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Prove: $$\cot x=\sin x\,\sin\left(\frac{\pi}{2}-x\right)+\cos^2x\,\cot x$$

Hi there! So this problem asks to prove this trigonometric identity. I am not sure how to approach these problems other than needing to know the quotient,p ythagorean, and reciprocal identities. From here I can see that $\cot x$ can be changed to $1/\tan x$, but is it really necessary? If someone could help with this, it'd be very appreciated!

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  • $\begingroup$ Hint: Transform everyhing with $\sin x$ and $\cos x$ and factor out what you can. $\endgroup$ – Bernard Apr 4 '15 at 21:19
  • $\begingroup$ so change the cos^2x to 1-sin^2x and sin to csc? $\endgroup$ – Ash Apr 4 '15 at 21:32
  • $\begingroup$ Take common $\cot x$ from RHS and use the identity $\cos^2\theta+\sin^2\theta=1$ to get the LHS form. $\endgroup$ – Prasun Biswas Apr 4 '15 at 21:41
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$$\textrm{RHS}=\sin x\sin\left(\frac{\pi}{2}-x\right)+\cos^2 x\cot x=\sin x\cos x+\cos^2 x\cot x\\ =\cot x(\sin^2 x + \cos^2 x)=\cot x=\textrm{LHS}$$

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$$\cot x = \sin x \sin(π/2−x)+\cos ^2 x \cot x $$ $$= \sin ⁡x \sin⁡(π/2−x) + \cos^2 ⁡x \cot ⁡x$$ $$= \sin x \cos x + \cos^2 x \left(\frac{\cos x} {\sin x}\right)$$ $$= \frac{\sin^2 x \cos x + \cos x (\cos^2 x)} {\sin x}$$ $$= \frac{\cos x (\sin^2 x + \cos^2 x)} {\sin x}$$ $$= \frac {\cos x (1)} {\sin x}$$ $$= \frac {1} {\tan x}$$ $$= \cot x$$

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