Question

(psinФ + qcosФ)² + (psinФ + qcosФ)²=p²+q²
prove that

Answer 1

Answer:

May be your question has some defect. Please check the question at first.

Answer 2

Answer:

cosθ=p2+q2q

squaring both sides

cos2θ=p2+q2q2⇒p2+q2=cos2θq2⇒p2=cos2θq2−q2⇒p2=cos2θq2−q2cos2θ

⇒p2=cos2θq2(1−cos2θ)⇒p2=cos2θq2sin2θ⇒