Find \(\displaystyle{\sum_{{{n}={1}}}^{{{89}}}}{{\sin}^{{6}}{\left({n}\right)}}={\frac{{{m}}}{{{n}}}}\)

Eddie Clarke

Eddie Clarke

Answered question

2022-04-13

Find n=189sin6(n)=mn

Answer & Explanation

Buizzae77t

Buizzae77t

Beginner2022-04-14Added 13 answers

Let
S=n=189sin6(n)
as you have used sin(90n)=cos(n)
So we get
S=n=189sin6(90n)=n=189cos6(n)
as you have used sin(90n)=cos(n)
Now add these two equation, we get
2S=n=189[sin6(n)+cos6(n)]
=n=189[13sin2(n)cos2(n)]
Above we have uesd the formula If sin2x+cos2x+(1)=0, then
sin6x+cos6x1=3sin2xcos2x1
So we get
sin6x+cos6x=13sin2xcos2x
So we get
2S=n=189134n=189sin2(2n)=8934T
Where
T=n=189sin2(2n)=2n=144sun2(n)+1

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