If cos - sin - - -2 sin - prove that sin - cos - - -2 cos -

Cos θ + sin θ = √(2) sin θ ⇒ (cos θ +sin θ)^2 = 2 sin^2 θ ⇒ cos^2 θ +sin^2 θ + 2 sin θ cos θ = 2 sin^2 θ ⇒ sin^2 θ cos^2 θ 2 sin θ cos θ = 0 ⇒ sin^2 θ +cos^ ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

If cos θ+ sin...

Question

If $(c o s θ + s i n θ) = \sqrt{2} s i n θ$ , prove that $(s i n θ - c o s θ) = \sqrt{2} c o s θ$ .

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Solution

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

$\Rightarrow (c o s θ + s i n θ)^{2} = 2 s i n^{2} θ$

$\Rightarrow c o s^{2} θ + s i n^{2} θ + 2 s i n θ c o s θ = 2 s i n^{2} θ$

$\Rightarrow s i n^{2} θ - c o s^{2} θ - 2 s i n θ c o s θ = 0$

$\Rightarrow s i n^{2} θ + c o s^{2} θ - 2 s i n θ c o s θ = 2 c o s^{2} θ$

$\Rightarrow (s i n θ - c o s θ)^{2} = 2 c o s^{2} θ$

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

Suggest Corrections

If (cosθ+sinθ)=√2sinθ, prove that (sinθ−cosθ)=√2cosθ.

cosθ+sinθ=√2sinθ ⇒(cosθ+sinθ)2=2sin2θ ⇒cos2θ+sin2θ+2sinθ cosθ=2sin2θ ⇒sin2θ−cos2θ−2sinθ cosθ=0 ⇒sin2θ+cos2θ−2sinθ cosθ=2cos2θ ⇒(sinθ−cosθ)2=2cos2θ ∴sinθ−cosθ=√2cosθ

If $(c o s θ + s i n θ) = \sqrt{2} s i n θ$ , prove that $(s i n θ - c o s θ) = \sqrt{2} c o s θ$ .