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Is it possible to treat it as a binomial?

rae306
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Kapooky Handy
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  • What about $\sin^2(2x+1)$? – rae306 Oct 21 '14 at 20:12
  • I need to expand if it's even possible. – Kapooky Handy Oct 21 '14 at 20:14
  • Would you accept $\sin^2(2x + 1) = 1 - \cos^2\left(2x + 1\right)$ ? – Nick Oct 21 '14 at 20:53
  • Yeah. I'm just trying to find the derivative of sine using limit defination. I'm also a bit curious to know where you derived cos2x=cossquared x - sinessquared x. – Kapooky Handy Oct 21 '14 at 20:57
  • @KapookyHandy: It can be easily derived using $\cos(A+B) = \cos A\cos B - \sin A\sin B$. Put $\theta = A = B$ . Then use $\cos^2x = 1 - \sin^2x$. Presto, you've derived an equation for $\cos2\theta$, I've put a link to a video in my answer which will hopefully be helpful. – Nick Oct 21 '14 at 21:06

1 Answers1

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We know that $\cos2\theta = 1 - 2\sin^2\theta$ $\implies \sin^2\theta = \frac{1}{2}\left(1-\cos2\theta\right)$

Putting $\theta = 2x + 1$, $$ \sin^2(2x + 1) = \frac{1}{2}\Big( 1 - \cos(4x + 2) \Big) $$

Nick
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