sinα+cisα=m then sin^6α+cos^6α
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Answer:
sin
6
α+cos
6
α
=(sin
2
α)
3
+(cos
2
α)
3
=(sin
2
α+cos
2
α)(sin
4
α+cos
4
α−sin
2
αcos
2
α)
Now
sin
2
α+cos
2
α=1
Hence
(sin
2
α+cos
2
α)(sin
4
α+cos
4
α−sin
2
αcos
2
α)
=(sin
4
α+cos
4
α−sin
2
αcos
2
α)
=((sin
2
α+cos
2
α)
2
−2sin
2
αcos
2
α−sin
2
αcos
2
α)
=(1−3sin
2
αcos
2
α) ....(i)
Now it is given that
sinα+cosα=m
(sin
2
α+cos
2
α)=1
sin
2
α+cos
2
α+2sinα.cosα=m
2
1+2sinα.cosα=m
2
m
2
−1=2sinα.cosα
2
m
2
−1
=sinα.cosα
(sinα.cosα)
2
=
4
(1−m
2
)
2
Hence from i
(1−3sin
2
αcos
2
α)
=1−
4
3(1−m
2
)
2
=
4
4−3(1−m
2
)
2
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