Question

Urgent please answer


$if \: cot( \frac{3\pi}{2} - b) = \sqrt{3} \\ then \: prove \: that \\ \: cos3b = 4{cos}^{3}b \: - \: 3cosb$
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Answer 1

Answer:

$\displaystyle{\cos 3b = 4\cos^3 b - 3 \cos b \ \text{Proved}}$

Step-by-step explanation:

Given :

$\displaystyle{\cot\left(\frac{3\pi}{2} -b\right)=\sqrt{3}}$

We know :

$\displaystyle{\cot\left(\frac{3\pi}{2} -\theta\right) = \tan \theta}$

$\displaystyle{\cot\left(\frac{3\pi}{2} -b\right)=\sqrt{3}}\\\\\\\displaystyle{\rightarrow \tan b=\sqrt{3}}$

$\display\text{Rewrite \sqrt{3} as \tan\dfrac{\pi}{3} }$

$\displaystyle{\tan b = \tan\dfrac{\pi}{3} }\\\\\\\displaystyle{\implies b= \dfrac{\pi}{3} }$

$\displaystyle{\cos 3b = 4\cos^3 b - 3 \cos b}$

Now putting value of b here :

$\displaystyle{\text{L.H.S.}=\cos 3b}\\\\\\\displaystyle{\implies \cos 3\times\frac{\pi}{3} }\\\\\\\displaystyle{\implies \cos \pi}\\\\\\\displaystyle{\implies -1.}$

$\displaystyle{\text{R.H.S.}= 4\cos^3 b - 3 \cos b}\\\\\\\displaystyle{\implies 4\cos^3 \frac{\pi}{3} - 3 \cos \frac{\pi}{3}}\\\\\\\displaystyle{\implies 4\left(\frac{1}{8}\right) -3\left(\frac{1}{2} \right)}\\\\\\\displaystyle{\implies \frac{1}{2} -\frac{3}{2}}\\\\\\\displaystyle{\implies -1}$

$\display \large \text{Since L.H.S. = R.H.S. , proved.}$