# Tough question

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• Apr 2nd 2010, 10:47 AM
OmegaCenturion
Tough question
I'm not strong with this, i need some help:

2(1-sin(b)sin(c)) = cos^2(b) + cos^(c) when b+c = 180
• Apr 2nd 2010, 10:58 AM
Archie Meade
Quote:

Originally Posted by OmegaCenturion
I'm not strong with this, i need some help:

2(1-sin(b)sin(c)) = cos^2(b) + cos^(c) when b+c = 180

Your clue lies in "b+c=180 degrees", as now c=180-b.

sin(angle) gives the vertical co-ordinate in a circle of radius 1, centre (0,0).

Hence sin(b)=sin(180-b).

Therefore your equation becomes

$\displaystyle 2\left(1-sin^2(b)\right)=cos^2b+cos^2c$

Using similar logic

$\displaystyle cos(c)=-cos(b)$

allows you to write the entire equation in terms of any one of the four of
sinc, sinb, cosc, cosb

with $\displaystyle cos^2b+sin^2b=1$ etc
• Apr 2nd 2010, 11:08 AM
harish21
Quote:

Originally Posted by OmegaCenturion
I'm not strong with this, i need some help:

2(1-sin(b)sin(c)) = cos^2(b) + cos^(c) when b+c = 180

$\displaystyle 2(1-sinb. sinc) = cos^{2}b + cos^{2}c$

write this as:

$\displaystyle 2 - 2.sinb.sinc = (1-sin^{2}b) + (1-sin^{2}c)$

or,

$\displaystyle 2 - 2.sinb.sinc = 2 - sin^{2}b - sin^{2}c$

$\displaystyle sin^{2}b - 2.sinb.sinc + sin^{2}c = 0$

$\displaystyle (sinb -sinc)^2 = 0$

Note that:

$\displaystyle b+c = 180 \rightarrow c=180-b$

$\displaystyle \therefore (sinb -sinc) = (sinb - sin(\pi-b)) = (sinb-sinb) = 0$

You should know that $\displaystyle sin(180-b) = sin(\pi-b) = sinb$
• Apr 2nd 2010, 11:36 AM
OmegaCenturion
thanks guys!