Double Angle Formulas for Sine and Cosine

The double-angle formulas for sine and cosine tell how to find the sine and cosine of twice an angle ($\,2x\,$),
in terms of the sine and cosine of the original angle ($\,x\,$):

double-angle formula for the sine: $$\sin 2x = 2\sin x\,\cos x$$
double-angle formulas for the cosine: three versions: $$ \begin{align} \cos 2x &=\cos^2 x - \sin^2 x\cr\cr \cos 2x &= 1 - 2\sin^2 x\cr\cr \cos 2x &= 2\cos^2 x - 1 \end{align} $$

Notes on the Double-Angle Formulas:

Derivations of the Double-Angle Formulas

The double-angle formulas are simple to prove, once the Addition Formulas for Sine and Cosine are in place.
For all real numbers $\,x\,$:

$\sin 2x$ $=$ $\sin (x + x)$ (rename $\,2x\,$ as $\,x + x\,$)
  $=$ $\sin x\,\cos x + \cos x\,\sin x$ (the sine addition formula)
  $=$ $2\sin x\,\cos x$ (add like terms)
$\cos 2x$ $=$ $\cos (x + x)$ (rename $\,2x\,$ as $\,x + x\,$)
  $=$ $\cos x\,\cos x - \sin x\,\sin x$ (the cosine addition formula)
  $=$ $(\cos x)^2 - (\sin x)^2$ (exponent notation)
  $=$ $\cos^2 x - \sin^2 x$ (shorthand notation)

By the Pythagorean Identity,   $\,\cos^2 x = 1 - \sin^2 x\,$   and   $\,\sin^2 x = 1 - \cos^2 x\,$.
Thus, we get two alternate versions of the cosine double angle formula: $$\cos 2x \quad =\quad \cos^2 x - \sin^2 x \quad =\quad (1 - \sin^2 x) - \sin^2 x \quad =\quad 1 - 2\sin^2 x$$ and $$\cos 2x \quad =\quad \cos^2 x - \sin^2 x \quad =\quad \cos^2 x - (1 - \cos^2 x) \quad =\quad \cos^2 x - 1 + \cos^2 x \quad =\quad 2\cos^2 x - 1$$

A Geometric Proof of the Double-Angle Formulas, for Small Angles

For the sketch and derivation below, assume $\,x\,$ is measured in degrees and $\,2x < 90^\circ\,$.
This is a different ‘stacking’ of triangles than in the prior lesson, to show alternative proof approaches.

  • Start with right triangle $\,\triangle ABC\,$ (yellow).
    It has acute angle $\,x\,$ and hypotenuse of length $\,1\,$.
    • bottom leg: $\,\cos x\,$
    • side leg: $\,\sin x\,$
  • Stack a right triangle $\,\triangle ACD\,$ (mostly green)
    on the hypotenuse of the yellow triangle.
    It also has acute angle $\,x\,$.
    The bottom leg has length $\,1\,$.
    • Since $\,\cos x = \frac{\text{ADJ}}{\text{HYP}} = \frac{1}{\text{HYP}}\,$,
      the hypoteuse has length $\,\frac{1}{\cos x}\,$.
    • Thus, the side leg is $\,\frac{1}{\cos x}\cdot \sin x = \frac{\sin x}{\cos x}\,$.
  • the blue right triangle:
    • since $\,\overline{EC}\, ||\, \overline{AB}\,$, $\,\angle ACE = x\,$
      (alternate interior angles)
    • thus, $\,\angle ECD = 90^\circ - x\,$
    • thus, $\,\angle EDC = x\,$
    • Thus:
      • bottom leg: $\,\frac{\sin x}{\cos x}(\sin x) = \frac{\sin^2 x}{\cos x}\,$
      • side leg: $\,\frac{\sin x}{\cos x}(\cos x) = \sin x\,$
  • Using right triangle $\,\triangle AFD\,$,
    we get the cosine double-angle formula: $$ \begin{align} \cos 2x &= \frac{\text{ADJ}}{\text{HYP}}\cr &= \frac{\cos x - \frac{\sin^2 x}{\cos x}}{1/\cos x}\cr\cr &= \left(\cos x - \frac{\sin^2 x}{\cos x}\right)(\cos x)\cr\cr &= \cos^2 x - \sin^2 x\cr\cr \end{align} $$

Again using right $\,\triangle AFD\,$, we get the sine double-angle formula: $$ \sin 2x \ =\ \frac{\text{OPP}}{\text{HYP}} \ =\ \frac{\sin x +\sin x}{1/\cos x} \ =\ 2\sin x\,\cos x $$

Review:   Does $\,\cos x\,$ Uniquely Determine $\,\sin x\,$?   No!

Recall that every nonzero real number is uniquely identified by knowing two things:

This fact is often crucial when working with trigonometric expressions.

Often, knowing the value of a trigonometric expression (like $\,\cos x\,$)
allows you to correctly determine the size of other trigonometric expressions, but not the sign.
Additional information must be provided to get the correct sign.

For example, suppose you know that $\,\cos x = \frac 45\,$, and nothing else.
Does this allow us to uniquely determine $\,\sin x\,$?
It gives the size of $\,\sin x\,$, but not the sign, as follows:
  • Knowing $\,\cos x = \frac 45\,$ puts the terminal point for $\,x\,$ in either quadrant I (point $\,A\,$)
    or quadrant IV (point $\,B\,$), as shown at right.
  • Finding the size of $\,\sin x\,$ is routine.
    For example, you can use the Pythagorean Identity: $$\sin^2 x \ =\ 1 - \cos^2 x \ =\ 1 - \left(\frac 45\right)^2 \ =\ \frac{25}{25} - \frac{16}{25}\ =\ \frac{9}{25}$$ Thus, $\sin x = \pm \frac{3}{5}\,$.
    The size of $\,\sin x\,$ is $\,\frac 35\,$. The sign is unknown (without additional information).
There is a variety of additional information that might be given to determine the sign of $\,\sin x\,$.
Here are some examples:

information: consequence: information: consequence:
$\,x\,$ is in Quadrant I$\,\sin x\,$ is positive $\,x\,$ is in Quadrant IV$\,\sin x\,$ is negative
$\csc x > 0$cosecant is the reciprocal of the sine;
if cosecant is positive, so is sine;
$\,\sin x\,$ is positive
$\csc x < 0$cosecant is the reciprocal of the sine;
if cosecant is negative, so is sine;
$\,\sin x\,$ is negative
$\tan x > 0$tangent is sine over cosine;
if cosine is positive
and tangent is positive, then:
$\,\sin x\,$ is positive
$\tan x < 0$tangent is sine over cosine;
if cosine is positive
and tangent is negative, then:
$\,\sin x\,$ is negative

Example: Using the Double-Angle Formulas

Suppose that $\displaystyle\,\cos x = \frac 45\,$ and $\,\csc x < 0\,$.
Find $\,\sin 2x\,$, $\,\cos 2x\,$, and $\,\tan 2x\,$.


Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Trying to Undo Trigonometric Functions
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