1. Prove the following trigonometric identities. (a) [1...

Transcript

00:01 When we approach this problem, the cosine of x plus y times the cosine of x minus y.

00:07 And we have been asked to prove that it's equal to 2x minus sine of 2y.

00:16 Based on some of the work that i've done down below here, i can tell you that this is not actually a trick identity.

00:23 These two things are not equal.

00:25 I think there's a typo in what was provided.

00:28 However, within that typo, if we look at some of the things that we can do, help ourselves get close or to start to see where that, what options you have in that, i would start with using this cosine of x plus y times cosine of x minus y.

00:46 I used the sum formula and i used the difference formula to rewrite.

01:00 Then after you've used a trig identity, it's wise to use an algebraic step.

01:06 So i went ahead and i multiplied.

01:08 And when i multiply these difference of perfect squares, you're going to have those middle terms that end up adding to zero.

01:16 So i really only had to multiply the firsts and the lasts, the first terms and the last terms.

01:22 I ended up with cosine squared of x times cosine square of y minus sine square of x times sine squared of y.

01:31 And this is when i noticed that there had to be an error in the way this question was presented.

01:36 But i went ahead and i tried a couple of things just to see where i ended up to see if perhaps i was missing something.

01:43 I went ahead and i used the half angle formulas here on each of these x's and y's...

Question

1. Prove the following trigonometric identities. (a) [1 point] \cos^2 y - \sin^2 x = \cos^2 x - \sin^2 y. (b) [2 points] \cos(x - y)\cos(x + y) = \cos^2 x - \sin^2 y.

Question

1. Prove the following trigonometric identities. (a) [1 point] \cos^2 y - \sin^2 x = \cos^2 x - \sin^2 y. (b) [2 points] \cos(x - y)\cos(x + y) = \cos^2 x - \sin^2 y.

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