E6.3 - Exact trigonometric values

E6.3 - Exact trigonometric values

A calculator gives decimals for trigonometric ratios, but a carefully chosen triangle gives their exact values. You will derive the values for special angles, use surds in exact calculations, and explain what happens at zero and ninety degrees. This lesson is for Extended.

The forty-five-degree triangle

Can a triangle give you an exact answer that a long decimal cannot? Draw a right-angled isosceles triangle with two legs of length 11. Its two acute angles are equal and add to 90∘90^\circ, so each is 45∘45^\circ.

Pythagoras gives the hypotenuse 12+12=2\sqrt{1^2+1^2}=\sqrt2. The sine and cosine ratios are both 1/21/\sqrt2. Multiplying numerator and denominator by 2\sqrt2 writes this as 2/2\sqrt2/2, with no approximation.

[DIAGRAM: asset_name: The exact special triangles; asset_slug: exact_triangles; caption: Pythagoras fixes the side ratios for 45 degrees and for 30 and 60 degrees.; description: Two exact right triangles side by side. The first has legs 1 and 1, hypotenuse square root 2 and two 45 degree angles. The second is half of an equilateral triangle of side 2, with short leg 1, long leg square root 3, hypotenuse 2, and angles 30 and 60 degrees. The remaining half of the equilateral triangle is shown faintly. All right angles are marked.]
Pythagoras fixes the side ratios for 45 degrees and for 30 and 60 degrees.

Exact values at forty-five degrees

sin⁡45∘=cos⁡45∘=22,tan⁡45∘=1\sin45^\circ=\cos45^\circ=\frac{\sqrt2}{2}, \qquad \tan45^\circ=1

These are ratios, so enlarging the triangle does not change them.

Not given in the List of formulas: learn them.

For a length problem, still choose opposite, adjacent and hypotenuse first. An exact ratio replaces a calculator key; it does not replace the triangle.

Worked example 1: finding a leg

Question. A right-angled triangle has a 45∘45^\circ angle and hypotenuse 12 cm12\,\mathrm{cm}. Find the exact length of each leg.

Plan. Use cosine for one leg; isosceles symmetry gives the other.

x12=cos⁡45∘adjacent over hypotenusex=12×22=62simplify exactly\begin{aligned} \frac{x}{12}&=\cos45^\circ && \text{adjacent over hypotenuse}\\ x&=12\times\frac{\sqrt2}{2}\\ &=6\sqrt2 && \text{simplify exactly} \end{aligned}

Answer. Each leg is 62 cm6\sqrt2\,\mathrm{cm}.

Check. (62)2+(62)2=72+72=144=122(6\sqrt2)^2+(6\sqrt2)^2=72+72=144=12^2. This verifies the original hypotenuse without using decimals.

Now the given length is a leg instead of the hypotenuse. The unknown moves to the denominator of the ratio.

Worked example 2: finding the hypotenuse

Question. A right-angled isosceles triangle has legs of length 12 cm12\,\mathrm{cm}. Find its exact hypotenuse hh.

Plan. Use the same 45∘45^\circ ratio, but solve for the whole hypotenuse.

12h=22adjacent over hypotenuseh=242=2422rationalise the denominator=122\begin{aligned} \frac{12}{h}&=\frac{\sqrt2}{2} && \text{adjacent over hypotenuse}\\ h&=\frac{24}{\sqrt2}\\ &=\frac{24\sqrt2}{2} && \text{rationalise the denominator}\\ &=12\sqrt2 \end{aligned}

Answer. 122 cm12\sqrt2\,\mathrm{cm}.

Check. Pythagoras gives h2=122+122=288h^2=12^2+12^2=288, and (122)2=288(12\sqrt2)^2=288. Also, the hypotenuse is longer than either leg.

Watch out

2/2\sqrt2/2 is not 0.70.7 exactly. If asked for an exact value, keep the surd. Replacing it by a rounded decimal loses the exact relationship.

The thirty-sixty-degree triangle

Split an equilateral triangle of side 22 down its line of symmetry, as in the second diagram. The perpendicular bisects both its base and its top angle. One half has hypotenuse 22, short leg 11, and angles 30∘30^\circ and 60∘60^\circ.

Its remaining leg has length 22−12=3\sqrt{2^2-1^2}=\sqrt3. Looking from 30∘30^\circ and then 60∘60^\circ swaps the opposite and adjacent legs.

Exact values at thirty and sixty degrees

sin⁡30∘=12cos⁡30∘=32tan⁡30∘=33sin⁡60∘=32cos⁡60∘=12tan⁡60∘=3\begin{aligned} \sin30^\circ&=\frac12& \cos30^\circ&=\frac{\sqrt3}{2}& \tan30^\circ&=\frac{\sqrt3}{3}\\ \sin60^\circ&=\frac{\sqrt3}{2}& \cos60^\circ&=\frac12& \tan60^\circ&=\sqrt3 \end{aligned}

For 30∘30^\circ, tangent is 1/3=3/31/\sqrt3=\sqrt3/3. For 60∘60^\circ, it is 3/1\sqrt3/1.

Not given in the List of formulas: learn them.

The short leg is opposite 30∘30^\circ; the long leg is opposite 60∘60^\circ. This gives a useful independent check on any calculation.

Worked example 3: viewing from thirty degrees

Question. A right-angled triangle has hypotenuse 14 cm14\,\mathrm{cm} and an angle of 30∘30^\circ. Find the exact opposite and adjacent lengths relative to that angle.

Plan. Multiply the hypotenuse by sine and cosine.

opposite=14sin⁡30∘use the opposite ratio=14×12=7adjacent=14cos⁡30∘use the adjacent ratio=14×32=73\begin{aligned} \text{opposite}&=14\sin30^\circ && \text{use the opposite ratio}\\ &=14\times\frac12=7\\ \text{adjacent}&=14\cos30^\circ && \text{use the adjacent ratio}\\ &=14\times\frac{\sqrt3}{2}=7\sqrt3 \end{aligned}

Answer. Opposite 7 cm7\,\mathrm{cm}; adjacent 73 cm7\sqrt3\,\mathrm{cm}.

Check. Their ratio is 7/(73)=3/3=tan⁡30∘7/(7\sqrt3)=\sqrt3/3=\tan30^\circ. Also, 72+(73)2=196=1427^2+(7\sqrt3)^2=196=14^2.

Keep the same triangle but view it from its other acute angle. Only the reference angle changes.

Continue this lesson.

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