The Double Angle Formulas

Put B = A: sin 2A = 2 sin A cos A

The compound angle formula sin(A + B) = sin A cos B + cos A sin B holds for every A and B, so it holds when B is equal to A. Then A + B is A + A = 2A, and the formula becomes sin 2A = sin A cos A + cos A sin A.

The two products are the same, so they add to twice one of them: sin 2A = 2 sin A cos A. This is the double angle formula for sine.

Check it at A = 30°: $2\sin 30^{\circ }\cos 30^{\circ }=2\times \frac{1}{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$2 sin 30° cos 30° = 2 × ½ × √3/2 = √3/2, which is sin 60°. In radians, at A = 0.3, 2 sin 0.3 cos 0.3 = 2 × 0.2955 × 0.9553 = 0.5646, which is sin 0.6; at A = 1.1, 2 × 0.8912 × 0.4536 = 0.8085, which is sin 2.2.

Doubling an angle does not double its sine. 2 sin 30° = 1, but sin 60° is $\frac{\sqrt{3}}{2}$√3/2, about 0.87. The cos A in the formula is what corrects the doubling.

The gold curve is y = sin 2x and the plain curve is y = 2 sin x, over one turn with one square across for every 90°. Doubling the angle squeezes the wave into half the width and leaves its height at 1; doubling the sine stretches it to a height of 2. The dots at 45° show the difference: sin 90° = 1, but $2\sin 45^{\circ }=\sqrt{2}$2 sin 45° = √2, about 1.41.

The same move in cosine

Put B = A in cos(A + B) = cos A cos B − sin A sin B. It becomes cos 2A = cos A cos A − sin A sin A, which is $\cos 2A={\cos }^{2}A-{\sin }^{2}A$cos 2A = cos²A − sin²A.

Check it at A = 30°: ${\cos }^{2}30^{\circ }-{\sin }^{2}30^{\circ }=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$cos²30° − sin²30° = 3/4 − 1/4 = ½, which is cos 60°. In radians, ${\cos }^{2}0.3-{\sin }^{2}0.3=0.8253$cos²0.3 − sin²0.3 = 0.8253 to 4 decimal places, which is cos 0.6, and ${\cos }^{2}1.1-{\sin }^{2}1.1=-0.5885$cos²1.1 − sin²1.1 = −0.5885, which is cos 2.2. The answer is negative there because 2.2 radians is more than a quarter turn.

Three forms of cos 2A

The identity ${\sin }^{2}A+{\cos }^{2}A=1$sin²A + cos²A = 1 lets either square be replaced by the other. Replace ${\cos }^{2}A$cos²A with $1-{\sin }^{2}A$1 − sin²A: $\cos 2A=\left(1-{\sin }^{2}A\right)-{\sin }^{2}A=1-2{\sin }^{2}A$cos 2A = (1 − sin²A) − sin²A = 1 − 2sin²A.

Replace ${\sin }^{2}A$sin²A with $1-{\cos }^{2}A$1 − cos²A instead: $\cos 2A={\cos }^{2}A-\left(1-{\cos }^{2}A\right)=2{\cos }^{2}A-1$cos 2A = cos²A − (1 − cos²A) = 2cos²A − 1.

So there are three forms: $\cos 2A={\cos }^{2}A-{\sin }^{2}A=1-2{\sin }^{2}A=2{\cos }^{2}A-1$cos 2A = cos²A − sin²A = 1 − 2sin²A = 2cos²A − 1. At A = 30°, all three give ½: $\frac{3}{4}-\frac{1}{4}$3/4 − 1/4, then $1-2\times \frac{1}{4}$1 − 2 × 1/4, then $2\times \frac{3}{4}-1$2 × 3/4 − 1. Choose the form whose square matches the other terms of the problem, so that only one function is left.

Turned around: ${\sin }^{2}A$sin²A and ${\cos }^{2}A$cos²A

Start from $\cos 2A=1-2{\sin }^{2}A$cos 2A = 1 − 2sin²A. Add $2{\sin }^{2}A$2sin²A to both sides and subtract cos 2A: $2{\sin }^{2}A=1-\cos 2A$2sin²A = 1 − cos 2A. Divide by 2: ${\sin }^{2}A=\frac{\left(1-\cos 2A\right)}{2}$sin²A = (1 − cos 2A)/2. Here the equation is being rearranged to make ${\sin }^{2}A$sin²A the subject, not proved, so working on both sides is allowed.

The same steps on $\cos 2A=2{\cos }^{2}A-1$cos 2A = 2cos²A − 1 give ${\cos }^{2}A=\frac{\left(1+\cos 2A\right)}{2}$cos²A = (1 + cos 2A)/2. The minus sign goes with ${\sin }^{2}A$sin²A and the plus sign with ${\cos }^{2}A$cos²A.

