If cosθ + sinθ = root 2 cosθ, show that cosθ sinθ = root 2 sinθ.
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58
cosθ+sinθ=√2cosθ
or, sinθ=√2cosθ-cosθ
or, sinθ=cosθ(√2-1)
or, sinθ=cosθ[(√2-1)(√2+1)/(√2+1)]
or, sinθ=cosθ(2-1)/(√2+1)
or, √2sinθ+sinθ=cosθ
or, sinθ-cosθ=-√2sinθ
or, cosθ-sinθ=√2sinθ
or, sinθ=√2cosθ-cosθ
or, sinθ=cosθ(√2-1)
or, sinθ=cosθ[(√2-1)(√2+1)/(√2+1)]
or, sinθ=cosθ(2-1)/(√2+1)
or, √2sinθ+sinθ=cosθ
or, sinθ-cosθ=-√2sinθ
or, cosθ-sinθ=√2sinθ
Answered by
21
cosθ+sinθ=√2cosθ
or, sinθ=√2cosθ-cosθ
or, sinθ=cosθ(√2-1)
or, sinθ=cosθ[(√2-1)(√2+1)/(√2+1)]
or, sinθ=cosθ(2-1)/(√2+1)
or, √2sinθ+sinθ=cosθ
or, sinθ-cosθ=-√2sinθ
or, cosθ-sinθ=√2sinθ
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