Exact Values and Reference Angles | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Two triangles give the exact values

means no rounding: keep a fraction or square root instead of a decimal approximation.

Pythagoras supplies the missing length in each triangle.

An isosceles right triangle has two equal angles: each is $(180^\circ-90^\circ)/2=45^\circ$ . With both short sides 1, Pythagoras gives the hypotenuse $\sqrt{1^2+1^2}=\sqrt2$ .

Split an equilateral triangle of side 2 down the middle. Each half has hypotenuse 2, base 1 and height $\sqrt{2^2-1^2}=\sqrt3$ . The angles are 30°, 60° and 90°. Divide the appropriate sides:

Angle	sin	cos	tan
$30^\circ=\pi/6$	$1/2$	$\sqrt3/2$	$\sqrt3/3$
$45^\circ=\pi/4$	$\sqrt2/2$	$\sqrt2/2$	$1$
$60^\circ=\pi/3$	$\sqrt3/2$	$1/2$	$\sqrt3$

The forms $1/\sqrt2=\sqrt2/2$ and $1/\sqrt3=\sqrt3/3$ are equal: multiply top and bottom by the same square root. These exact values are not tabulated in MF19.

A short memory pattern

For 0°, 30°, 45°, 60°, 90°, sine follows $\sqrt0/2$ , $\sqrt1/2$ , $\sqrt2/2$ , $\sqrt3/2$ , $\sqrt4/2$ . Cosine uses the list backwards. Divide sine by cosine for tangent; at 90° the denominator is zero.

Reference angle first, sign second

The is the acute angle between the radius and the nearest horizontal axis. It gives the size of the ratio. The quadrant gives the sign.

Reference angle: 60°. Sine is positive; cosine and tangent are negative.

CAST records which ratios are positive: all in I, sine in II, tangent in III, cosine in IV. On the axes, use the coordinates directly rather than an acute reference angle.

Worked example

Find the values exactly

$120^\circ$ is in II, with reference angle $180^\circ-120^\circ=60^\circ$ . Sine is positive:

\sin120^\circ=\sin60^\circ=\frac{\sqrt3}{2}

$7\pi/6$ is in III, with reference angle $7\pi/6-\pi=\pi/6$ . Cosine is negative:

\cos\frac{7\pi}{6}=-\cos\frac{\pi}{6}=-\frac{\sqrt3}{2}

For a negative or large angle, first remove complete turns:

\sin(-30^\circ)=\sin330^\circ=-\frac12

\tan405^\circ=\tan45^\circ=1

In a product such as $\sin30^\circ\cos45^\circ$ , evaluate each function first, then multiply. The angles are not multiplied together.

Your turn

Your turnPractice

Find exactly: $\cos150^\circ$ , $\sin(-5\pi/6)$ and $\tan225^\circ$ . Explain each sign.

Show worked answer

150° is in II with reference angle 30°. Add $2\pi$ to $-5\pi/6$ to get $7\pi/6$ , in III. 225° is in III with reference angle 45°.

Answer

\cos150^\circ=-\frac{\sqrt3}{2}

\sin(-5\pi/6)=-\frac12

\tan225^\circ=1