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Looking for some help with an explanation of the following simplification in trig,

$$\cos\theta \sin\theta (\sin^2\theta -\cos^2\theta)=-\frac{\sin4\theta}{4}$$

In simplifying this do they take the negative out of $(\sin^2\theta -\cos^2\theta)$ to get

$$-\cos\theta \sin\theta =-\frac{\sin4\theta}{4}$$

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Heres a complete answer $$\cos\theta \sin\theta (\sin^2\theta -\cos^2\theta)=\frac{\sin(2\theta)}{2}*-\cos(2\theta)$$

Again, $$\frac{-\sin(2\theta)\cos(2\theta)}{2}=\frac{-\sin(4\theta)}{4}$$

I mainly used these two relation $$\cos^2 (x)-\sin^2 (x)=\cos (2x)$$ which gives $$\sin^2 (x)-\cos^2 (x)=-\cos (2x)$$ and $$\sin(2\theta)=2\sin(\theta)\cos(\theta)$$ which gives $$\frac{\sin(2\theta)}{2}=\sin(\theta)\cos(\theta)$$ Similarly $$\frac{\sin(4\theta)}{2}=\sin(2\theta)\cos(2\theta)$$

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  • $\begingroup$ shouldn't it be divded by 4 in the end not 2? $\endgroup$ – user123 Mar 12 '17 at 14:45
  • $\begingroup$ Edited the answer $\endgroup$ – user35508 Mar 12 '17 at 14:46
  • $\begingroup$ awesome explanation, thanks your your help $\endgroup$ – user123 Mar 12 '17 at 14:52
  • $\begingroup$ You're welcome :) $\endgroup$ – user35508 Mar 12 '17 at 14:55
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No use $\cos^2 (x)-\sin^2 (x)=\cos (2x) $. Then use $\cos (a)sin (a)=2\sin (2a) $

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  • $\begingroup$ so since $sin^2x-cos^2x$ it becomes $-cos(2x)$? $\endgroup$ – user123 Mar 12 '17 at 14:37
  • $\begingroup$ and where does the divide by 4 part come in? $\endgroup$ – user123 Mar 12 '17 at 14:40
  • $\begingroup$ See the edit now $\endgroup$ – Archis Welankar Mar 12 '17 at 14:44
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HINT

Hope short hand helps.. non-unity arguments written to the right

$$ s^2 -c^2 = -c2 $$

given

$$ \dfrac12 \, s2 \cdot -c2 = - s4 /4 $$

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