Question

given: sinA + 2cosA = 1
to prove: 2sinA - cosA = 2​

Answer 1

Answer:

$2 \sin( \alpha ) - \cos( \alpha ) = - 2$

Answer 2

Answer:

Heya army!

Step-by-step explanation:

Given that Sin A + 2 cos A = 1

Squaring on both sides, we get

(sin A + 2 cos A)^2 = 1

We know that (a+b)^2 = a^2 + b^2 + 2ab.

(sin^2 A + 4 cos^2 A + 4 sin A cos A) = 1

4 cos^2 A + 4 sin A cos A = 1 - sin^2 A

4 cos^2 A + 4 sin A cos A = cos^2 A

3 cos^2 A + 4 sin A cos A = 0    

3 cos^2 A = - 4 sin A cos A  ---- (1).

Given 2 sin A - cos A  

Squaring on both sides, we get

(2 sin A - cos A)^2 = 4 sin^2 A + cos^2 A - 4 sin A cos A

                              = 4 sin^2 A + cos^2 A + 3 cos^2 A

                              = 4 sin^2 A + 4 cos^2 A

                              = 4(sin^2 A + cos^2 A)

                              = 4.

2 sin A - cos A = 2.

LHS  = RHS.

Hope this helps!

Have a day like smooth like butter :)