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Question Number 86490 by ram roop sharma last updated on 29/Mar/20

If \sin θ + \cos θ = \sqrt{2} \cos θ then

\cos θ - \sin θ is equal to

Commented by john santu last updated on 29/Mar/20

\sin θ = \sqrt{2} \cos θ - \cos θ

\tan θ = \sqrt{2} - 1

let \cos θ - \sin θ = k

\sin θ - \cos θ = - k

(\sin θ - \cos θ) (\sin θ + \cos θ) = - k \sqrt{2} \cos θ

- \cos 2 θ = - k \sqrt{2} \cos θ

2 \cos^{2} θ - 1 = k \sqrt{2} \cos θ

k = \frac{2 \cos^{2} θ - 1}{\sqrt{2} \cos θ} = \sqrt{2} \cos θ - \frac{1}{\sqrt{2}} \cos θ

= \sqrt{2} (\frac{1}{\sqrt{2} (\sqrt{2 - \sqrt{2}})}) - \frac{1}{2 \sqrt{2 - \sqrt{2}}}

= \frac{1}{\sqrt{2 - \sqrt{2}}} - \frac{1}{2 \sqrt{2 - \sqrt{2}}} = \frac{1}{2 \sqrt{2 - \sqrt{2}}}

Answered by som(math1967) last updated on 29/Mar/20

s i n θ = (\sqrt{2} - 1) c o s θ

(\sqrt{2} + 1) s i n θ = (\sqrt{2} + 1) (\sqrt{2} - 1) c o s θ

\sqrt{2} s i n θ + s i n θ = 1 . c o s θ

\sqrt{2} s i n θ = c o s θ - s i n θ

∴ c o s θ - s i n θ = \sqrt{2} s i n θ a n s

Commented by peter frank last updated on 29/Mar/20

t h a n k y o u