# Math Help - Trigonometry problem ; Help?

1. ## Trigonometry problem ; Help?

I'm trying to find the value of this question :

Cos β / (sinα.sinγ) + cos α / (sin β. sin γ)+ cos γ/ (sin α. Sin β)
Provided that α, β, γ is the angle in a triangle

The answer in the book says that the value is 2, but how on earth are we going to find "2"

= Cos β / (sinα.sinγ) + cos α / (sin β. sin γ)+ cos γ/ (sin α. Sin β)
= cos β sin β + sin α cos α + sin γ cos γ / sin α sin β sin γ
= sin 2β + sin 2α + sin 2 γ/ 2 (sin α sin β sin γ)

Using identities sin A + sin B = 2 sin (A+B/2) cos (A-B/2)
And 2 sin A sin B =sin (A+B) – sin (A-B)
We have

= 2 sin β cos 0 + 2 sin α cos 0 + 2 sin γ cos 0 / (sin ( α+β ) – sin ( α- β)) sin γ
= 2 ( sin α + sin β + sin γ) / (sin γ( 2 sin β cos α))

Then 2 sin β sin γ becomes 2 sin γ cos β

= 2 (sin α + sin β + sin γ) / 2 cos α cos β sin γ
= sin α + sin β + sin γ / cos α cos β sin γ
= ( (sin α + sin β / sin γ) x 1/(cos α cos β) + 1/ cos α cos β
= 1/cos α cos β((sin α + sin β / sin γ) +1)
2cos A cos B = cos (A+B) + cos (A-B)
= (2 / (cos (α + β) + cos (α- β) ) ((sin α + sin β / sin γ) +1)

2. ## Re: Trigonometry problem ; Help?

first add all the fractions to get one : with common denominator sinasinbsinc and numerator cosbsinb+cosasina+coscsinc
then multiply the fraction by two both numerator and denominator...to get 2sinasinbsinc as denominator and sin(2a)+sin(2b)+sin(2c) as numerator.
then leave the denominator as it is and try to factorize the numerator using sum to product formulas .Dont forget that 2a+2b+2c=360 now...and you have to prove that
sin2a+sin2b+sin2c =4sinasinbsinc this proves your relation.....
so start the job right now....good luck...

3. ## Re: Trigonometry problem ; Help?

What do you mean by sum to product formula? is it