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Question Number 99321 by Ar Brandon last updated on 20/Jun/20

$$\left[\begin{array}{ccc}\mathbf{Angle}\left(\mathit{\theta }\right)& \mathbf{sin}\left(\mathit{\theta }\right)& \mathbf{cos}\left(\mathit{\theta }\right)[\\ 0^{°}& 0& 1\\ 15°& \frac{\sqrt{6}-\sqrt{2}}{4}& \frac{\sqrt{6}+\sqrt{2}}{4}\\ 18°& \frac{\sqrt{5}-1}{4}& \frac{\sqrt{5+\sqrt{5}}}{2\sqrt{2}}\\ 30°& \frac{1}{2}& \frac{\sqrt{3}}{2}\\ 36°& \frac{\sqrt{5-\sqrt{5}}}{2\sqrt{2}}& \frac{\sqrt{5}+1}{4}\\ 45°& \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ 54°& \frac{\sqrt{5}+1}{4}& \frac{\sqrt{5-\sqrt{5}}}{2\sqrt{2}}\\ 60°& \frac{\sqrt{3}}{2}& \frac{1}{2}\\ 72°& \frac{\sqrt{5+\sqrt{5}}}{2\sqrt{2}}& \frac{\sqrt{5}-1}{4}\\ 75°& \frac{\sqrt{6}+\sqrt{2}}{4}& \frac{\sqrt{6}-\sqrt{2}}{4}\\ 90°& 1& 0\end{array}\right]$$$$\mathbf{sin}\left(\mathit{\theta }\right)=\mathbf{cos}(90°-\mathit{\theta })$$

Commented by bobhans last updated on 20/Jun/20

$$\mathrm{waw}...\mathrm{cooll}$$

Commented by mahdi last updated on 20/Jun/20

$$:)$$

Answered by MWSuSon last updated on 20/Jun/20

$$\mathrm{This}\mathrm{is}\mathrm{really}\mathrm{helpful}\mathrm{sir}.$$

Commented by bobhans last updated on 20/Jun/20

$$\mathrm{yes}\mathrm{sir}$$

Commented by Ar Brandon last updated on 20/Jun/20

Thanks��

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