How do you express sin3θ in terms of trigonometric functions of θ?
Solution
We know the trigonometric identity:
sin(A + B) = sinA cosB + cosA sinB
Step 1: Rewrite sin³θ
Write sin 3θ as sin(2θ + θ):
sin 3θ = sin(2θ + θ)
Applying the identity:
sin 3θ = sin 2θ cos θ + cos 2θ sin θ
Step 2: Substitute Standard Identities
We know that:
sin 2θ = 2 sin θ cos θ
cos 2θ = cos²θ - sin²θ
Substitute these values:
sin 3θ = (2 sin θ cos θ) cos θ + (cos²θ - sin²θ) sin θ
Step 3: Simplify
= 2 sin θ cos²θ + sin θ cos²θ - sin³θ
= 3 sin θ cos²θ - sin³θ
Factor out sin θ:
sin 3θ = sin θ (3 cos²θ - sin²θ)
Step 4: Express in Final Standard Form
Using the identity:
cos²θ = 1 - sin²θ
Substitute into the expression:
sin 3θ = 3 sin θ - 4 sin³θ
Final Answer
sin 3θ = 3 sin θ - 4 sin³θ