Question

use compound angle formulae to find sin 15 and cos 15 in form of [ surd a (surd b + c) ] . Hence, find tan 15 in the simplest surd form​

Answer 1

Answer:

see explanation

Step-by-step explanation:

Using the difference formulae for sine and cosine

sin(x - y) = sinxcosy - cosxsiny

cos(x - y) = cosxcosy + sinxsiny

sin15° = sin(45 - 30)°

sin(45 - 30)°

= sin45°cos30° - cos45°sin30°

= $\frac{\sqrt{2} }{2}$ × $\frac{\sqrt{3} }{2}$ - $\frac{\sqrt{2} }{2}$ × $\frac{1}{2}$

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

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

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cos15° = cos(45 - 30)°

cos(45 - 30)°

= cos45°cos30° + sin45°sin30°

= $\frac{\sqrt{2} }{2}$ × $\frac{\sqrt{3} }{2}$ + $\frac{\sqrt{2} }{2}$ × $\frac{1}{2}$

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

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

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tan15° = $\frac{sin15}{cos15}$

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

= $\frac{\sqrt{6}-\sqrt{2} }{\sqrt{6}+\sqrt{2} }$

Rationalise the denominator by multiplying numerator/ denominator by the conjugate of the denominator, that is ($\sqrt{6}$ - $\sqrt{2}$ )

= $\frac{(\sqrt{6}-\sqrt{2})^2 }{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2}) }$

= $\frac{6-4\sqrt{3}+2 }{6-2}$

= $\frac{8-4\sqrt{3} }{4}$

= 2 - $\sqrt{3}$