Check at 1.1 radians: $\frac{\left(1-\cos 2.2\right)}{2}=\frac{\left(1+0.5885\right)}{2}=0.7943$(1 − cos 2.2)/2 = (1 + 0.5885)/2 = 0.7943, which is ${\sin }^{2}1.1$sin²1.1. These forms turn a square of a trigonometric function into a plain cosine, and that is how ${\sin }^{2}x$sin²x and ${\cos }^{2}x$cos²x are integrated.

The gold curve is $y={\sin }^{2}x$y = sin²x and the plain curve is y = cos 2x, over one turn with one square across for every 90°; the plain straight line is y = ½. The gold wave is the plain wave turned upside down, halved in height and raised to the level ½: where cos 2x is 1, ${\sin }^{2}x$sin²x is 0, and where cos 2x is −1, ${\sin }^{2}x$sin²x is 1. That is ${\sin }^{2}x=\frac{1}{2}-\frac{1}{2}\cos 2x$sin²x = ½ − ½ cos 2x.

Using the formulas

Suppose A is acute and $\sin A=\frac{3}{5}$sin A = 3/5. Then ${\cos }^{2}A=1-\frac{9}{25}=\frac{16}{25}$cos²A = 1 − 9/25 = 16/25, and $\cos A=\frac{4}{5}$cos A = 4/5 because A is acute. So $\sin 2A=2\times \frac{3}{5}\times \frac{4}{5}=\frac{24}{25}$sin 2A = 2 × 3/5 × 4/5 = 24/25 and $\cos 2A=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$cos 2A = 16/25 − 9/25 = 7/25. Check: ${\left(\frac{24}{25}\right)}^2+{\left(\frac{7}{25}\right)}^2=\frac{\left(576+49\right)}{625}=1$(24/25)² + (7/25)² = (576 + 49)/625 = 1. On a calculator, A is 36.87°, and sin 73.74° = 0.96 and cos 73.74° = 0.28, which are $\frac{24}{25}$24/25 and $\frac{7}{25}$7/25.

The ${\cos }^{2}A$cos²A form gives exact values for half of a known angle. With A = 15°, ${\cos }^{2}15^{\circ }=\frac{\left(1+\cos 30^{\circ }\right)}{2}=\frac{\left(1+\frac{\sqrt{3}}{2}\right)}{2}=\frac{\left(2+\sqrt{3}\right)}{4}$cos²15° = (1 + cos 30°)/2 = (1 + √3/2)/2 = (2 + √3)/4. So $\cos 15^{\circ }=\frac{\sqrt{2+\sqrt{3}}}{2}$cos 15° = √(2 + √3)/2, which is 0.9659 to 4 decimal places, the same as a calculator gives.

Solving an equation with a double angle

Solve sin 2x = sin x for $0^{\circ }\le x\le 360^{\circ }$0° ≤ x ≤ 360°. The equation mixes the angles 2x and x, so first rewrite sin 2x: 2 sin x cos x = sin x.

Bring every term to one side and factor: 2 sin x cos x − sin x = 0, so sin x (2 cos x − 1) = 0. Either sin x = 0, which gives x = 0°, 180° or 360°, or cos x = ½, which gives x = 60° or 300°.

Do not divide both sides by sin x: that throws away the three solutions where sin x = 0. Check x = 60°: $\sin 120^{\circ }=\frac{\sqrt{3}}{2}$sin 120° = √3/2 and $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$sin 60° = √3/2. Check x = 300°: $\sin 600^{\circ }=\sin 240^{\circ }=-\frac{\sqrt{3}}{2}$sin 600° = sin 240° = −√3/2, and $\sin 300^{\circ }=-\frac{\sqrt{3}}{2}$sin 300° = −√3/2.

The usual mistakes

Writing sin 2A = 2 sin A. At A = 30° that gives 1, but sin 60° is $\frac{\sqrt{3}}{2}$√3/2. The formula is 2 sin A cos A.

Writing $\cos 2A={\cos }^{2}A+{\sin }^{2}A$cos 2A = cos²A + sin²A. That sum is the Pythagorean identity, and it equals 1, not cos 2A; the cosine formula has a minus sign.

Mixing up the signs when rearranging. ${\sin }^{2}A=\frac{\left(1-\cos 2A\right)}{2}$sin²A = (1 − cos 2A)/2 and ${\cos }^{2}A=\frac{\left(1+\cos 2A\right)}{2}$cos²A = (1 + cos 2A)/2. Check with A = 0: ${\sin }^{2}0=0$sin²0 = 0 and cos 0 = 1, so the sine form needs the minus.

Forgetting to halve. 1 − cos 2A is $2{\sin }^{2}A$2sin²A, so ${\sin }^{2}A$sin²A is half of it.

The launch angle for the greatest range

In the application below, a ball thrown at $20\frac{m}{s}$20 m/s at an angle $\theta$θ lands $80\sin \theta \cos \theta m$80 sin θ cos θ m away. The double angle formula turns that into $40\sin 2\theta$40 sin 2θ, and a sine is greatest, 1, when $2\theta$2θ is a right angle, so the best angle is 45°.