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!=)

Printable View

- January 18th 2009, 05:42 AMImKoProving sin 15 deg
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!=) - January 18th 2009, 05:59 AMhmmmm
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. - January 18th 2009, 06:30 AMILoveMaths07
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

---> Q.E.D.

I hope that helps. :)

ILoveMaths07.