FG19, HG21 - Exact trigonometric values | Eduqas GCSE Mathematics

Calculate with special angles without replacing their ratios by rounded decimals. FG19 and HG21 require exact sine and cosine values at $0^\circ,30^\circ,45^\circ,60^\circ,90^\circ$ , and exact tangent values at $0^\circ,30^\circ,45^\circ,60^\circ$ . You will derive the acute-angle values from simple triangles and use them in exact calculations.

Special triangles

A right triangle has hypotenuse $14\,\mathrm{cm}$ and angle $30^\circ$ . Can you find the opposite side exactly without a calculator? A special triangle shows that the side is always half the hypotenuse.

Two simple constructions generate the acute-angle values: a right isosceles triangle, and half an equilateral triangle. The ratios do not change when the triangles are enlarged.

Here $\sqrt2$ is the positive number whose square is $2$ , and $\sqrt3$ is the positive number whose square is $3$ . A fraction such as $\sqrt2/2$ means $\sqrt2$ divided by $2$ ; a length such as $7\sqrt3$ means $7\times\sqrt3$ . In particular, $\sqrt2\times\sqrt2=2$ and $\sqrt3\times\sqrt3=3$ .

[DIAGRAM: asset_name: Special right triangles; asset_slug: exact-special-triangles; caption: Unit isosceles legs give the 45-degree values; half an equilateral triangle gives 30 and 60 degrees.; description: Exact right isosceles triangle with legs 1, 1, hypotenuse square root 2 and two 45-degree angles; beside it an equilateral triangle of side 2 split by its perpendicular altitude into right triangles with sides 1, square root 3, 2 and angles 30, 60, 90 degrees.]
Unit isosceles legs give the 45-degree values

Worked example 1: deriving the 45-degree values

Question. A right isosceles triangle has both shorter sides $1$ . Find its hypotenuse and derive the exact sine, cosine and tangent of $45^\circ$ .

Plan. Use Pythagoras for the hypotenuse, then form the side ratios.

The two acute angles are equal and sum to $90^\circ$ , so each is $45^\circ$ .

\begin{aligned} c^2&=1^2+1^2=2,\qquad c=\sqrt2\\ \sin45^\circ&=\frac1{\sqrt2}=\frac{\sqrt2}{2}\\ \cos45^\circ&=\frac1{\sqrt2}=\frac{\sqrt2}{2}\\ \tan45^\circ&=\frac11=1 \end{aligned}

Answer. Sine and cosine are $\sqrt2/2$ ; tangent is $1$ .

Check. The fraction $\sqrt2/2$ times hypotenuse $\sqrt2$ gives $(\sqrt2)^2/2=1$ , recovering either shorter side. This also verifies that $1/\sqrt2$ and $\sqrt2/2$ are equal.

Both $1/\sqrt2$ and $\sqrt2/2$ mean the number which becomes $1$ when multiplied by $\sqrt2$ . Likewise, $1/\sqrt3=\sqrt3/3$ , since multiplying either form by $\sqrt3$ gives $1$ . These two equalities explain the alternative fraction forms used below.

Change to an equilateral triangle of side $2$ . Its perpendicular altitude splits the base into two lengths of $1$ and the top $60^\circ$ angle into two $30^\circ$ angles.

The two right triangles have equal hypotenuses and a shared height. They are congruent by RHS, so their base halves and top angles are equal.

Worked example 2: deriving 30 and 60 degrees

Question. Use half an equilateral triangle of side $2$ to find the exact trigonometric ratios at $30^\circ$ and $60^\circ$ .

Plan. Find the altitude, then choose opposite and adjacent relative to each acute angle.

h^2=2^2-1^2=3,\qquad h=\sqrt3

At the top $30^\circ$ angle, opposite is $1$ , adjacent $\sqrt3$ , hypotenuse $2$ . At the bottom $60^\circ$ angle, the shorter-side roles swap.

\begin{aligned} \sin30^\circ&=\tfrac12\\ \cos30^\circ&=\tfrac{\sqrt3}{2}\\ \tan30^\circ&=\tfrac1{\sqrt3}=\tfrac{\sqrt3}{3}\\ \sin60^\circ&=\tfrac{\sqrt3}{2}\\ \cos60^\circ&=\tfrac12\\ \tan60^\circ&=\sqrt3 \end{aligned}

Answer. The ratios are those shown above.

Check. The three side lengths satisfy $1^2+(\sqrt3)^2=4=2^2$ . Also $(\sqrt3/3)\times\sqrt3=1$ , confirming the alternative exact form of $\tan30^\circ$ .

The sine and cosine columns swap between $30^\circ$ and $60^\circ$ , since the opposite and adjacent sides swap.

Exact acute-angle values

Angle	$\sin\theta$	$\cos\theta$	$\tan\theta$
$30^\circ$	$1/2$	$\sqrt3/2$	$\sqrt3/3$
$45^\circ$	$\sqrt2/2$	$\sqrt2/2$	$1$
$60^\circ$	$\sqrt3/2$	$1/2$	$\sqrt3$

These exact values are not supplied on the formula lists: learn them, or reconstruct them from the special triangles. Equivalent forms $1/\sqrt2$ and $1/\sqrt3$ are also exact.

Keep the root symbols when an exact answer is requested.

Watch out

$\sin30^\circ=1/2$ , while $\cos30^\circ=\sqrt3/2$ . First identify the chosen angle and its opposite/adjacent sides; do not choose a table entry by familiarity alone.

Use the half-valued ratios in new calculations.

Axis angles

At $0^\circ$ and $90^\circ$ , an ordinary right triangle collapses onto an axis. A radius of length $1$ gives a geometric way to extend the sine and cosine values to these endpoints.

Measure the angle anticlockwise from the positive horizontal axis. For a radius of length $1$ in the first quadrant, sine is the endpoint's vertical coordinate and cosine its horizontal coordinate: the ratio denominator is $1$ . At $0^\circ$ , the endpoint is $(1,0)$ ; at $90^\circ$ , it is $(0,1)$ .

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Number

Algebra

Ratio and proportion

Geometry and measures

Higher graphs

Probability

Statistics

FG19, HG21 - Exact trigonometric values

Special triangles

Axis angles

Continue this lesson.

Learn faster with Nova