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Question : 20 of 160
Marks:
+1,
-0
Solution:
Consider the equation.
A(n)=sinnα+cosnα
Let I=A(1)A(4)+A(2)A(5)
Thus,
I=[(sin‌α+cos‌α)(sin4α+cos4α) +(sin2α+cos2α)(sin5α+cos5α)]
=[(sin‌α+cos‌α)[(sin2α+cos2α)3 −2sin2αcos2α]+](sin5α+cos5α)]
=[(sin‌α+cos‌α)[(sin6α+cos6α) +sin2αcos2α]+](sin5α+cos5α)]
=[(sin‌α+cos‌α)(sin6α+cos6α) +sin2αcos2α(sin‌α+cos‌α)+(sin5α+cos5α)]
Solve further,
I=[(sin‌alpha+cos‌α)(sin6α+cos6α) +sin3αcos2αsin5α+sin2αcos3αcos5α]
=[(sin‌α+cos‌α)(sin6α+cos6α) +(sin2α+cos2α)(sin3α+cos3α)]
=A(1)A(6)+A(2)A(3)
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