Find cos θ + sin θ in terms of p and q.

Given sin θ = p/q, find cos θ + sin θ in terms of p and q.

Answer :

Given that sin θ = p/q

AB/AC = p/q

Let AB = px

And then AC = qx

In right angled triangle ABC

AC² = AB² + BC²

⇒ BC² = AC² – AB²

⇒ BC² = q²x² – p²x²

⇒ BC² = (q² – p²)x²

⇒ BC = √( q² – p²)x

In right angled triangle ABC,

cos θ = base/hypotenuse

= BC/AC

= √( q² – p²)x/qx

= √( q² – p²)/q

Now,

Sin θ + cos θ = p/q + √(q²–p²)/q

= [p + √(q²–p²)]/q

More Solutions:

• If tan θ = p/q
• If 3 cot θ = 4, find the value of (5 sinθ – 3 cosθ)/(5 sinθ + 3 cosθ).
• If 5 cosθ – 12 sinθ = 0.
• If 5 sin θ = 3
• Find the value of 2 tan2 θ + sin2 θ – 1.
• Prove the following: cos θ tan θ = sin θ
• Prove that sin A cos B + cos A sin B = 1.
• ∆ABC is right-angled at B
• Its diagonal AC = 15 cm and ∠ACD = α. If cot α = 3/2.
• Using the measurements given in the figure