Usually when you need to calculate trigonometry values such as sin15 and cos15 then you can use: \[sin(\theta\pm \phi) = sin(\theta)cos(\phi)\pm sin(\phi)cos(\theta) \] However, I found an alternative method using geometry directly.

Firstly you draw an isosceles triangle ABC (A is the top angle) with A = 30, B=C = 75. Now draw a line from B that is perpendicular to AC at a point D. This means that \(\angle\)ABD = 60 and \(\angle\)DBC = 15 \(\Rightarrow\) \(\angle\)BCD = 75. Also, without loss of generality, let AB = 2

By using special triangle AB = 2, BD = 1 and AD = \(\sqrt{3}\) \(\Rightarrow\) DC = 2 - \(\sqrt{3}\). Now by using pythagoras' theorem: \[BC = \sqrt{1 + (2-\sqrt{3})^2 } = \sqrt{8-4\sqrt{3}} = \sqrt{(\sqrt{6} - \sqrt{2})^2} = \sqrt{6} - \sqrt{2}\] So now we have a rightangle triangle BCD where all the sides are known.

Finally; \[cos75 = sin15 = \frac{opposite}{Hypotenuse}\ = \frac{2-\sqrt{3}}{\sqrt{6} - \sqrt{2}}\times\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} + \sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}\] \[sin75 = cos15 = \frac{1}{\sqrt{6} - \sqrt{2}}\ = \frac{\sqrt{6} + \sqrt{2}}{4}\] \[tan15 = \frac{opposite}{adjacent} = 2 - \sqrt{3} \ and \ tan75 = \frac{1}{2-\sqrt{3}} = 2+\sqrt{3} \]

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TopNewestNice work !!!!!

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Thankyou :)

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