Answer to Question #86063 in Trigonometry for Ankur

Question

if a+b+c=pie then prove that cosa*cosa+cosb*cosb+cosc*cosc=1-2cosa*cosb*cosc

Answer 1

$$c=pie-(a+b)$$

$$cosc=-cos(a+b)$$

Therefore

$$cos^2(a)+cos^2(b)+(cosa*cosb-sina*sinb)^2=1+2cosa*cosb*(cosa*cosb-sina*sinb)$$

$$cos^2(a)+cos^2(b)+cos^2(a)*cos^2(b)+sin^2(a)*sin^2(b)-2cosa*cosb*sina*sinb=1+2cos^2(a)*cos^2(b)-2cosa*cosb*sina*sinb$$

After simplifying:

$$cos^2(a)+cos^2(b)+sin^2(a)*sin^2(b)=1+cos^2(a)*cos^2(b)$$

$$sin^2(x)=1-cos^2(x)$$

Therefore

$$cos^2(a)+cos^2(b)+(1-cos^2(a))(1-cos^2(b))=1+cos^2(a)*cos^2(b)$$

$$cos^2(a)+cos^2(b)+1+cos^2(a)*cos^2(b)-cos^2(a)-cos^2(b)=1+cos^2(a)*cos^2(b)$$

Proved.