1. ## Identities

It's been a while since my last math class, and my current teacher isnt the greatest. We're assigned a fair amount of homework everynight, and it just takes me forever to do it, and its taking all of my time, leaving no time left for other classes. I have some problems and if anyone can help me with them, I'd greatly appreciate it. I learn best by seeing an example, but our teacher tends not to give many examples, if any.

2) If sin x = −3/5,x in quadrant III, then what are the following ?

sin2x
cos2x
tan2x

4)If sin4 x = A+Bcos2x+Ccos4x, then what are A, B, C ?

5If sin(15Degrees) = 1/2 sqrt(A - sqrt(B))
B, then, by using a half-angle formula, find A , B

I have some more, but help with these would be amazing.

2. Hi

2) Knowing $\sin x$ you can compute $\cos^2x = 1 - \sin^2 x$
Then x being in quadrant III, you can compute $\cos x$

Then $\sin 2x = 2 \sin x \cos x$

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

$\tan 2x = \frac{\sin 2x}{\cos 2x}$

4) 2 possibilities (at least)

a) Chose some values (x=0, x=pi/2, ...) to get 3 linear equations with unknowns A, B and C

b) $\cos 4x = \cos 2(2x) = 2 \cos^2 2x -1 = 2(1-2\sin^2 x)^2-1 = 8 \sin^4 x - 8 \sin^2 x + 1$
$\cos 2x = 1- 2 \sin^2 x$

Therefore you can find the expression of $A + B \cos 2x + C \cos 4x$ and find A, B and C such that $A + B \cos 2x + C \cos 4x = \sin^4 x$

5) $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$
and $\cos \frac{\pi}{6} = 1 - 2 \sin^2 \frac{\pi}{12} = 1 - \frac12\A-\sqrt{B})" alt="\cos \frac{\pi}{6} = 1 - 2 \sin^2 \frac{\pi}{12} = 1 - \frac12\A-\sqrt{B})" />