Double Angle Formulas

Double angle formulas are used to express the trigonometric ratios of double angles (2θ) in terms of trigonometric ratios of single angle (θ). The double angle formulas are the special cases of (and hence are derived from) the sum formulas of trigonometry and some alternative formulas are derived by using the Pythagorean identities. Let us recall the sum formulas of trigonometry.

• sin (A + B) = sin A cos B + cos A sin B
• cos (A + B) = cos A cos B - sin A sin B
• tan (A + B) = (tan A + tan B) / (1 - tan A tan B)

What Are Double Angle Formulas?

We will derive the double angle formulas of sin, cos, and tan by substituting A = B in each of the above sum formulas. Also, we will derive some alternative formulas are derived using the Pythagorean identities. Here are the double angle formulas followed by the derivation of each formula.

Double Angle Formulas

The double angle formulas of sin, cos, and tan are,

• sin 2A = 2 sin A cos A (or) (2 tan A) / (1 + tan²A)
• cos 2A = cos²A - sin²A (or) 2cos²A - 1 (or) 1 - 2sin²A (or) (1 - tan²A) / (1 + tan²A)
• tan 2A = (2 tan A) / (1 - tan²A)

Let us derive the double angle formula(s) of each of sin, cos, and tan one by one.

Double Angle Formulas of Sin

The sum formula of sine function is,

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

When A = B, the above formula becomes,

sin (A + A) = sin A cos A + cos A sin A

sin 2A = 2 sin A cos A

Let us derive an alternate formula for sin 2A in terms of tan using the Pythagorean identity sec²A = 1 + tan²A.

\( \begin{align} \sin 2A&= 2 \sin A \cos A \\[0.2cm] &= \dfrac{2 \sin A \cos^2 A}{\cos A}\\[0.2cm] &= \dfrac{2 \sin A}{\cos A} \cdot \cos^2A\\[0.2cm] &= 2 \tan A \cdot \dfrac{1}{ \sec^2A}\\[0.2cm] &= \dfrac{2 \tan A}{1+ \tan^2A} \end{align}\)

Thus, the double angle formulas of sine function are

sin 2A = 2 sin A cos A (or) (2 tan A) / (1 + tan²A)

Double Angle Formulas of Cos

The sum formula of cosine function is,

cos (A + B) = cos A cos B - sin A sin B

When A = B, the above formula becomes,

cos (A + A) = cos A cos A - sin A sin A

cos 2A = cos²A - sin²A

Let us use this as a base formula to derive two other formulas of cos 2A using the Pythagorean identity sin²A + cos²A = 1.

(i) cos 2A = cos²A − (1 − cos²A) = 2cos²A - 1

(ii) cos 2A = (1- sin²A) - sin²A = 1 - 2sin²A

Now, we will derive the formula of cos 2A in terms of tan using the base formula.

\( \begin{align} \cos 2A &= \cos^2A - \sin^2A \\[0.2cm] &= \cos^2A \left( 1- \dfrac{\sin^2A}{\cos^2A} \right)\\[0.2cm] &= \dfrac{1}{\sec^2A} (1- \tan^2A)\\[0.2cm] &= \dfrac{1}{1+\tan^2A} (1-\tan^2A)\\[0.2cm] &= \dfrac{1- \tan^2A}{1+\tan^2A} \end{align} \)

Thus, the double angle formulas of the cosine function are:

cos 2A = cos²A - sin²A (or) 2cos²A - 1 (or) 1 - 2sin²A (or) (1 - tan²A) / (1 + tan²A)

Double Angle Formulas of Tan

The sum formula of tangent function is,

tan (A + B) = (tan A + tan B) / (1 - tan A tan B)

When A = B, the above formula becomes,

tan (A + A) = (tan A + tan A) / (1 - tan A tan A) =(2 tan A) / (1 - tan²A)

Thus, the double angle formula of tan function is,

tan 2A = (2 tan A) / (1 - tan²A)

Let us see the applications of the double angle formulas in the section below.

Examples Using Double Angle Formulas

Example 1: If tan A = 3 / 4, find the values of sin 2A, cos 2A, and tan 2A.

Solution:

Since the value of tan A is given, we use the double angle formulas for finding each of sin 2A, cos 2A, and tan 2A in terms of tan.

\( \begin{align}\sin 2A &= \dfrac{2 \tan A }{1+\tan^2A}\\[0.2cm] &= \dfrac{2 \left(\dfrac{3}{4} \right)}{1+\left(\dfrac{3}{4} \right)^2}\\[0.2cm]&=\dfrac{24}{25} \end{align}\)

\( \begin{align} \cos 2A &= \dfrac{1-\tan ^{2} A}{1+\tan ^{2} A}\\[0.2cm] &= \dfrac{1-\left(\dfrac{3}{4} \right)^2}{1+\left(\dfrac{3}{4} \right)^2}\\[0.2cm] &=\frac{7}{25} \end{align}\)

\( \begin{align}\tan 2A &= \dfrac{2 \tan A }{1-\tan^2A}\\[0.2cm] &= \dfrac{2 \left(\dfrac{3}{4} \right)}{1-\left(\dfrac{3}{4} \right)^2}\\[0.2cm]&=\dfrac{24}{7} \end{align}\)

Answer: sin 2A = \(\dfrac{24}{25}\), cos 2A = \(\dfrac{7}{25}\), and tan 2A = \(\dfrac{24}{7}\)

Example 2: Prove the following identity \( \dfrac{1- \cos 2x}{1+ \cos 2x}= \tan^2 x\).

Solution:

The double angle formula of cos is,

cos 2A = 2cos²A - 1 (or) 1 - 2sin²A

We will use these formulas to prove the given identity.

\( \begin{align} \dfrac{1- \cos 2x}{1+ \cos 2x} &= \dfrac{1-(1-2 \sin^2x)}{1+(2\cos^2x-1)}\\[0.2cm] &= \dfrac{1-1+2 \sin^2x}{1+ 2\cos^2x -1}\\[0.2cm] &= \dfrac{2 \sin^2x}{2 \cos^2x}\\[0.2cm] &= \dfrac{\sin^2x}{\cos^2x}\\[0.2cm] &= \tan^2x \end{align} \)

Answer: The given identity is proved.

Example 3: Use the double angle formulas to derive the formula for sin 3x.

Solution:

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

= sin 2x cos x + cos 2x sin x (using sin (A + B) = sin A cos B + cos A sin B)

= (2 sin x cos x) cos x + (1 - 2 sin²x) sin x (using the double angle formulas)

= 2 sin x cos²x + sin x - 2 sin³x

= 2 sin x (1 - sin²x) + sin x - 2 sin³x

= 2 sin x - 2 sin³x + sin x - 2 sin³x

= 3 sin x - 4 sin³x

Answer: sin 3x = 3 sin x - 4 sin³x.