Question #86063
if a+b+c=pie then prove that cosa*cosa+cosb*cosb+cosc*cosc=1-2cosa*cosb*cosc
1
Expert's answer
2019-03-08T10:13:22-0500
c=pie(a+b)c=pie-(a+b)

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

Therefore


cos2(a)+cos2(b)+(cosacosbsinasinb)2=1+2cosacosb(cosacosbsinasinb)cos^2(a)+cos^2(b)+(cosa*cosb-sina*sinb)^2=1+2cosa*cosb*(cosa*cosb-sina*sinb)

cos2(a)+cos2(b)+cos2(a)cos2(b)+sin2(a)sin2(b)2cosacosbsinasinb=1+2cos2(a)cos2(b)2cosacosbsinasinbcos^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:


cos2(a)+cos2(b)+sin2(a)sin2(b)=1+cos2(a)cos2(b)cos^2(a)+cos^2(b)+sin^2(a)*sin^2(b)=1+cos^2(a)*cos^2(b)

sin2(x)=1cos2(x)sin^2(x)=1-cos^2(x)

Therefore


cos2(a)+cos2(b)+(1cos2(a))(1cos2(b))=1+cos2(a)cos2(b)cos^2(a)+cos^2(b)+(1-cos^2(a))(1-cos^2(b))=1+cos^2(a)*cos^2(b)





cos2(a)+cos2(b)+1+cos2(a)cos2(b)cos2(a)cos2(b)=1+cos2(a)cos2(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.


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Comments

Assignment Expert
08.03.19, 23:29

The last formula is 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)

Ankur
08.03.19, 20:26

but which is the last formula

Assignment Expert
08.03.19, 18:16

If you simplify the left-hand side in the last formula, then you get 1+cos^2(a)*cos^2(b), which equals to the right-hand side in the last formula. Thus, the initial identity was proved.

Ankur
08.03.19, 17:49

the answer is not done fully

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