Trigonometric Values of Specific Angles

Lesson · Intermediate

Trigonometry

$\theta$	$\sin\theta$	$\cos\theta$	$\tan\theta$	$\cot\theta$
$0^\circ$	$0$	$1$	$0$	Undefined
$30^\circ$	$\frac12$	$\frac{\sqrt3}{2}$	$\frac{\sqrt3}{3}$	$\sqrt3$
$45^\circ$	$\frac{\sqrt2}{2}$	$\frac{\sqrt2}{2}$	$1$	$1$
$60^\circ$	$\frac{\sqrt3}{2}$	$\frac12$	$\sqrt3$	$\frac{\sqrt3}{3}$
$90^\circ$	$1$	$0$	Undefined	$0$
$180^\circ$	$0$	$-1$	$0$	Undefined
$270^\circ$	$-1$	$0$	Undefined	$0$

Angle	$\sin\theta$	$\cos\theta$	$\tan\theta$	$\cot\theta$
$15^\circ$	$\frac{\sqrt6-\sqrt2}{4}$	$\frac{\sqrt6+\sqrt2}{4}$	$2-\sqrt3$	$2+\sqrt3$
$75^\circ$	$\frac{\sqrt6+\sqrt2}{4}$	$\frac{\sqrt6-\sqrt2}{4}$	$2+\sqrt3$	$2-\sqrt3$

Angle	$\sin\theta$	$\cos\theta$	$\tan\theta$	$\cot\theta$
$18^\circ$	$\frac{\sqrt5-1}{4}$	$\frac{\sqrt{10+2\sqrt5}}{4}$	$\frac{1}{\sqrt{5+2\sqrt5}}$	$\sqrt{5+2\sqrt5}$
$36^\circ$	$\frac{\sqrt{10-2\sqrt5}}{4}$	$\frac{1+\sqrt5}{4}$	$\sqrt{5-2\sqrt5}$	$\frac{1}{\sqrt{5-2\sqrt5}}$
$54^\circ$	$\frac{1+\sqrt5}{4}$	$\frac{\sqrt{10-2\sqrt5}}{4}$	$\frac{1}{\sqrt{5-2\sqrt5}}$	$\sqrt{5-2\sqrt5}$
$72^\circ$	$\frac{\sqrt{10+2\sqrt5}}{4}$	$\frac{\sqrt5-1}{4}$	$\sqrt{5+2\sqrt5}$	$\frac{1}{\sqrt{5+2\sqrt5}}$

\begin{aligned} \sin15^\circ &= \sin(45^\circ-30^\circ)\\[6pt] &= \sin45^\circ\cos30^\circ - \cos45^\circ\sin30^\circ\\[6pt] &= \frac{\sqrt2}{2}\left(\frac{\sqrt3}{2}\right) - \frac{\sqrt2}{2}\left(\frac12\right)\\[6pt] &= \frac{\sqrt6-\sqrt2}{4}. \end{aligned}

\begin{aligned} \cos15^\circ &= \cos(45^\circ-30^\circ)\\[6pt] &= \cos45^\circ\cos30^\circ + \sin45^\circ\sin30^\circ\\[6pt] &= \frac{\sqrt2}{2}\left(\frac{\sqrt3}{2}\right) + \frac{\sqrt2}{2}\left(\frac12\right)\\[6pt] &= \frac{\sqrt6+\sqrt2}{4}. \end{aligned}

\begin{aligned} \tan15^\circ &= \frac{\sin15^\circ}{\cos15^\circ}\\[6pt] &= \frac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}\\[6pt] &= \frac{(\sqrt6-\sqrt2)^2}{6-2}\\[6pt] &= \frac{8-2\sqrt{12}}{4}\\[6pt] &= 2-\sqrt3. \end{aligned}

\cot15^\circ = \frac1{\tan15^\circ} = \frac1{2-\sqrt3} = 2+\sqrt3.

\sin75^\circ = \cos15^\circ = \frac{\sqrt6+\sqrt2}{4}

\cos75^\circ = \sin15^\circ = \frac{\sqrt6-\sqrt2}{4}

\tan75^\circ = \cot15^\circ = 2+\sqrt3

\cot75^\circ = \tan15^\circ = 2-\sqrt3

For $\sin18^\circ$ , assume $\alpha=18^\circ$ .

5\alpha=90^\circ

2\alpha+3\alpha=90^\circ

2\alpha=90^\circ-3\alpha

\sin2\alpha = \sin(90^\circ-3\alpha) = \cos3\alpha

2\sin\alpha\cos\alpha = 4\cos^3\alpha-3\cos\alpha

2\sin\alpha\cos\alpha - 4\cos^3\alpha + 3\cos\alpha = 0

\cos\alpha \left( 2\sin\alpha - 4\cos^2\alpha + 3 \right) = 0

2\sin\alpha - 4\cos^2\alpha + 3 = 0

2\sin\alpha - 4(1-\sin^2\alpha) + 3 = 0

4\sin^2\alpha+2\sin\alpha-1=0

\sin\alpha = \frac{-2\pm\sqrt{4-4(4)(-1)}}{8}

\sin\alpha = \frac{-2\pm\sqrt{20}}{8} = \frac{-1\pm\sqrt5}{4}.

\sin18^\circ = \frac{\sqrt5-1}{4}.

\begin{aligned} \cos^218^\circ &= 1-\sin^218^\circ\\[6pt] &= 1- \left( \frac{\sqrt5-1}{4} \right)^2\\[6pt] &= 1-\frac{6-2\sqrt5}{16}\\[6pt] &= \frac{10+2\sqrt5}{16}. \end{aligned}

\cos18^\circ = \frac{\sqrt{10+2\sqrt5}}{4}.

\begin{aligned} \tan18^\circ &= \frac{\sin18^\circ}{\cos18^\circ}\\[6pt] &= \frac{\sqrt5-1}{\sqrt{10+2\sqrt5}}\\[6pt] &= \frac1{\sqrt{5+2\sqrt5}}. \end{aligned}

\cot18^\circ = \sqrt{5+2\sqrt5}.

\begin{aligned} \sin36^\circ &= 2\sin18^\circ\cos18^\circ\\[6pt] &= 2 \left( \frac{\sqrt5-1}{4} \right) \left( \frac{\sqrt{10+2\sqrt5}}{4} \right)\\[6pt] &= \frac{\sqrt{10-2\sqrt5}}{4}. \end{aligned}

\begin{aligned} \cos^236^\circ &= 1-\sin^236^\circ\\[6pt] &= 1-\frac{10-2\sqrt5}{16}\\[6pt] &= \frac{6+2\sqrt5}{16}\\[6pt] &= \frac{(1+\sqrt5)^2}{16}. \end{aligned}

\cos36^\circ = \frac{1+\sqrt5}{4}.

\tan36^\circ = \frac{\sin36^\circ}{\cos36^\circ} = \sqrt{5-2\sqrt5}.

\cot36^\circ = \frac1{\tan36^\circ} = \frac1{\sqrt{5-2\sqrt5}}.

The trigonometric values of $54^\circ$ and $72^\circ$ can be obtained from those of $36^\circ$ and $18^\circ$ , respectively.