Trigonometric identities: Problem 01

We want to show that:

$$\sin(3x) = (3\cos^2(x) - \sin^2(x))\sin(x)$$

11. Use the addition formula for

$$\sin(3x) = \sin(2x + x)$$

We use the addition formula:

$$\sin(A + B) = \sin A \cos B + \cos A \sin B$$

Let

$$A = 2x, \quad B = x$$

Then:

$$\sin(3x) = \sin(2x + x) = \sin(2x)\cos(x) + \cos(2x)\sin(x)$$

22. Use the double-angle formulas for

$$\sin(2x) \quad \text{and} \quad \cos(2x)$$

We use the known identities:

$$\sin(2x) = 2\sin(x)\cos(x)$$

$$\cos(2x) = \cos^2(x) - \sin^2(x)$$

Substitute these into the equation:

$$\sin(3x) = (2\sin(x)\cos(x))\cos(x) + (\cos^2(x) - \sin^2(x))\sin(x)$$

33. Simplify the expression

First term:

$$(2\sin(x)\cos(x))\cos(x) = 2\sin(x)\cos^2(x)$$

Second term:

$$(\cos^2(x) - \sin^2(x))\sin(x) = \cos^2(x)\sin(x) - \sin^3(x)$$

Add both terms:

$$\sin(3x) = 2\sin(x)\cos^2(x) + \cos^2(x)\sin(x) - \sin^3(x)$$

Group terms:

$$= (2\cos^2(x) + \cos^2(x))\sin(x) - \sin^3(x)$$

$$= (3\cos^2(x))\sin(x) - (\sin^2(x))\sin(x)$$

$$= (3\cos^2(x) - \sin^2(x))\sin(x)$$

44. Conclusion

We have now shown that:

$$\sin(3x) = (3\cos^2(x) - \sin^2(x))\sin(x)$$

55. Bonus: Multiple-Angle Formulas

The identity we just proved for $\sin(3x)$ is a specific case of what’s known as multiple-angle formulas in trigonometry. These formulas express $\sin(nx)$ and $\cos(nx)$ in terms of powers of $\sin(x)$ and $\cos(x)$.

For example:

• $\sin(2x) = 2\sin(x)\cos(x)$
• $\sin(3x) = 3\sin(x) - 4\sin^3(x)$
• $\cos(3x) = 4\cos^3(x) - 3\cos(x)$

See: Multiple-Angle Trigonometric Formulas