cot2θ=tan4θ where 2θ and 4θ are acute angles,find the value of sin3θ
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cot2Θ=tan4Θ
tan(90-2Θ)=tan4Θ [Complementary angles tan(90-Θ)=cotΘ]
90-2Θ=tan4Θ
tan
90-2Θ=4Θ
4Θ+2Θ=90
6Θ=90
Θ=90
6
Θ=15
So sin3Θ=sin(3·15)
=sin45
=1/⌡2
= 1
⌡2
tan(90-2Θ)=tan4Θ [Complementary angles tan(90-Θ)=cotΘ]
90-2Θ=tan4Θ
tan
90-2Θ=4Θ
4Θ+2Θ=90
6Θ=90
Θ=90
6
Θ=15
So sin3Θ=sin(3·15)
=sin45
=1/⌡2
= 1
⌡2
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