# 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
3 years, 8 months ago

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

- 3 years, 8 months ago

Yes, sir. I got it from my teacher

- 3 years, 8 months ago