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​

Respuesta :

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°

= [tex]\frac{\sqrt{2} }{2}[/tex] × [tex]\frac{\sqrt{3} }{2}[/tex] - [tex]\frac{\sqrt{2} }{2}[/tex] × [tex]\frac{1}{2}[/tex]

= [tex]\frac{\sqrt{6} }{4}[/tex] - [tex]\frac{\sqrt{2} }{4}[/tex]

= [tex]\frac{\sqrt{6}-\sqrt{2}  }{4}[/tex]

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cos15° = cos(45 - 30)°

cos(45 - 30)°

= cos45°cos30° + sin45°sin30°

= [tex]\frac{\sqrt{2} }{2}[/tex] × [tex]\frac{\sqrt{3} }{2}[/tex] + [tex]\frac{\sqrt{2} }{2}[/tex] × [tex]\frac{1}{2}[/tex]

= [tex]\frac{\sqrt{6} }{4}[/tex] + [tex]\frac{\sqrt{2} }{4}[/tex]

= [tex]\frac{\sqrt{6}+\sqrt{2}  }{4}[/tex]

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tan15° = [tex]\frac{sin15}{cos15}[/tex]

= [tex]\frac{\sqrt{6}-\sqrt{2}  }{4}[/tex] × [tex]\frac{4}{\sqrt{6}+\sqrt{2}  }[/tex]

= [tex]\frac{\sqrt{6}-\sqrt{2}  }{\sqrt{6}+\sqrt{2}  }[/tex]

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

= [tex]\frac{(\sqrt{6}-\sqrt{2})^2  }{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})    }[/tex]

= [tex]\frac{6-4\sqrt{3}+2 }{6-2}[/tex]

= [tex]\frac{8-4\sqrt{3} }{4}[/tex]

= 2 - [tex]\sqrt{3}[/tex]