Sine and Cosine of Complementary Angles

The two acute angles of a right triangle add to $90^\circ$, so they are complementary. Look at them with the ratios from Sine, Cosine and Tangent and something nice happens.

Por qué funciona

In right $\triangle ABC$ with the right angle at $C$, the side opposite $A$ is $\overline{BC}$, and that same side is adjacent to $B$. So $\sin A = \frac{BC}{AB} = \cos B$. Since $B = 90^\circ - A$, this says $\sin A = \cos(90^\circ - A)$. That is where the name comes from: the cosine is the sine of the complement.

Switch between angle $A$ and angle $B$: the sine of one is always the cosine of the other.

Idea clave

$\sin A = \cos(90^\circ - A)$ and $\cos A = \sin(90^\circ - A)$. Also $\tan A = \frac{1}{\tan(90^\circ - A)}$, because opposite and adjacent swap.

Ejemplo resuelto: Using the identity

$\sin 25^\circ \approx 0.4226$. Find $\cos 65^\circ$ without a calculator.

$25^\circ + 65^\circ = 90^\circ$, so they are complementary.
$\cos 65^\circ = \sin 25^\circ \approx 0.4226$.

Respuesta: $\cos 65^\circ \approx 0.4226$

Ejemplo resuelto: Solving an equation

Both angles are acute and $\sin(2x + 10)^\circ = \cos(3x)^\circ$. Find $x$.

Sine equals cosine for complementary acute angles: $(2x + 10) + 3x = 90$.
$5x + 10 = 90$, so $x = 16$.
Check: the angles are $42^\circ$ and $48^\circ$, which add to $90^\circ$.

Respuesta: $x = 16$

Ejemplo resuelto: Tangents of the two angles

In right $\triangle ABC$ (right angle at $C$), $\tan A = \frac{3}{4}$. Find $\tan B$.

$\tan A = \frac{BC}{AC} = \frac{3}{4}$. From $B$, the legs trade roles: $\tan B = \frac{AC}{BC}$.

Respuesta: $\tan B = \frac{4}{3}$

You can see the identity in the exact values from the special right triangles: $\sin 30^\circ = \frac{1}{2} = \cos 60^\circ$, and $\sin 60^\circ = \frac{\sqrt{3}}{2} = \cos 30^\circ$. It also explains why $\sin 45^\circ = \cos 45^\circ$: a $45^\circ$ angle is its own complement. In practice, the identity lets you switch between sine and cosine whenever that makes an equation easier to solve.

The identity shows up in word problems too. A ladder of length $L$ that makes a $70^\circ$ angle with the ground makes a $20^\circ$ angle with the wall, so the height it reaches is $L \sin 70^\circ$, which is the same number as $L \cos 20^\circ$. Either way of writing it is correct.

Error común

Writing $\sin A = \cos A$. That is true only when $A = 45^\circ$. The identity pairs an angle with its complement: $\sin 20^\circ = \cos 70^\circ$, not $\cos 20^\circ$. Check: the two angles in the equation should add to $90^\circ$.

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