Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am stuck on 7, 9

steps please, thank you

enter image description here

share|cite|improve this question
up vote 2 down vote accepted


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

share|cite|improve this answer
To prove for example 7. replace in the LHS $1$ by $\cos^2\theta+\sin^2\theta$ and you find the RHS. – user63181 Mar 16 '14 at 0:19
got it, thank you – Mandy Quan Mar 16 '14 at 0:22

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

Second is similar.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.