Question

prove that cos 7 alpha + cos 3 alpha - cis 5 alpha - cos alpha ÷sin 7 alpha -sin 3 aloha - sin 5 alpha + sin alpha = cot 2 alpha

Answer 1

$\underline{\textsf{To prove:}}$

$\mathsf{\dfrac{cos7\alpha+cos3\alpha-cos5\alpha-cos\alpha}{sin7\alpha-sin3\alpha-sin5\alpha+sin\alpha}=cot2\alpha}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{\dfrac{cos7\alpha+cos3\alpha-cos5\alpha-cos\alpha}{sin7\alpha-sin3\alpha-sin5\alpha+sin\alpha}}$

$\mathsf{=\dfrac{cos7\alpha+cos3\alpha-(cos5\alpha+cos\alpha)}{sin7\alpha-sin3\alpha-(sin5\alpha-sin\alpha)}}$

$\textsf{Using the identities,}$

$\boxed{\begin{minipage}{7cm} \\\mathsf{cosC+cosD=2\,cos(\dfrac{C+D}{2})\,cos(\dfrac{C-D}{2})}\\\\\mathsf{sinC-sinD=2\,cos(\dfrac{C+D}{2})\,sin(\dfrac{C-D}{2})} \end{minipage}}$

$\mathsf{=\dfrac{2\,cos5\alpha\,cos2\alpha-2\,cos3\alpha\,cos2\alpha}{2\,cos5\alpha\,sin2\alpha-2\,cos3\alpha\,sin2\alpha}}$

$\mathsf{=\dfrac{2\,cos2\alpha(cos5\alpha-cos3\alpha)}{2\,sin2\alpha(cos5\alpha-cos3\alpha)}}$

$\mathsf{=\dfrac{cos2\alpha}{sin2\alpha}}$

$\mathsf{=cot\,2\alpha}$

$\implies\boxed{\mathsf{\dfrac{cos7\alpha+cos3\alpha-cos5\alpha-cos\alpha}{sin7\alpha-sin3\alpha-sin5\alpha+sin\alpha}=cot\,2\alpha}}$

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Answer 2

\textsf{Consider,}Consider,

\mathsf{\dfrac{cos7\alpha+cos3\alpha-cos5\alpha-cos\alpha}{sin7\alpha-sin3\alpha-sin5\alpha+sin\alpha}}sin7α−sin3α−sin5α+sinαcos7α+cos3α−cos5α−cosα

\mathsf{=\dfrac{cos7\alpha+cos3\alpha-(cos5\alpha+cos\alpha)}{sin7\alpha-sin3\alpha-(sin5\alpha-sin\alpha)}}=sin7α−sin3α−(sin5α−sinα)cos7α+cos3α−(cos5α+cosα)

\textsf{Using the identities,}

prove that cog7 alpha + cos 3 alpha - cos 5 alpha - cos alpha / sin 7 alpha - sin 3 alpha - sin 5 alpha + sin alpha = cot 2 alpha

hi

=2cos5αsin2α−2cos3αsin2α2cos5αcos2α−2cos3αcos2α

\mathsf{=\dfrac{2\,cos2\alpha(cos5\alpha-cos3\alpha)}{2\,sin2\alpha(cos5\alpha-cos3\alpha)}}=2sin2α(cos5α−cos3α)2cos2α(cos5α−cos3α)

\mathsf{=\dfrac{cos2\alpha}{sin2\alpha}}=sin2αcos2α

\mathsf{=cot\,2\alpha}=cot2α

\implies\boxed{\mathsf{\dfrac{cos7\alpha+cos3\alpha-cos5\alpha-cos\alpha}{sin7\alpha-sin3\alpha-sin5\alpha+sin\alpha}=cot\,2\alpha}}⟹sin7α−sin3α−sin5α+sinαcos7α+cos3α−cos5α−cosα=cot2α

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