No need for compound angle formulae.....

Usually when you need to calculate trigonometry values such as sin15 and cos15 then you can use: sin(θ±ϕ)=sin(θ)cos(ϕ)±sin(ϕ)cos(θ)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 \angleABD = 60 and \angleDBC = 15 \Rightarrow \angleBCD = 75. Also, without loss of generality, let AB = 2

By using special triangle AB = 2, BD = 1 and AD = 3\sqrt{3} \Rightarrow DC = 2 - 3\sqrt{3}. Now by using pythagoras' theorem: BC=1+(23)2=843=(62)2=62BC = \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=oppositeHypotenuse =2362×6+26+2=624cos75 = 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=162 =6+24sin75 = cos15 = \frac{1}{\sqrt{6} - \sqrt{2}}\ = \frac{\sqrt{6} + \sqrt{2}}{4} tan15=oppositeadjacent=23 and tan75=123=2+3tan15 = \frac{opposite}{adjacent} = 2 - \sqrt{3} \ and \ tan75 = \frac{1}{2-\sqrt{3}} = 2+\sqrt{3}

Note by Curtis Clement
6 years, 4 months ago

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Nice work !!!!!

A Former Brilliant Member - 6 years, 4 months ago

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Thankyou :)

Curtis Clement - 6 years, 4 months ago

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