Understand · explore · practise

Double angle formulae

Derive sin 2x, cos 2x and tan 2x, choose the right cosine form and calculate exact values. Explore angle doubling, circle geometry and parameter elimination.

Before you startCompound angles, sin² x + cos² x = 1 and quadrant signs

01 / Double the angle inside the function

sin 2θ means sin(2θ), not 2sin θ.

Doubling an angle moves a unit-circle point twice as far around the circle. Its new height is not generally twice its old height: the horizontal coordinate matters too.

sin 2θ = 2sin θ cos θ

Choose θ yourself. The blue point marks θ and the gold point marks 2θ. At θ = 90°, the original sine is 1 but the doubled-angle sine is 0. At θ = 180°, the gold point has already made one full turn.

One turn becomes twoExplore

Angle θ

Blue: θ. Gold: 2θ. Coordinates are (cosine, sine). Decimals are rounded to 4 places.

θ = 30°; 2θ = 60°. sin θ = 0.5000; cos θ = 0.8660.

sin 2θ = 2sin θ cos θ = 0.8660.

cos 2θ = cos² θ − sin² θ = 1 − 2sin² θ = 2cos² θ − 1 = 0.5000.

tan 2θ = 1.7321.

The tangent quotient is valid here.

Watch one angular turn become two

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Set the two compound angles equal

The double-angle identities come directly from addition.

Derive sine and cosine of twice an angleWorked example

sin(θ + θ) = sin θ cos θ + cos θ sin θ

The two terms are equal.

sin 2θ = 2sin θ cos θ

This holds for every real θ.

cos(θ + θ) = cos θ cos θ − sin θ sin θ

Multiply each pair.

cos 2θ = cos² θ − sin² θ

This also holds for every real θ.

The notation cos² θ means (cos θ)². It is different from cos 2θ, whose input is twice the angle.

03 / Choose among three cosine forms

Use Pythagoras to remove the ratio you do not want.

cos 2θ = cos² θ − sin² θ
= 1 − 2sin² θ
= 2cos² θ − 1

Rewrite 7 − 4sin² θ using cos 2θWorked example

2sin² θ = 1 − cos 2θ

Rearrange the sine-only form.

7 − 4sin² θ = 7 − 2(1 − cos 2θ)

Replace the whole squared term.

7 − 4sin² θ = 5 + 2cos 2θ

The frequency changes when squared trig terms are rewritten.

Use 1 − 2sin² θ if sine is given or you want an equation only in sine. Use 2cos² θ − 1 for the corresponding cosine task. Use cos² θ − sin² θ when both ratios are already available.

04 / Double tangent with its domain attached

A rational formula can have a narrower domain than the original function.

tan 2θ = 2tan θ/(1 − tan² θ)

Use this quotient only where cos θ ≠ 0 and tan² θ ≠ 1.

Divide sin 2θ by cos 2θ, then divide numerator and denominator by cos² θ. That final division requires cos θ ≠ 0. If tan² θ = 1, cos 2θ = 0 and tan 2θ really is undefined.

Compare θ = 45° with θ = 90°Worked example

At 45°, tan 2θ = tan 90° is undefined

The quotient has a zero denominator.

At 90°, tan θ is undefined but tan 2θ = tan 180° = 0

Use sine and cosine of the doubled angle instead of the tangent quotient.

05 / Calculate exact values without finding θ

Recover the missing signed ratio first.

sin θ = 5/13, with 90° < θ < 180°. Find sin 2θ, cos 2θ and tan 2θ.Worked example

cos θ = −12/13

Choose the negative square root in quadrant two.

sin 2θ = 2(5/13)(−12/13) = −120/169

Keep the sign through multiplication.

cos 2θ = 1 − 2(25/169) = 119/169

No sign ambiguity arises from squaring here.

tan 2θ = −120/119

Divide the combined sine by the combined cosine.

The signs place 2θ in quadrant four. The original information gives 180° < 2θ < 360°, so that location is consistent.

