Question

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Answer 1

By taking LHS

= sin11x/4 sinx/4+ sin7x/4 sin3x/4

= (1/2)[2sin11x/4 sinx/4+ 2sin7x/4 sin3x/4]           {multiply and divide by 2}

= (1/2)[cos{(11x-x)/4}-cos{(11x+x)/4} +cos{(7x-3x)/4}-cos{(7x+3x)/4}]

[Using 2sinAsinB = cos(A-B) - cos(A+B)]

= (1/2)[cos{(5x)/2}-cos(3x) + cos(x)-cos{(5x)/2}]

= (1/2)[cos(x)-cos(3x)]

= (1/2)[2sin{(x+3x)/2} sin{(3x-x)/2}]   (∵ cosA-cosB = 2sin{(A+B)/2} sin{(B-A)/2})

= sin2x sinx

=RHS(Proved)

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Answer 2

Hi there

Here is your answer (sorry for late)

To Prove sin11x/4 sinx/4+ sin7x/4 sin3x/4== sin2x sinx

Plan: In this type of questions  our plan is to convert the question into the  

following formulaes :

2 sin A cos B =sin(A + B) + sin(A − B)

2 cos A cos B =cos(A − B) + cos(A + B)

2 sin A sin B =cos(A − B) − cos(A + B)

Now,

By taking LHS

= sin11x/4 sinx/4+ sin7x/4 sin3x/4

= (1/2)[2sin11x/4 sinx/4+ 2sin7x/4 sin3x/4]           {multiply and divide by 2}(to convert it into above formula)

= (1/2)[cos{(11x-x)/4}-cos{(11x+x)/4} +cos{(7x-3x)/4}-cos{(7x+3x)/4}]

[Using 2sinAsinB = cos(A-B) - cos(A+B)]

= (1/2)[cos{(5x)/2}-cos(3x) + cos(x)-cos{(5x)/2}]

= (1/2)[cos(x)-cos(3x)]

= (1/2)[2sin{(x+3x)/2} sin{(3x-x)/2}]   (∵ cosA-cosB = 2sin{(A+B)/2} sin{(B-A)/2})

= sin2x sinx

Hence Proved

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