1. ## trigo problem again

trigo problem again

2. Hello, afeasfaerw23231233!

You showed very few steps.
I assume you used the product-to-sum identities.

. . $\displaystyle \sin A\sin B \:=\:\frac{1}{2}\left[\cos(A-B) - \cos(A+B)\right]$
. . . $\displaystyle \cos A\cos B \:=\:\frac{1}{2}\left[\cos(A-B) + \cos(A + B)\right]$

29 (a) Prove that if $\displaystyle \tan\theta\tan(\theta + \alpha) \:=\:k$

. . then: .$\displaystyle (k+1)\cos(2\theta+\alpha) \;=\;(1-k)\cos\alpha$

We have: .$\displaystyle \frac{\sin\theta}{\cos\theta}\cdot\frac{\sin(\thet a+\alpha)}{\cos(\theta+\alpha)} \;=\;k$ . . $\displaystyle \Rightarrow\quad\sin\theta\!\cdot\!\sin(\theta +\alpha) \;=\;k\cdot\cos\theta\!\cdot\!\cos(\theta + \alpha)$

Then: .$\displaystyle \frac{1}{2}\left[\cos\alpha - \cos(2\theta+\alpha)\right] \;=\;k\cdot\frac{1}{2}\left[\cos\alpha + \cos(2\theta + \alpha)\right]$

. . . . . $\displaystyle \cos\alpha - \cos(2\theta +\alpha) \;=\;k\!\cdot\!\cos\alpha + k\!\cdot\!\cos(2\theta+\alpha)$

. . . . . $\displaystyle k\!\cdot\!\cos(2\theta+\alpha) + \cos(2\theta+\alpha) \;=\;\cos\alpha - k\!\cdot\!\cos\alpha$

Factor: . $\displaystyle (k+1)\cos(2\theta + \alpha) \;=\;(1-k)\cos\alpha$