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(1+cos2θ)/sin2θ = cotθ

Maplesnowy  May 14, 2018
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\(\dfrac{1+\cos2\theta}{\sin2\theta}=\cot\theta\)

 

Let's turn the left side into the right side.

 

\(\phantom{=\,}\ \dfrac{1+\cos2\theta}{\sin2\theta}\)

                                           First let's use the double-angle identity for sin:    \(\sin2θ=2 \sinθ \cosθ\)

\(=\,\dfrac{1+\cos2\theta}{2\sin\theta\cos\theta}\)

                                           Now let's use this double-angle identity for cos:  \(\cos2\theta=2\cos^2\theta-1\)

\(=\,\dfrac{1+2\cos^2\theta-1}{2\sin\theta\cos\theta}\)

                                           Add together the 1 and the -1  in the numerator to get 0 .

\(=\,\dfrac{2\cos^2\theta}{2\sin\theta\cos\theta}\)

                                           Divide the numerator and denominator by  2 .

\(=\,\dfrac{\cos^2\theta}{\sin\theta\cos\theta}\)

                                           Divide the numerator and denominator by  \(\cos\theta\)  .

\(=\,\dfrac{\cos\theta}{\sin\theta}\)

                                           And by the quotient identity for cotangent,  \(\cot\theta=\frac{\cos\theta}{\sin\theta}\)

\(=\,\cot\theta\)

hectictar  May 15, 2018

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