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Question

Let fn(x)=(1/n)(sinnx+cosnx) for n=1,2,3,.... then f4(3π/8)f6(3π/8) is equal to

A
2/(2+1)
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B
1/12
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C
4/(31)
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D
1/8
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Solution

The correct option is A 2/(2+1)
fn(x)=(1n)(sinnx+cosnx)
f4(x)f6(x)=(14)(cos4x+sin4x)16(cos6x+sin6x) ........(1)
using a3+b3=(a+b)(a2+b2ab)
(cos2x)3+(sin2x)3=(sin2x+cos2x)(sin4+cos4xsin2xcos2x) ..........(2)
sin2x+cos2x=1 ........(3)
put eq (2) & eq (1)
f4(x)f6(x)=14(cos4x+sin4x)16(sin4x+cos4xsin2xcos2x)
=224(cos4x+sin4x)+16sin2xcos2x
=112(cos4x+sin4x+2sin2xcos2x)
=112(sin2x+cos2x)2
=112
Independent of x.

1157366_699537_ans_2b2bc440cdb045158ac53f7e892b47cf.jpg

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