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Question: Prove that \[\cot \theta - \cot 2\theta = \cos ec2\theta \]...

Prove that cot⁡θ−cot⁡2θ=cos⁡ec2θ\cot \theta - \cot 2\theta = \cos ec2\theta

Explanation

Solution

Use the various general trigonometric formulas as cot⁡θ=cos⁡θsin⁡θ\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}, and also the formula of cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x and sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x. We just need to expand the given term and if possible convert it to the similar denominator so that the numerator can be easily calculated.

Complete step-by-step answer:
As given thatcot⁡θ−cot⁡2θ\cot \theta - \cot 2\theta ,
cot⁡θ−cot⁡2θ\cot \theta - \cot 2\theta
On using cot⁡θ=cos⁡θsin⁡θ\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}, we get,
=cos⁡θsin⁡θ−cos⁡2θsin⁡2θ= \dfrac{{\cos \theta }}{{\sin \theta }} - \dfrac{{\cos 2\theta }}{{\sin 2\theta }}
On further simplification, use sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos xand cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x putting it in the above equation, we get,
=cos⁡θsin⁡θ−cos⁡2θ−sin⁡2x2sin⁡θcos⁡θ= \dfrac{{\cos \theta }}{{\sin \theta }} - \dfrac{{{{\cos }^2}\theta - {{\sin }^2}x}}{{2\sin \theta \cos \theta }}
Now multiply the denominator and numerator of first term with 2cos⁡θ2\cos \theta in order to make them similar,
=2cos⁡θcos⁡θ2cos⁡θsin⁡θ−cos⁡2θ−sin⁡2x2sin⁡θcos⁡θ= \dfrac{{2\cos \theta \cos \theta }}{{2\cos \theta \sin \theta }} - \dfrac{{{{\cos }^2}\theta - {{\sin }^2}x}}{{2\sin \theta \cos \theta }}
On further simplification we get,

=2cos⁡2θ−(cos⁡2θ−sin⁡2θ)2cos⁡θsin⁡θ =cos⁡2θ+sin⁡2θ2cos⁡θsin⁡θ  = \dfrac{{2{{\cos }^2}\theta - ({{\cos }^2}\theta - {{\sin }^2}\theta )}}{{2\cos \theta \sin \theta }} \\\ = \dfrac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{2\cos \theta \sin \theta }} \\\

Using the identitycos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1
=12cos⁡θsin⁡θ= \dfrac{1}{{2\cos \theta \sin \theta }}
Now as sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x, we get,
=1sin⁡2θ= \dfrac{1}{{\sin 2\theta }}
As, cos⁡ecθ=1sin⁡θ\cos ec\theta = \dfrac{1}{{\sin \theta }},
=cos⁡ec2θ= \cos ec2\theta
Hence, cot⁡θ−cot⁡2θ=cos⁡ec2θ\cot \theta - \cot 2\theta = \cos ec2\theta
Hence, proved.

Additional information:
Manufacturing Industry. Trigonometry plays a major role in industry, where it allows manufacturers to create everything from automobiles to zigzag scissors. Engineers rely on trigonometric relationships to determine the sizes and angles of mechanical parts used in machinery, tools and equipment.

Note: There are six trigonometric ratios, sine, cosine, tangent, cosecant, secant and cotangent. These six trigonometric ratios are used in a short form as sin, cos, tan, csc, sec, cot. These are referred to as ratios since they can be expressed in terms of the sides of a right-angled triangle for a specific angle θ\theta .