, times the cosine of pi over 3, which is 1 over 2, plus the cosine of pi over 4, which is the square root of 2 over 2, times the sine of pi over 3, which is the square root of 3 over 2. Therefore, we get 4 as denominator. And we have the square root of 2 plus the square root of 6. OK. Now, when we have to compute now the sine of the difference between pi over 4 and pi over 3, OK, you have not to think to a different, a new law.
But you can apply exactly the same law as before, thinking at this like what? To the sine of pi over 4 plus the angle minus pi over 3. And then, what you get? Now, the sine of the first angle times the cosine of the second angle plus the cosine of the first angle times the sine of the second angle.
OK. And now, the sine of pi over 4 is the square root of 2 over 2. The cosine is an even function. Therefore, the cosine of minus pi over 3 is equal to the cosine of pi over 3, which is 1 over 2, plus the cosine of pi over 4, which is the square root of 2 over 2 times– now, the sine is an odd function. Therefore, the sine of minus pi over 3 is minus the sine of pi over 3. The sine of pi over 3 is the square root of 3 over 2. Therefore, here we get minus the square root of 3 over 2.
Then our computation ends with 4 at the denominator and the square root of 2 minus the square root of 6 at the numerator. OK. And now let us move to the second point of the exercise. OK. We have first to compute the cosine of pi over 6 plus 2 pi over 3. OK. This is equal to the cosine of pi over 6 times the cosine of the second angle minus the sine of the first angle times the sine of the second angle.
The cosine of pi over 6 is the square root of 3 over 2. Now, the cosine of 2 pi over 3 is equal to the cosine of pi minus pi over 3. And therefore, the cosine of 2 pi over 3 is equal to minus the cosine of pi over 3, which is minus 1 over 2.
And then I have minus the sine of pi over 6. The sine of pi over 6 is 1 over 2. And now the sine of 2 pi over 3 is the sine, again, of pi minus pi over 3, which is equal to the sine of pi over 3, and which is the square root of 3 over 2. OK. Therefore, we get again 4 at the denominator. And we have minus the square root of 3 minus the square root of 3. Therefore, minus 2 times the square root of 3. That is, minus the square root of 3 over 2.
And now, let us compute the cosine of the difference of these two angles, pi over 6 minus 2 times pi over 3. Again, as before, we have not to remember a new rule, but we can just apply the same rule as before for the cosine of a sum of the angles to what? OK. We can think to the cosine of pi over 6 plus the angle minus 2 pi over 3. OK. Then this is equal to the cosine of the first angle times the cosine of the second angle minus the sine of the first angle times the sine of the second angle.
OK. And we get the cosine of pi over 6, which is the square root of 3 over 2.
Then, the cosine of minus 2 pi over 3. OK. The cosine is an even function. Therefore, this is equal to the cosine of 2 pi over 3, which is, as before, the cosine of pi minus pi over 3, which is minus the cosine of pi over 3. And then we get minus 1 over 2. OK. Then minus the sine of pi over 6, which is 1 over 2, times the sine of minus 2 pi over 3. OK. The sine is an odd function. Therefore, this is equal to minus the sine of 2 pi over 3.
And the sine of 2 pi over 3 is equal to the sine of pi minus pi over 3, which is the square root of 3 over 2. Therefore, we get minus the square root of 3 over 2. And then we can conclude. And then we get 4 at the denominator. And on the top, what we have? Minus the square root of 3 minus minus, therefore plus, the square root of 3. And we get 0. Observe, we could immediately realize that the result will be 0, because pi over 6 minus 2 pi over 3 is equal to what? 6 pi minus 4 times pi over 6. That is minus 3 pi over 6, which is minus pi over 2.
And we know that in the angle minus pi over 2, the cosine is equal to 0. OK. Thank you very much for your attention.
Do your best in trying to solve the following problems. In any case some of them are solved in the video and all of them are solved in the pdf file below.
Exercise 1.
Compute the following values [1) sin(pi/4+pi/3) text{ and } sin(pi/4-pi/3);] [2) cos(pi/6+2pi/3) text{ and } cos(pi/6-2pi/3).]
Exercise 2.
Simplify the following expressions [1) dfrac{sin^4 x-cos^4 x}{sin^2 x-cos^2 x};] [2) dfrac{1}{cos x(1+tan^2x)}.]
Exercise 3.
Prove the following identities [1) cos^4x-sin^4x=cos(2x);] [2) dfrac{tan x}{1+tan^2x}=sin xcos x;] [3) sin xcos xtan x=1-cos^2x;] [4) dfrac{sin x}{1-cos x}+dfrac{1-cos x}{sin x}=dfrac 2{sin x};] [5) dfrac{sin x-cos x}{sin x+cos x}=-dfrac{cos 2x}{1+sin 2x}.]
This article is from the online course:
Advanced Precalculus: Geometry, Trigonometry and Exponentials
Created by
This article is from the free online
Advanced Precalculus: Geometry, Trigonometry and Exponentials
Reach your personal and professional goals
Unlock access to hundreds of expert online courses and degrees from top universities and educators to gain accredited qualifications and professional CV-building certificates.
Join over 18 million learners to launch, switch or build upon your career, all at your own pace, across a wide range of topic areas.
