Proving sin 15 deg

Dec 2008
15
0
Philippines
Hello! Need help on this problem

prove that sin 15deg = (sqrt(6) - sqrt(2)) / 4

using

sinC= sinA cosB + cosA sinB
sinB= sinA cosC + cosA sinC
sinA= sinB cosC + cosB sinC

Hoping for your attention!=)
 
Dec 2008
288
13
what you want to use is the addintion formulae sin(45-30)
=sin45cos30-sin30cos45
substitute in your exact values of sin45=1/sqrt2 sin30=0.5 cos45=1/sqrt2 and cos30=sqrt3/2 and then rearrange the fraction you get by multiplying by sqrt2 and that should be it.
 
Aug 2008
58
14
Dubai, UAE
Recall that sin (A - B) = sin A cos B - cos A sin B

sin 15 = sin (45 - 30)

= sin 45 cos 30 - cos 45 sin 30

\(\displaystyle = \frac {\sqrt {2}}{2} \cdot \frac {\sqrt {3}}{2} - \frac {\sqrt {2}}{2} \cdot \frac {1}{2}\)

\(\displaystyle = \frac {\sqrt {6}}{4} - \frac {\sqrt {2}}{4} \)

\(\displaystyle = \frac {\sqrt {6} - \sqrt {2}}{4}.\) ---> Q.E.D.

I hope that helps. :)

ILoveMaths07.