Question

if (cosФ+sinФ)=√2 sin(90°-Ф), show that (sinФ-cosФ)= √2 cosФ

Answer 1

SEE THE VALUES ABOVE:

sin 0° = √(0/4) = 0

sin 30° = √(1/4) = ½

sin 45° = √(2/4) = 1/√2 = √2/2

sin 60° = √3/4 = √3/2;

cos 90° = √(4/4) = 1.

Answer 2

sin 0° = √(0/4) = 0

sin 30° = √(1/4) = ½

sin 45° = √(2/4) = 1/√2 = √2/2

sin 60° = √3/4 = √3/2;

cos 90° = √(4/4) = 1.