if cos theta +cos²theta=1, prove that sin¹²+3sin^10+3sin^8+sin^6+2sin⁴+2sin²-2=1
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If cosθ+cos
2
θ=1, prove that sin
12
θ+3sin
10
θ+3sin
8
θ+sin
6
θ+2sin
4
θ+2sin
2
θ - 2 = 1
Medium
Solution
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Given :
cosθ=1−cos
2
θ
⇒cosθ=sin
2
θ ..... [sin²x+cos²x=1]
Square on both sides ;
cos
2
θ=sin
4
θ
1−sin
2
θ=sin
4
θ ...... [sin²x+cos²x=1]
sin
4
θ+sin
2
θ=1 →equation (1)
Now cube on both sides ;
⇒sin
12
θ+sin
6
θ+3sin
4
θsin
2
θ(sin
4
θ+sin
2
θ)=1
⇒sin
12
θ+sin
6
θ+3sin
10
θ+3sin
8
θ=1
To obtain above result we add and subtract 2 on LHS side ;
⇒sin
12
θ+sin
6
θ+3sin
10
θ+3sin
8
θ+2(1)−2=1
From equation (1), 1=sin
4
θ+sin
2
θ
⇒sin
12
θ+sin
6
θ+3sin
10
θ+3sin
8
θ+2(sin
4
θ+sin
2
θ)−2=1
⇒sin
12
θ+3sin
10
θ+3sin
8
θ+sin
6
θ+2sin
4
θ+2sin
2
θ−2=1
Hence proved