1. ## sine and cosine

I am stuck on these two problems. I have absolutely no idea how to get them. any help would be greatly appreciated.

1. $\displaystyle 1-4sin^2xcos^2x$
2. $\displaystyle cos^22x-cos^2x=sin^2xtan^2x$ I just think you have to solve the identity

the answer has to be simplified to a single term.

Thank you.

2. Originally Posted by OnMyWayToBeAMathProffesor
I am stuck on these two problems. I have absolutely no idea how to get them. any help would be greatly appreciated.

1. $\displaystyle 1-4sin^2xcos^2x$

For the 1st one

$\displaystyle 1-4sin^2xcos^2x=1-(2 \sin(x) \cos(x))^2=1-\sin^2(2x)=\cos^{2}(2x)$

3. could you please explain this step.
$\displaystyle 1-(2 \sin(x) \cos(x))^2=1-\sin^2(2x)=\cos^{2}(2x)$

4. Originally Posted by OnMyWayToBeAMathProffesor
could you please explain this step.
$\displaystyle 1-(2 \sin(x) \cos(x))^2=1-\sin^2(2x)=\cos^{2}(2x)$
I used this identity $\displaystyle \sin(2x)=2\sin(x) \cos(x)$

and this one

$\displaystyle 1-\sin^2(y)=cos^2(y)$

5. Thanks a lot. makes sense now.

6. any idea on the second one?

7. 2nd one isn't an identity because it doesn't work for all values of x. For example, if we let $\displaystyle x = \frac{\pi}{5}$, then the left hand side is about -.559 while the right hand side is about .182.

8. I apologize for the inconvenience but number 2 is NOT correct. these are the ones i meant to type in.

2) $\displaystyle cos(3x)=4cos^3x-3cosx$

3) $\displaystyle cos^22x-cos^2x=sin^2x-sin^22x$

I would like to thank everybody who tried to help me. Once again I apologize. These are both identities. I did some work on number 2 but then I got stuck. I do not understand number 3.

Thank You.

9. 2. Use the double angle formula for cosine using 3x = 2x + x and convert all your sin terms into cos terms.

3. Ignore the angles and just convert the cosine terms into sine terms.