# Value of Trigonometry of Special Angles

Let us start from the value of sin.

$\sin{0} = \frac{1}{2}\sqrt{0}$

$\sin{30} = \frac{1}{2}\sqrt{1}$

$\sin{45} = \frac{1}{2}\sqrt{2}$

$\sin{60} = \frac{1}{2}\sqrt{3}$

$\sin{90} = \frac{1}{2}\sqrt{4}$

$\sin{120} = \frac{1}{2}\sqrt{3}$

$\sin{135} = \frac{1}{2}\sqrt{2}$

$\sin{150} = \frac{1}{2}\sqrt{1}$

$\sin{180} = \frac{1}{2}\sqrt{0}$

$\sin{210} = -\frac{1}{2}\sqrt{1}$

$\sin{225} = -\frac{1}{2}\sqrt{2}$

$\sin{240} =- \frac{1}{2}\sqrt{3}$

$\sin{270} = -\frac{1}{2}\sqrt{4}$

$\sin{300} = -\frac{1}{2}\sqrt{3}$

$\sin{315} = -\frac{1}{2}\sqrt{2}$

$\sin{330} = -\frac{1}{2}\sqrt{1}$

$\sin{360} = -\frac{1}{2}\sqrt{0}$

Let us continue with cos.

$\cos{0} = \frac{1}{2}\sqrt{4}$

$\cos{30} = \frac{1}{2}\sqrt{3}$

$\cos{45} = \frac{1}{2}\sqrt{2}$

$\cos{60} = \frac{1}{2}\sqrt{1}$

$\cos{90} = \frac{1}{2}\sqrt{0}$

$\cos{120} = -\frac{1}{2}\sqrt{1}$

$\cos{135} = -\frac{1}{2}\sqrt{2}$

$\cos{150} = -\frac{1}{2}\sqrt{3}$

$\cos{180} = -\frac{1}{2}\sqrt{4}$

$\cos{210} = -\frac{1}{2}\sqrt{3}$

$\cos{225} = -\frac{1}{2}\sqrt{2}$

$\cos{240} = -\frac{1}{2}\sqrt{1}$

$\cos{270} = \frac{1}{2}\sqrt{0}$

$\cos{300} = \frac{1}{2}\sqrt{1}$

$\cos{315} = \frac{1}{2}\sqrt{2}$

$\cos{330} = \frac{1}{2}\sqrt{3}$

$\cos{360} = \frac{1}{2}\sqrt{4}$

For the value of tan, cot, sec, and csc, just remember that

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

$\cot \theta = \frac{1}{\tan \theta}$

$\sec \theta = \frac{1}{\cos \theta}$

$\csc \theta = \frac{1}{\sin \theta}$

$\sin{15} = \frac{\sqrt{6}-\sqrt{2}}{4}$

$\sin{75} = \frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin{105} = \frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin{165} = \frac{\sqrt{6-}\sqrt{2}}{4}$

$\sin{195} = \frac{-\sqrt{6}+\sqrt{2}}{4}$

$\sin{255} = -\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin{285} = -\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin{345} = \frac{-\sqrt{6}+\sqrt{2}}{4}$

$\cos{15} = \frac{\sqrt{6}+\sqrt{2}}{4}$

$\cos{75} = \frac{\sqrt{6}-\sqrt{2}}{4}$

$\cos{105} = \frac{-\sqrt{6}+\sqrt{2}}{4}$

$\cos{165} = -\frac{\sqrt{6}+\sqrt{2}}{4}$

$\cos{195} = -\frac{\sqrt{6}+\sqrt{2}}{4}$

$\cos{255} = \frac{-\sqrt{6}+\sqrt{2}}{4}$

$\cos{285} = \frac{\sqrt{6}-\sqrt{2}}{4}$

$\cos{345} = \frac{\sqrt{6}+\sqrt{2}}{4}$ Note by Jonathan Christianto
5 years, 4 months ago

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Very nice! I'd never seen them laid out that way before.

- 5 years, 4 months ago

Yes, sir. I got it from my teacher

- 5 years, 4 months ago