# Math Help - A lot of trigonometric identities

1. ## A lot of trigonometric identities

You should know the following trigonometric identities.
(A) $sin(-x)=-sinx$
(B) $cos(-x)=cosx$
(C) $cos(x+y)=cosxcosy-sinxsin y$
(D) $sin(x+y)=sinxcosy+cosxsin y$
Use these to derive the following important identities, which you should also know.
(a) $sin^2x+cos^2x=1$ (use C and $cos0=1$
(b) $sin2x=2sinxcosx$
(c) $cos2x=cos^2x-sin^2x$
(d) $cos2x=2cos^2x-1$
(e) $cos2x=1-2sin^2x$
(f) $|cos\frac{x}{2}|=\sqrt{\frac{1+cosx}{2}}$
(g) $|sin\frac{x}{2}|=\sqrt{\frac{1-cosx}{2}}$

If anybody can finish these would be a tremendous help to me
took me over 30min just to write this in latex form... sigh...
i really hope someone will reply on how to do these with 1 or 2 as a example so i can do the rest

2. Originally Posted by saar4ever
You should know the following trigonometric identities.
(A) $sin(-x)=-sinx$
(B) $cos(-x)=cosx$
(C) $cos(x+y)=cosxcosy-sinxsin y$
(D) $sin(x+y)=sinxcosy+cosxsin y$
Use these to derive the following important identities, which you should also know.
(a) $sin^2x+cos^2x=1$ (use C and $cos0=1$
(b) $sin2x=2sinxcosx$
For (a), note that if you let $y=-x$ in C, we have

$\cos\left(x+(-x)\right)=\cos x\cos\left(-x\right)-\sin x\sin\left(-x\right)\implies \cos 0=\cos x\cos x-\sin x\left(-\sin x\right)$ $\implies 1=\cos^2x+\sin^2x$.

For (b), use D and let $y=x$.

Can you try the second part?

3. thanks sooo much now i get how to solve these probloms just finished a b c d e now trying f and g

thx again

4. Originally Posted by saar4ever
You should know the following trigonometric identities.
(A) $sin(-x)=-sinx$
(B) $cos(-x)=cosx$
(C) $cos(x+y)=cosxcosy-sinxsin y$
(D) $sin(x+y)=sinxcosy+cosxsin y$
Use these to derive the following important identities, which you should also know.

(f) $|cos\frac{x}{2}|=\sqrt{\frac{1+cosx}{2}}$

If anybody can finish these would be a tremendous help to me
took me over 30min just to write this in latex form... sigh...
i really hope someone will reply on how to do these with 1 or 2 as a example so i can do the rest
You need to use exercise (d) ( $\cos\left(2x\right)= 2\cos^2x-1$) to solve this one.

Note that $\cos\left(2x\right)=2\cos^2x-1\implies\cos^2x=\frac{\cos\left(2x\right)+1}{2}$.

Taking the square root, we have $\cos x=\pm\sqrt{\frac{1+\cos\left(2x\right)}{2}}$.

Taking the absolute value, we have $\left|\cos x\right|=\left|\pm\sqrt{\frac{1+\cos\left(2x\right )}{2}}\right|=\sqrt{\frac{1+\cos\left(2x\right)}{2 }}$.

The final step is to let $x=\frac{y}{2}$.

Therefore, $\left|\cos\left(\tfrac{y}{2}\right)\right|=\sqrt{\ frac{1+\cos\left(2\left(\tfrac{y}{2}\right)\right) }{2}}=\sqrt{\frac{1+\cos y}{2}}$.

I hope that helps.

5. thanks
you da man
helped me alot u are a math genius
all i got left is g and im done thanks again