1. (a) By writing sin 30° as sin (2θ + θ), show that sin 30° = 3sin θ – 4sin³θ - Edexcel - A-Level Maths Pure - Question 2 - 2007 - Paper 6

To show that $\sin 30° = 3\sin θ – 4\sin³θ$, we start by expressing ( heta ) such that ( heta = 10° ). Therefore, we have:

$$\sin(30°) = \sin(2θ + θ)$$

Using the sine addition formula, we can state:

$$\sin(2θ + θ) = \sin 2θ \cos θ + \cos 2θ \sin θ$$

We know from trigonometric identities:

• ( \sin 2θ = 2\sin θ \cos θ )
• ( \cos 2θ = 1 - 2\sin²θ )

Substituting these identities, we get:

$$\sin(30°) = (2\sin θ \cos θ)(\cos θ) + (1 - 2\sin²θ)(\sin θ)$$

This simplifies to:

$$\sin(30°) = 2\sin θ \cos^2 θ + \sin θ - 2\sin³ θ$$

Factoring out ( \sin θ ):

$$\sin(30°) = \sin θ (2\cos² θ + 1 - 2\sin² θ)$$

Now by recognizing that ( 2\cos² θ + 1 - 2\sin² θ = 3 - 4\sin² θ ), we can rewrite it as:

$$\sin(30°) = 3\sin θ - 4\sin³ θ$$

Thus, we've shown the required expression.