1. ## Verify identity

am terrible with identity and this book proposes no answers or how to for the HW.

I can do the examples given okay but when it comes to the homework it's like wtf is all this. The problem is:

(sin@+cos@)^2=1+sin2@

i need to verify that one side equals the other.

only thing i know is that sin2@=2sin@cos@.

2. The two sides do NOT equal each other.

$\sin^2{\theta} + \cos^2{\theta} \equiv 1$ is a well known Identity known as the Pythagorean Identity.

Since $\sin{2\theta}$ is NOT identically equal to $0$, (though it does equal $0$ for CERTAIN values of $\theta$).

That means that the LHS can not possibly equal the RHS for ALL $\theta$.

3. sorry i wrote the problem wrong! I started this a day ago and i typed it from my page not the book.

It's (sin@+cos@)^2=1+sin2@

4. $(\sin{\theta} + \cos{\theta})^2 = \sin^2{\theta} + 2\sin{\theta}\cos{\theta} + \cos^2{\theta}$

$= \sin^2{\theta} + \cos^2{\theta} + 2\sin{\theta}\cos{\theta}$

$= 1 + 2\sin{\theta}\cos{\theta}$

$= 1 + \sin{2\theta}$.

5. Can i get help with this other one?

sin2t-tant=tantcost2t

2sintcost-sint/cost

(2sintcost^2/cost) - (sint/cost)

(2sintcost^2 - sint)/cost

6. $\frac{2sintcos^2t - sint}{cost}$

= $\frac{sint(2cos^2t - 1)}{cost}$

= tant cos2t

7. thanks guys, that helped me solving these annoying problems, i'm understanding the more complex ones a little better now. I have an exam on Tuesday >_< gotta keep studying.