1
$\begingroup$

I am having problems understanding how to verify this identity. I am quite sure that it is to be solved using the Pythagorean identities but, alas, I'm not seeing what might otherwise be obvious.

I need to verify the identity

$$\cos^2x-\sin^2x = 2\cos^2x-1$$

Thank you for your help.

$\endgroup$
  • $\begingroup$ You know that $\cos^2x+\sin^2x=1$, then write $\sin^2x$ in terms of $\cos^2x$, then substitute your expression in $\cos^2x-\sin^2x$ and simplify. $\endgroup$ – Hakim Nov 17 '14 at 13:02
  • $\begingroup$ find some way of getting rid $\sin^2x$ on the LHS. $\endgroup$ – John Joy Nov 17 '14 at 15:00
3
$\begingroup$

Yes, indeed, we can use the Pythagorean Identity:

$$\cos^2 x + \sin^2 x = 1 \iff \color{blue}{\sin^2 x = 1-\cos^2 x}$$


$$\begin{align}\cos^2x-\color{blue}{\sin^2x} & = \cos^2 x - \color{blue}{(1-\cos^2 x)}\\\\ &= \cos^2 x - 1 + \cos^2 x \\ \\ & = 2\cos^2 x - 1\end{align}$$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Remember that $sin^2(x) = 1 - cos^2(x)$.

Then $cos^2x - sin^2x = ?$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

we have $\cos(x)^2-\sin(x)^2=\cos(x)^2-(1-\cos(x)^2)=2\cos(x)^2-1$ since $\sin(x)^2+\cos(x)^2=1$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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