if cosA+sinA=2^1/2*cosA, then show that cosA-sinA=2^1/2*sinA.

Question

please give the simplest way of solving it...i know the complicated way.....pls hurry.....

Answer 1

cos A + sin A = 2^1/2 cos A

Subtract cos A from each side

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

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

Take out common factor

sin A = cos A(2^1/2  - 1)

Divide each side by cos A

sin A / cos A = {cos A(2^1/2  - 1)}/ cos A

sin A / cos A = (2^1/2  - 1)

tan A = (2^1/2  - 1)

Apply reciprocal identity: cot A = 1/ tan A

cot A = 1/(2^1/2  - 1)

Multiply numarator and denominator by (2^1/2 + 1)

cot A = {1(2^1/2  + 1)/(2^1/2  - 1)(2^1/2  + 1)}

cot A = {(2^1/2  + 1)/{(2^1/2)^2-1^2}}

cot A = {(2^1/2  + 1)/{2-1}}

cot A = (2^1/2  + 1)

Trigonometry identity: cot A = cos A / sin A

cos A / sin A = (2^1/2  + 1)

Multiply each side by sin A

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

cos A  = (2^1/2  + 1)sin A

cos A  = 2^1/2 sin A + sin A

Subtract sin A from each side

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

cos A - sin A  = 2^1/2 sin A

The solution is cos A - sin A  = 2^1/2 sin A.