# 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

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

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

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

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

These are some additional values

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Math master must remember this value, but I'd made it simple for you. If you have additional information about this note, just write it in comment, amd I'll add it on my note.

Note by Jonathan Christianto
6 years, 6 months ago

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## Comments

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

- 6 years, 6 months ago

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Yes, sir. I got it from my teacher

- 6 years, 6 months ago

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