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8)To find : sin5θGiven: sin2θ=cos3θsin2θ=sin90°-3θ     sin(90°-x)=cosx2θ=90°-3θ5θ=90°θ=18°So, sin5θ=sin5×18°=sin90°=1

 
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8) Sin 2A = Cos 3A
​Sin 2A = Sin ( 90 - 3A )
2A = 90 - 3A
​2A + 3A = 90
5A = 90.

Sin 90 = 1.

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