Math, asked by ankesh111, 11 months ago

Answer this trigonometry problem in the pic

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Answered by karannnn43

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$\small\underline\bold\red{Given-}$

cosθ + sinθ = $\sqrt{2}$ cosθ

$\small\underline\bold\red{To \:prove-}$

cosθ - sinθ = $\sqrt{2}$ sinθ

_____________________________________

$\small\underline\bold\red{Proof:-}$

(cosθ + sinθ )² = ( $\sqrt{2}$ cosθ )²

$\implies$ cos²θ + sin²θ + 2cosθ.sinθ = 2cos²θ

$\implies$ 1 + 2cosθ.sinθ = 2cos²θ

$\implies$ 2cosθ.sinθ = 2cos²θ -1

$\implies$ 2cosθ.sinθ = cos2θ ........(i)

Now,

(cosθ - sinθ )²

= cos²θ + sin²θ - 2cosθ.sinθ

= 1 - 2cosθ.sinθ

= 1 - cos2θ ; [ from (i) ]

= 2sin²θ

∵ (cosθ - sinθ )² = 2sin²θ

$\implies$ cosθ - sinθ = √(2sin²θ)

$\implies$ cosθ - sinθ = $\sqrt{2}$ sinθ

$\small\fcolorbox{red}{white}{Hence, \:proved}$

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Answered by Anonymous

Given−

cosθ + sinθ = {2}

cosθ

To prove

cosθ - sinθ = {2}

sinθ

_____________________________________

Proof:−

(cosθ + sinθ )² = {2}

cosθ )²

cos²θ + sin²θ + 2cosθ.sinθ = 2cos²θ

1 + 2cosθ.sinθ = 2cos²θ

2cosθ.sinθ = 2cos²θ -1

2cosθ.sinθ = cos2θ ........(i)

Now,

(cosθ - sinθ )²

= cos²θ + sin²θ - 2cosθ.sinθ

= 1 - 2cosθ.sinθ

= 1 - cos2θ ; [ from (i) ]

= 2sin²θ

∵ (cosθ - sinθ )² = 2sin²θ

cosθ - sinθ = √(2sin²θ)

cosθ - sinθ = {2}

sinθ

Hence,proved

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