# Thread: [SOLVED] my last two double angle questions

1. ## [SOLVED] my last two double angle questions

I am pretty sure I'm almost done the first one.
#1
Use the identity for sin2x and an identity for cos2x to verify that

but when I use the formula for double angles like where with a
I think it is

And
#2
how do you find roots/solve something like
$4sinxcosx-1=0$ or $cos^x-sin^2x=1$ I algebraically? I have done questions where you have to do proofs, but I have no examples like this from my teacher, thanks.

sorry about the multiple questions but the site was down lol.

2. Originally Posted by brentwoodbc
I am pretty sure I'm almost done the first one.
#1
Use the identity for sin2x and an identity for cos2x to verify that

but when I use the formula for double angles like where with a
I think it is

$\sin ^2 2x + \cos ^2 2x \hfill \\$

$= \left( {2\sin x\cos x} \right)^2 + \left( {\cos ^2 x - \sin ^2 x} \right)^2 \hfill \\$

$= 4\sin ^2 x\cos ^2 x + \cos ^4 x - 2\cos ^2 x\sin ^2 x + \sin ^4 x \hfill \\$

$= \sin ^4 x + 2\cos ^2 x\sin ^2 x + \cos ^4 x \hfill \\$

$= \left( {\sin ^2 x + \cos ^2 x} \right)^2 = 1^2 = 1 \hfill \\$

3. Originally Posted by brentwoodbc
#2
how do you find roots/solve something like
$4sinxcosx-1=0$ or $cos^x-sin^2x=1$ I algebraically? I have done questions where you have to do proofs, but I have no examples like this from my teacher, thanks.

sorry about the multiple questions but the site was down lol.
$4 \sin x \cos x - 1 = 0$

$2(2\sin x \cos x) - 1 = 0$

$2\sin 2x -1=0$

$\sin 2x =\frac{1}{2}$

$2x = \frac{\pi}{6}, \frac{5\pi}{6}$

$x = \frac{\pi}{12}, \frac{5\pi}{12}$

Now,

$\cos x - sin^2 x =1$

Change $sin^2 x$ into $1- \cos ^2 x$, then solve the quadratic eqn that is formed.

4. thanks, very much.
ya on the first one factoring the
$= \sin ^4 x + 2\cos ^2 x\sin ^2 x + \cos ^4 x \hfill \\$
was confusing me because of all the $sin^4$ 's etc.