Question

if cos theta +sin theta =root 2 cos theta prove that cos theta -sin theta =root 2 sin theta

Answer 1

Answer:-

Given:-

cos θ + sin θ = √2 cos θ

Dividing both sides by cos θ we get,

⟹ (cos θ + sin θ) / cos θ = √2 × cos θ/cos θ

⟹ (cos θ/cos θ) + (sin θ/cos θ) = √2

using sin θ/cos θ = tan θ we get,

⟹ 1 + tan θ = √2

⟹ tan θ = (√2 - 1) / 1

We know,

tan θ = Opposite side/Adjacent side

So,

• Opposite side = √2 - 1
• Adjacent side = 1

By using Pythagoras Theorem,

⟹ (Hypotenuse)² = (Opposite side)² + (Adjacent side)²

⟹ (Hypotenuse)² = (√2 - 1)² + (1)²

• (a - b)² = a² + b² - 2ab

⟹ (Hypotenuse)² = (√2)² + (1)² - 2(√2)(1) + 1

⟹ Hypotenuse = √ [ 2 + 1 - 2√2 + 1 ]

⟹ Hypotenuse = $\sf \sqrt{4 - 2\sqrt{2}}$

Now,

We have to prove that:

cos θ - sin θ = √2 sin θ

We know,

sin θ = Opposite side/Hypotenuse

$\implies \sf \sin \theta = \dfrac{ \sqrt{2} - 1 }{ \sqrt{4 - 2 \sqrt{2} } }$

cos θ = Adjacent side/Hypotenuse

$\implies \sf \cos \theta = \dfrac{1}{ \sqrt{4 - 2 \sqrt{2} } }$

Putting the values we get,

$\implies \sf \: \frac{1}{ \sqrt{4 - 2 \sqrt{2} } } - \frac{ \sqrt{2} - 1 }{ \sqrt{4 - 2 \sqrt{2} } } = \sqrt{2} \times \frac{ \sqrt{2} - 1}{ \sqrt{4 - 2 \sqrt{2} } } \\ \\ \\ \implies \sf \: \frac{1 - ( \sqrt{2} - 1) }{ \sqrt{4 - 2 \sqrt{2} } } = \frac{ \sqrt{2}( \sqrt{2} - 1) }{ \sqrt{4 - 2 \sqrt{2} } } \\ \\ \\ \implies \sf \: \frac{1 - \sqrt{2} + 1}{ \sqrt{4 - 2 \sqrt{2} } } = \frac{2 - \sqrt{2} }{ \sqrt{4 - 2 \sqrt{2} } } \\ \\ \\ \implies \sf \: \frac{2 - \sqrt{2} }{ \sqrt{4 - 2 \sqrt{2} } } = \frac{2 - \sqrt{2} }{ \sqrt{4 - 2 \sqrt{2} } }$

Hence Proved.

Answer 2

$\bigstar \underline{\underline{\sf Given:-}}$

• Cosθ + sinθ=√2cosθ

$\bigstar \underline{\underline{\sf To\ prove:-}}$

• Cosθ - sinθ=√2sinθ

$\bigstar \underline{\underline{\sf Proof:-}}$

We need to know :

$\mapsto \sf cos^2\theta +sin^2\theta =1$

$\mapsto \sf 1-cos^2\theta = sin^2\theta$

$\mapsto \sf (a+b)^2= a^2+b^2+2ab$

$\mapsto \sf (a-b)^2=a^2+b^2 -2ab$

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Given,

• Cosθ + sinθ=√2cosθ

Squaring on both sides,

$\\ :\implies \sf (cos\theta + sin\theta )^2 =(\sqrt{2} cos\theta )^2\\\\:\implies \sf cos^2\theta +sin^2\theta + 2sin\theta cos\theta = 2cos^2\theta \\\\:\implies \sf 1 + 2sin\theta cos\theta = 2cos^2\theta \\\\ :\implies \sf 2sin\theta cos\theta = 2cos^2\theta -1.....(1)$

And,

$:\implies \sf (cos\theta - sin\theta )^2= cos^2\theta + sin^2\theta -2sin\theta cos\theta$

From (1) :

$:\implies \sf (cos\theta -sin\theta )^2 = 1 - 2cos^2\theta +1 \\\\:\implies \sf (cos\theta -sin\theta )^2 = 2-2cos^2\theta\\\\:\implies \sf (cos\theta -sin\theta )^2= 2(1-cos^2\theta)\\\\:\implies \sf cos\theta -sin\theta =\sqrt{2sin^2\theta} =\sqrt{2} sin\theta$

Therefore,

$\longmapsto \ \boxed{\boxed{\bold{cos\theta - sin\theta = \sqrt{2} sin\theta }}}$

Hence proved !

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Know More :-

→ Sinθ = opposite/hypotenuse

→ Cosθ = adjacent/hypotenuse

→ Tanθ = opposite/adjacent

→ Tan θ =sinθ /cosθ

→ 1/sinθ =cosecθ

→ 1/cosθ =sec θ

→ cosec²θ - cot²θ=1

→ tan²θ+1=sec²θ

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HOPE IT HELPS !