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]