06 / Start from a tangent ratio

Either use the quotient or reconstruct a signed triangle.

tan θ = −3/4 and θ is in quadrant four. Find all three doubled-angle ratios.Worked example

sin θ = −3/5 and cos θ = 4/5

A 3–4–5 triangle gives the magnitudes.

sin 2θ = −24/25; cos 2θ = 7/25

Use sine doubling and the cosine difference of squares.

tan 2θ = [2(−3/4)]/[1 − 9/16] = −24/7

The tangent quotient agrees with dividing the doubled sine by cosine.

If only tan θ is specified, the original angle is ambiguous by a half-turn. Doubling changes that ambiguity into a whole turn, which leaves all three doubled-angle ratios unchanged wherever defined.

07 / See the sine formula in a triangle

Compare two ways of measuring the same area.

Take unit-circle points P = (cos θ, sin θ) and Q = (cos θ, −sin θ), with 0° < θ < 90°. Triangle OPQ has equal sides OP = OQ = 1, included angle 2θ, vertical base PQ = 2sin θ and horizontal height cos θ.

Area = (1/2)sin 2θ
Area = (1/2)(2sin θ)(cos θ)

Equating the areas gives sin 2θ = 2sin θ cos θ. This area argument proves the acute-θ case directly; the signed compound-angle derivation proves the identity for all real θ.

Use the same chord to derive a cosine formWorked example

PQ² = (2sin θ)² = 4sin² θ

Square the vertical chord length.

The cosine rule gives PQ² = 1² + 1² − 2cos 2θ

The included angle is 2θ.

cos 2θ = 1 − 2sin² θ

Rearrange and compare with the algebraic derivation.

Two equal unit radiiGeometry

08 / Eliminate a parameter and retain its interval

A Cartesian equation alone may describe too much of the curve.

x = 2sin θ, y = 1 + 3cos 2θ, for −π/2 ≤ θ ≤ π/2. Eliminate θ.Worked example

sin θ = x/2

Use the x-equation to replace sine.

y = 1 + 3(1 − 2sin² θ)

Choose the sine-only cosine form.

y = 4 − (3/2)x²

Substitute sin² θ = x²/4.

−2 ≤ x ≤ 2, with −2 ≤ y ≤ 4

The stated θ interval sweeps the whole x interval.

If the original parameter is restricted further to −π/6 ≤ θ ≤ π/6, the same equation holds, but only −1 ≤ x ≤ 1 occurs. Its y-range then becomes 5/2 ≤ y ≤ 4.

09 / Apply double-angle identities again

The input can itself be a multiple angle.

Express cos 4θ using powers of cos θWorked example

cos 4θ = 2cos² 2θ − 1

Treat 2θ as the input angle.

= 2(2cos² θ − 1)² − 1

Use the cosine-only form again.

= 8cos⁴ θ − 8cos² θ + 1

Expand the square carefully.

Express sin 4θ using sin θ and cos θWorked example

sin 4θ = 2sin 2θ cos 2θ

Double 2θ.

= 4sin θ cos θ(cos² θ − sin² θ)

Replace both doubled-angle ratios.

10 / Track the interval when the angle doubles

The new angle may go around the circle more than once.

Solve 2sin θ cos θ = 1/2 for 0° ≤ θ < 360°Worked example

sin 2θ = 1/2

The substitution is an identity for every original input.

0° ≤ 2θ < 720°

Double both interval endpoints.

2θ = 30°, 150°, 390°, 510°

List every sine root in the doubled interval.

At x = 60°, compare sin 2x, 2sin x and sin² x.

Hint

These mean three different operations.

Worked solution

sin 120° = √3/2; 2sin 60° = √3; (sin 60°)² = 3/4. Doubling the input, doubling the output and squaring the output are different.

12 / Recap

Choose the form that matches the information.

Set A = B in compound-angle formulae to derive double angles.
Use the cosine form that removes an unwanted ratio.
Keep quadrant signs when calculating sin 2θ.
Check both tangent domain conditions before using its quotient.
Retain parameter intervals after eliminating θ.
Double the interval when solving in 2θ.

Section 1 of 12 · Double the angle inside the function