# Thread: Solving and Proving Tri IDs

1. ## Solving and Proving Tri IDs

I'm given these two questions from a friend today, and i don't know how to solve them at all.

4 cos4xsin4x+1=0
where the range is -3pi ≤ x ≤ 2 pi

All i know is that you're suppose to subsitute the 4x as x
and thats basically it. any help?

The second question i got was proofing Ids

(Root(2)/2) ( cos pi/12 - sin pi/12) = 1/2

How am i suppose to solve that?

Thank you so much for helping me :]

2. Originally Posted by TeriyakiDonnQ
I'm given these two questions from a friend today, and i don't know how to solve them at all.

4 cos4xsin4x+1=0
where the range is -3pi ≤ x ≤ 2 pi

All i know is that you're suppose to subsitute the 4x as x
and thats basically it. any help?

The second question i got was proofing Ids

(Root(2)/2) ( cos pi/12 - sin pi/12) = 1/2

How am i suppose to solve that?

Thank you so much for helping me :]
I'd use the double angle identity for sin

$\sin(2ax) = 2\sin(ax) \cos(ax)$

$4\cos(4x) \sin(4x) +1 = 2\sin(8x) + 1 = 0$

$\sin(8x) = -\frac{1}{2}$

$8x = \arcsin \left(-\frac{1}{2}\right)$

$x = \frac{1}{8} \arcsin \left(-\frac{1}{2}\right)$

3. Hello, TeriyakiDonnQ!

The second question i got was proofing Ids

. . $\frac{\sqrt{2}}{2}\left(\cos\frac{\pi}{12} - \sin\frac{\pi}{12}\right) \:=\: \frac{1}{2}$

How am i suppose to solve that?
. .
You don't solve it . . . you "proof" it.

On the left, we have: . $\frac{\sqrt{2}}{2}\cos\frac{\pi}{12} - \frac{\sqrt{2}}{2}\sin\frac{\pi}{12}$

. . Note that: . $\sin\tfrac{\pi}{4} \:=\:\cos\tfrac{\pi}{4}\:=\:\tfrac{\sqrt{2}}{2}$

The expression becomes: . $\sin\frac{\pi}{4}\cos\frac{\pi}{12} - \cos\frac{\pi}{4}\sin\frac{\pi}{12}$

And we have: . $\sin\left(\frac{\pi}{4} - \frac{\pi}{12}\right) \;=\;\sin\frac{\pi}{6} \;=\;\frac{1}{2}$

4. Thank youu e^(i*pi) and Soroban