E7.3 - Exact trigonometric values (Extended)

E7.3 - Exact trigonometric values (Extended)

A calculator displays a decimal for the height of a triangle, but can you give the height without rounding at all? Special triangles give exact sine, cosine and tangent values. You will derive the values, recognise their patterns and use them in calculations that keep square roots exact.

Special triangles

A right triangle has hypotenuse 2 and an angle of 60∘60^\circ. Its height is 3\sqrt3, rather than a rounded decimal. Two familiar shapes explain where this number comes from.

You already know

In a right triangle, sin⁡θ=oppositehypotenuse\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}, cos⁡θ=adjacenthypotenuse\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}} and tan⁡θ=oppositeadjacent\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}. Pythagoras gives the third side. A surd such as 3\sqrt3 keeps an irrational length exact.

Cut a square along a diagonal to make two equal angles of 45∘45^\circ. Cut an equilateral triangle down its line of symmetry to make angles of 30∘30^\circ, 60∘60^\circ and 90∘90^\circ. Similar triangles have the same side ratios, whatever their size.

[DIAGRAM: asset_name: Two special right triangles; asset_slug: special_right_triangles; caption: A square diagonal gives 45 degrees; half an equilateral triangle gives 30 and 60 degrees.; description: Left, an exact right isosceles triangle has perpendicular sides 1 and 1 and hypotenuse square root 2, with both acute angles 45 degrees. Right, an equilateral triangle of side 2 is bisected perpendicularly into two right triangles with short side 1, height square root 3 and hypotenuse 2; angles 30 and 60 degrees are labelled.]
A square diagonal gives 45 degrees

To remove a square root from a denominator, multiply both numerator and denominator by that root. For example, 1/2=(1×2)/(2×2)=2/21/\sqrt2=(1\times\sqrt2)/(\sqrt2\times\sqrt2)=\sqrt2/2; multiplying by 2/2=1\sqrt2/\sqrt2=1 preserves the value.

Worked example 1: a square diagonal

Question. Derive the exact sine and cosine of 45∘45^\circ from a right triangle with two sides of length 1 meeting at the right angle.

Plan. Find the hypotenuse, then divide each equal leg by it.

h2=12+12=2Pythagorash=2positive lengthsin⁡45∘=cos⁡45∘=12=22rationalise\begin{aligned} h^2&=1^2+1^2=2 &&\text{Pythagoras}\\ h&=\sqrt2 &&\text{positive length}\\ \sin45^\circ=\cos45^\circ&=\frac1{\sqrt2}=\frac{\sqrt2}{2} &&\text{rationalise} \end{aligned}

Answer. Both values are 22\dfrac{\sqrt2}{2}.

Check. Symmetry makes the opposite and adjacent sides equal. Also (22)2+(22)2=1\left(\dfrac{\sqrt2}{2}\right)^2+\left(\dfrac{\sqrt2}{2}\right)^2=1, agreeing with the squared side ratios in Pythagoras.

The second shape changes the equal-sided triangle before the cut: start with three equal sides, so the resulting right triangle has unequal legs.

Worked example 2: half an equilateral triangle

Question. An equilateral triangle has side 2. Derive sin⁡30∘\sin30^\circ, cos⁡30∘\cos30^\circ, sin⁡60∘\sin60^\circ and cos⁡60∘\cos60^\circ.

Plan. Bisect the triangle through its top vertex, then use the resulting right triangle.

The cut halves the base and the top angle. The new triangle has hypotenuse 2, short leg 1 and height hh.

h2=22−12=3Pythagorash=3sin⁡30∘=12cos⁡30∘=32sin⁡60∘=32cos⁡60∘=12\begin{aligned} h^2&=2^2-1^2=3 &&\text{Pythagoras}\\ h&=\sqrt3\\ \sin30^\circ&=\frac12 & \cos30^\circ&=\frac{\sqrt3}{2}\\ \sin60^\circ&=\frac{\sqrt3}{2} & \cos60^\circ&=\frac12 \end{aligned}

For 30∘30^\circ, the side of length 1 is opposite. For 60∘60^\circ, it is adjacent: changing the reference angle swaps the roles of the legs.

Answer. The four exact values are 12, 32, 32, 12\dfrac12,\ \dfrac{\sqrt3}{2},\ \dfrac{\sqrt3}{2},\ \dfrac12, in the order asked.

Check. The angles add to 180∘180^\circ. The longer leg is opposite 60∘60^\circ, so sin⁡60∘>sin⁡30∘\sin60^\circ>\sin30^\circ, as the results show.

Scaling changes the lengths but leaves these ratios unchanged.

Watch out

3/2\sqrt3/2 is exact; 0.866 is an approximation. Keep the surd when an exact value is required.

Use the same construction at a different scale.

Sine and cosine values

The two triangles give the interior values in the table. To include 0∘0^\circ and 90∘90^\circ, imagine a line of length 1 rotating from horizontal to vertical. Its horizontal distance is cos⁡θ\cos\theta; its vertical height is sin⁡θ\sin\theta. At the horizontal position these are 1 and 0; at the vertical position they are 0 and 1. The endpoint positions are not ordinary right triangles, so we use this interpretation there.

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