Sine and cosine of complementary angles

Three quick questions from earlier lessons.

In a right triangle with acute angles $A$ and $B$ , the side opposite $A$ is adjacent to $B$ . So $\sin A$ equals…

Co-functions of complementary angles.

The acute angles of a right triangle add to

90^\circ

, and

\sin x = \cos(90^\circ - x)

,

\cos x = \sin(90^\circ - x)

.

$\sin 25^\circ = \cos ?$

90 - 25 = 65

Complement.

\sin 25^\circ = \cos 65^\circ

Co-function.

Your turn: $\cos 72^\circ = \sin y^\circ$ with $y$ acute.

y = 18

90 − 72.

Find the angle $y$ between $0^\circ$ and $90^\circ$ with $\sin 21^\circ = \cos y^\circ$ (degrees).
Find the angle $y$ between $0^\circ$ and $90^\circ$ with $\sin 69^\circ = \cos y^\circ$ (degrees).

In a right triangle, $\sin A = \dfrac{ 3 }{ 10 }$ . Find $\cos B$ , where $B$ is the other acute angle.

Why is $\sin 65^\circ = \cos 25^\circ$ ?
1. In one right triangle, the side opposite the $65^\circ$ angle is adjacent to the other acute angle
2. Because sin and cos always add to 1
3. Because sin and cos are the same function

Three questions to check you have got it.

Find the angle $y$ between $0^\circ$ and $90^\circ$ with $\sin 21^\circ = \cos y^\circ$ (degrees).
Find the angle $y$ between $0^\circ$ and $90^\circ$ with $\sin 69^\circ = \cos y^\circ$ (degrees).
Find the angle $y$ between $0^\circ$ and $90^\circ$ with $\sin 80^\circ = \cos y^\circ$ (degrees).

Complementary angles

$\sin x = \cos(90^\circ - x)$ .

Watch out: equal values need complementary angles, not the same angle.

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