Question

if cosA+sinA=2^(1/2) then find cosA -sinA.

Answer 1

Step-by-step explanation:

$cos \: A + sin \: A = {2}^{ \frac{1}{2} } \\squiring \: both \: sides : \: \\ (cos \: A + sin \: A)^{2} = ({2}^{ \frac{1}{2} })^{2} \\ cos^{2} \: A + sin^{2} \: A + 2cos \: A sin \: A = 2 \\ 1 + 2cos \: A sin \: A = 2 \\ 2sin \: A \: cos \: A = 2 - 1 \\ sin \: 2A = 1 \\ sin \: 2A = sin \: 90 \degree \\ 2A = 90 \degree \\ A = \frac{90 \degree}{2} \\ A = 45 \degree \\ cos \: A - sin \: A \\ =cos \: 45 \degree - sin \: 45 \degree \: \\ = \frac{1}{ \sqrt{2} } - \frac{1}{ \sqrt{2} } \\ = 0 \\ \therefore \: cos \: A - sin \: A = 0$