Question

If cos theta + sin theta=✓2 cos theta shw that cos theta-sin theta=✓2 sin theta

Answer 1

Solution :

Given :

cosθ + sinθ = √2cosθ

Squaring on both sides

⇒ (cosθ + sinθ)² = (√2cosθ)²

⇒ cos²θ + sin²θ + 2cosθ.sinθ = 2cos²θ

[ Because (a + b)² = a² + b² + 2ab ]

⇒ 1 + 2cosθ.sinθ = 2cos²θ

[ Because cos²θ + sin²θ = 1 ]

⇒ 2cosθ.sinθ = 2cos²θ - 1

Now, let us assume cosθ - sinθ = x

cosθ - sinθ = x

Squaring on both sides

⇒ (cosθ - sinθ)² = x²

⇒ cos²θ + sin²θ - 2cosθ.sinθ = x²

[ Because (a - b)² = a² + b² - 2ab ]

⇒ cos²θ + sin²θ - 2cosθ.sinθ = x²

⇒ cos²θ + sin²θ - (2cos²θ - 1) = x²

[ Proved, 2cosθ.sinθ = 2cos²θ - 1 ]

⇒ cos²θ + sin²θ - 2cos²θ + 1 = x²

⇒ - cos²θ + sin²θ + 1 = x²

Rearranging the terms

⇒ sin²θ + 1 - cos²θ = x²

[ Because 1 - cos²θ = sin²θ ]

⇒ sin²θ + sin²θ = x²

⇒ 2sin²θ = x²

Taking square root on both sides

⇒ √( 2sin²θ) = √x²

⇒ √2sinθ = x

⇒ x = √2sinθ

i.e cosθ - sinθ = √2sinθ

Hence shown.

Answer 2

Given:----

• cosA + sinA = √2cosA

To prove:----

• cosA - sinA = √2 sinA

Formula used:----

• (a+b)² =a²+ab+b²
• sin²A+cos²A=1
• a²+b²-2ab = (a-b)²

cos A +sin A = √2cos A

squaring on both sides we get

(cos A + sin A)² = (√2cos A)²

using (a+b)² =a²+ab+b² in LHS now we get,

a = cos A ,, b = sin A

cos²A + sin²A + 2sinAcosA = 2cos²A

now we know that sin²A+cos²A=1

or, sin²A = 1-cos²A

cos²A = 1-sin²A

putting these values in LHS now ,,

→ (1 - sin²A) + (1 - cos²A) + 2 sinAcosA = 2cos²A

Adding like terms and sin²A+cos²A -2sinA cosA -2cos²A on both sides we get,,

→ 2 - 2 cos²A = cos²A + sin²A - 2 sinAcosA

taking 2 as common in LHS and using (a²+b²-2ab = (a-b)² in RHS we get,

→ 2(1 - cos²A)= (cos A - sin A)²

putting (1-cos²A) = sin²A and square root both sides now,

→ (cosA - sinA) = √[2 sin²A]

→ (cosA-sinA) = √2 sinA

Hence , Proved .........

(Hope it helps you)