Prove the following trig identities

Question

I am stuck on 7, 9

steps please, thank you

Answer 1

Hint

Replace in the two equalities $1$ by $\cos^2\theta+\sin^2\theta$.

Answer 2

For the first one: note that $2\sin^2\theta - 1 = \sin^2\theta + \sin^2\theta -1$.

Second is similar.

Answer 3

Identity #$7$: $$2\sin^2 \theta-1=\sin^2 \theta - \cos^2 \theta$$ Let's make the right hand side equal to the left hand side (LHS-RHS proof). $$\sin^2 \theta - \cos^2 \theta$$ $$=1-\cos^2 \theta - \cos^2 \theta$$ $$=1-2\cos^2 \theta$$ $$=1-2(1-\sin^2 \theta)$$ $$=1-2+2\sin^2 \theta$$ $$=-1+2\sin^2 \theta$$ $$=2\sin^2 \theta-1$$ $$\displaystyle \boxed{\therefore 2\sin^2 \theta-1=\sin^2 \theta - \cos^2 \theta}$$ Identity #$9$: $$\cos^2 t=\sin^2 t +2\cos^2 t - 1$$ This time, let's ASSUME the identity is true. $$\cos^2 t=\sin^2 t +2\cos^2 t - 1$$ $$-\cos^2 t=\sin^2 t - 1$$ $$-\cos^2 t - \sin^2 t = -1$$ $$\cos^2 t + \sin^2 t = 1 \ \ \text{(This must be true)}$$ $$\displaystyle \boxed{\therefore \cos^2 t=\sin^2 t +2\cos^2 t - 1}$$ If you do not like my proof for the second one, here is a more... appropriate proof. Let's use LHS-RHS again. $$\cos^2 t=\sin^2 t +2\cos^2 t - 1$$ Let's make the right hand side equal to the left hand side. $$\sin^2 t +2\cos^2 t - 1$$ $$=\sin^2 t + \cos^2 t + \cos^2 t -1$$ $$=1 + \cos^2 t - 1$$ $$=\cos^2 t$$ $$\displaystyle \boxed{\therefore \cos^2 t=\sin^2 t +2\cos^2 t - 1}$$ I hope that this solves your problems.