Question

if sinA + 2cosA = 1 then prove that 2sinA - cosA=2​

Answer 1

sin A + 2cosA =1

Squaring on both sides

sin square + 4 cos square + 4sin.cos =1

1-cos square + 4- 4sin square + 4sin.cos =1

4sin square + cos square - 4sin.cos= 4

(2sin - cos ) square = 2 square

Taking square root

2sin - cos =2.

Answer 2

Step-by-step explanation:

Let's assume that $\angle A = \alpha$

We have been given that,

sinA + 2cosA = 1

Therefore, we get,

$= > \sin( \alpha ) + 2 \cos( \alpha ) = 1$

Squaring both the sides, We get

$= > {( \sin \alpha + 2 \cos \alpha ) }^{2} = {1}^{2} \\ \\ = > { \sin }^{2} \alpha + 4 { \cos }^{2} \alpha + 2(2 \sin \alpha \cos \alpha ) = 1 \\ \\ = > { \sin }^{2} \alpha + 4 { \cos }^{2} \alpha + 2 \sin(2 \alpha ) = 1 \\ \\ = > 1 - { \cos}^{2} \alpha +4 (1 - { \sin }^{2} \alpha ) + 4 \sin( \alpha ) \cos( \alpha ) = 1 \\ \\ = > 4 -4 { \sin }^{2} \alpha - { \cos}^{2} \alpha + 4 \sin( \alpha ) \cos( \alpha ) = 0 \\ \\ = > 4 { \sin }^{2} \alpha + { \cos }^{2} \alpha - 4 \sin( \alpha ) \cos( \alpha ) = 4 \\ \\ = > {(2 \sin \alpha ) }^{2} - 2(2 \sin\alpha ) ( \cos \alpha ) + {( \cos \alpha ) }^{2} = {(2)}^{2} \\ \\ = > {(2 \sin \alpha - \cos \alpha ) }^{2} = {2}^{2} \\ \\ = > 2 \sin( \alpha ) - \cos( \alpha ) = 2$

Hence, Proved.

Concept Map:-

• ${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}$

• ${(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}$

• ${ \sin }^{2} x + { \cos }^{2} x = 1$