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$\cos^2(\theta)-\sin^2(\theta)=1+\sin(\theta)$ 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.

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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 $\endgroup$ – Zach Effman May 11 '15 at 0:58
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    $\begingroup$ Find? No, you find. We help. What @ZachEffman said. $\endgroup$ – zahbaz May 11 '15 at 0:59
  • $\begingroup$ Oh right, also people really don't appreciate questions phrased as commands $\endgroup$ – Zach Effman May 11 '15 at 1:00
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    $\begingroup$ 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. $\endgroup$ – Johnny Appleseed May 11 '15 at 1:10
  • $\begingroup$ It is not an identity for all $\theta$ in $(0, 2\pi)$. You can check some values. Ex: Check $\theta=\frac{\pi}{2} $. $\endgroup$ – JimmyJP May 11 '15 at 1:48
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Hint: $\cos^2\theta - \sin^2\theta = 1-2\sin^2\theta$, so your equation becomes: $$2\sin^2\theta + \sin \theta = 0.$$

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  • $\begingroup$ I'm glad you answered this; I'm still trying to figure out where the question is! ;-) Endorsed!!! Cheers! $\endgroup$ – Robert Lewis May 11 '15 at 1:01
  • $\begingroup$ I guided myself by the title! $\endgroup$ – Ivo Terek May 11 '15 at 1:12
  • $\begingroup$ It is not an identity, we can check $\theta$ in $(0, 2\pi)$ with $\theta=\frac{\pi}{2}$ $\endgroup$ – JimmyJP May 11 '15 at 1:52
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    $\begingroup$ Yes, that expression stated in the question is not an identity. Most probably OP couldn't express himself very well. The only exercise that makes sense there is solving the equation. $\endgroup$ – Ivo Terek May 11 '15 at 1:53

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