Question

cos α+ 2 cos 3α+ cos 5α
_____________________
cos 3α + 2 cos 5α + cos 7 α

=cos 3α sec 5α.


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Answer 1

SOLUTION :

TO PROVE

$\displaystyle \sf{ \frac{ \cos \alpha + 2 \cos 3 \alpha + \cos 5 \alpha }{ \cos 3 \alpha + 2 \cos 5 \alpha + \cos 7 \alpha} \: }$

$\sf{ = \cos 3\alpha \sec 5\alpha}$

FORMULA TO BE IMPLEMENTED

$\displaystyle \sf{} \cos C + \cos D = 2\cos \bigg( \frac{C + D}{2} \bigg) \cos \bigg( \frac{C - D}{2} \bigg)$

PROOF

LHS

$= \displaystyle \sf{ \frac{ \cos \alpha + 2 \cos 3 \alpha + \cos 5 \alpha }{ \cos 3 \alpha + 2 \cos 5 \alpha + \cos 7 \alpha} \: }$

$= \displaystyle \sf{ \frac{ \cos \alpha + \cos 5 \alpha + 2 \cos 3 \alpha }{ \cos 3 \alpha + \cos 7 \alpha + 2 \cos 5 \alpha\: } \: }$

$= \displaystyle \sf{ \frac{ 2\cos \bigg( \frac{ \alpha + 5 \alpha }{2} \bigg) \cos \bigg( \frac{ \alpha - 5 \alpha }{2} \bigg)+ 2 \cos 3 \alpha }{ 2\cos \bigg( \frac{ 3\alpha + 7 \alpha }{2} \bigg) \cos \bigg( \frac{ 3\alpha - 7 \alpha }{2} \bigg) + 2 \cos 5 \alpha\: } \: }$

$= \displaystyle \sf{ \frac{ 2\cos \bigg( \frac{ 6 \alpha }{2} \bigg) \cos \bigg( \frac{ - 4 \alpha }{2} \bigg)+ 2 \cos 3 \alpha }{ 2\cos \bigg( \frac{ 10 \alpha }{2} \bigg) \cos \bigg( \frac{ - 4 \alpha }{2} \bigg) + 2 \cos 5 \alpha\: } \: }$

$= \displaystyle \sf{ \frac{2 \cos 3\alpha \cos2 \alpha + 2 \cos 3 \alpha }{ 2\cos 5 \alpha \cos 2 \alpha + 2 \cos 5 \alpha\: } \: }$

$= \displaystyle \sf{ \frac{2 \cos 3\alpha( \cos2 \alpha +1) }{ 2\cos 5 \alpha ( \cos 2 \alpha +1)} \: }$

$= \displaystyle \sf{ \frac{ \cos 3\alpha}{ \cos 5 \alpha } \: }$

$\sf{ = \cos 3\alpha \sec 5\alpha}$

= RHS

Hence proved

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