Trigonometric identities: Problem 01

We want to show that:

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

1. 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

$$A=2x,B=x$$

Then:

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

2. Use the double-angle formulas for

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

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:

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

3. 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:

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

4. Conclusion

We have now shown that:

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

5. 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