Question

if cosA+sinA=√2cosA, show that cosA- sinA = √2sinA.

Answer 1

Answer:

$cosA - sinA = 2sinA$

Step-by-step explanation:

Given : $cosA+sinA=\sqrt{2}cosA,$

To Show  : $cosA- sinA = \sqrt{2}sinA$

Solution :

$cosA+sinA=\sqrt{2}cosA$

Squaring both sides

$(cosA + sin A)^2= (\sqrt{2}cosA)^2$

$cos^2A + sin^2A + 2sinAcosA = 2cos^2A$

Using property : $Sin^2 x +Cos^2 x= 1$

$1 - sin^2A + 1 - cos^2A + 2sinAcosA = 2cos^2A$

$2 - 2cos^2A = cos^2A + sin^2A - 2sinAcosA$

$2(1 - cos^2A)= (cosA - sinA)^2$

$2(Sin^2A)= (cosA - sinA)^2$

$cosA - sinA =\sqrt{2sin^2A}$

$cosA - sinA = 2sinA$

Hence $cosA - sinA = 2sinA$

Answer 2

Step-by-step explanation:

cos A +sin A = √2cos A

squaring on both sides we get

(cos A + sin A)² = (√2cos A)²

it is in the form of (a+b)^2

formula for (a+b)^2=a^2+ab+b^2

by comparing here we have a=cos A and b=sin A

cos²A + sin²A + 2sinAcosA = 2cos²A

we know that

sin^2A+cos^2 A=1

from this we can write

sin^2 A=1-cos^2 A

and cos^2 A=1-sin^2 A

1 - sin²A + 1 - cos²A + 2 sinAcosA = 2 cos²A

adding like terms and sin^2 A+cos^2 A -2 sinA cosA -2 cos^2 A on both sides

2 - 2 cos²A = cos²A + sin²A - 2 sinAcosA  

taking 2 as common in left side of the equation

2(1 - cos²A)= (cos A - sin A)²

cos A - sin A = √[2 sin²A]

cos A-sin A = √2 sin A

hence it is proved