# $\cos^2(\theta)-\sin^2(\theta)=1+\sin(\theta)$ over the interval $(0,2\pi)$ [closed]

$\cos^2(\theta)-\sin^2(\theta)=1+\sin(\theta)$ over the interval $0<\theta<2\pi$

Find the trigonometric identity.

Apologize for the confusion, first time using this resource didnt read the instructions. i have tried manipulating the equation by substituting x^2 and y^2 in for the cos^2 and sin^2 and subtracting and adding the one but i could not find out what identities to use to make both side equal. Sorry again for the mistake.

## closed as unclear what you're asking by Chappers, ronno, Christopher, quid♦, sazMay 11 '15 at 18:13

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• What have you tried? You might not know, but on this site you're expected to show some of your own effort. That probably explains why you've gotten a downvote – Zach Effman May 11 '15 at 0:58
• Find? No, you find. We help. What @ZachEffman said. – zahbaz May 11 '15 at 0:59
• Oh right, also people really don't appreciate questions phrased as commands – Zach Effman May 11 '15 at 1:00
• Apologize for the confusion, first time using this resource didnt read the instructions. i have tried manipulating the equation by substituting x^2 and y^2 in for the cos^2 and sin^2 and subtracting and adding the one but i could not find out what identities to use to make both side equal. Sorry again for the mistake. – Gurnoor Singh May 11 '15 at 1:10
• It is not an identity for all $\theta$ in $(0, 2\pi)$. You can check some values. Ex: Check $\theta=\frac{\pi}{2}$. – JimmyJP May 11 '15 at 1:48

Hint: $\cos^2\theta - \sin^2\theta = 1-2\sin^2\theta$, so your equation becomes: $$2\sin^2\theta + \sin \theta = 0.$$
• It is not an identity, we can check $\theta$ in $(0, 2\pi)$ with $\theta=\frac{\pi}{2}$ – JimmyJP May 11 '15 at 1:52