Special Triangles

Section 9.3 Special Triangles

Certain nice angles appear in right triangles with pleasing geometry allowing us to compute their trigonometric ratios exactly. You can see their construction in the video or follow along in the text below.

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Example 9.3.1. The 45-45-90 Triangle.

Start with a square. You may label the side lengths any value you want, but \(1\) is convenient. All the angles are \(\pi/2 = 90^\circ\text{.}\)

Draw a diagonal with length \(\sqrt{1^2+1^2} = \sqrt{2}\text{.}\) The new angles are \(\pi/4 = 45^\circ\text{.}\)

This forms a “45-45-90” triangle. From this, we can compute exactly any trigonometric function value for \(\pi/4\) (radians) or equivalently \(45^\circ\text{.}\)

\begin{align*} &\sin(\pi/4) = \sin(45^\circ) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\\ &\cos(\pi/4) = \cos(45^\circ) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\\ &\tan(\pi/4) = \tan(45^\circ) = \frac{1}{1} = 1\\ &\csc(\pi/4) = \csc(45^\circ) = \frac{\sqrt{2}}{1} = \sqrt{2}\\ &\sec(\pi/4) = \sec(45^\circ) = \frac{\sqrt{2}}{1} = \sqrt{2}\\ &\cot(\pi/4) = \cot(45^\circ) = \frac{1}{1} = 1 \end{align*}

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Example 9.3.3. The 30-60-90 Triangle.

Start with an equilateral triangle, all sides and all angles are equal to \(\pi/3 = 60^\circ\text{.}\) You can label the side lengths whatever you like, but its convenient to start with length \(2\text{.}\)

Divide the triangle into two right triangles by bisecting one of the angles into two \(\pi/6 = 30^\circ\) angles. This divides the opposite side into two equal parts of length \(1\text{.}\)

Finally, the missing height can be found with the Pythagorean Theorem

\begin{equation*} (\text{height})^2 + 1^2 = 2^2 \end{equation*}

so that

\begin{equation*} (\text{height}) = \sqrt{4-1}=\sqrt{3}. \end{equation*}

This gives two “30-60-90” triangles which allows us to find exact values of trigonometric functions for \(30^\circ = \pi/6\) and its complementary angle \(60^\circ = \pi/3\text{.}\) We’ll summarize in radians:

\begin{align*} &\sin(\pi/6) = \cos(\pi/3) = \frac{1}{2}\\ &\cos(\pi/6) = \sin(\pi/3) = \frac{\sqrt{3}}{2}\\ &\tan(\pi/6) = \cot(\pi/3) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\\ &\csc(\pi/6) = \sec(\pi/3) = 2\\ &\sec(\pi/6) = \csc(\pi/3) = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\\ &\cot(\pi/6) = \tan(\pi/3) = \sqrt{3} \end{align*}

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