Cos^2A+cos^2B-sin^2C=-2cosA.cosB.sinC?

Question

8234 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-15T14:52:56

If #A+B+C=pi# then

#LHS=cos^2A+cos^2B-sin^2C#

#=1/2[2cos^2A+2cos^2B-2sin^2C]#

#=1/2[1+cos2A+1+cos2B-(1-cos2C)]#

#=1/2[1+cos2A+1+cos2B-1+cos2C]#

#=1/2[1+cos2A+cos2B+cos2C]#

#=1/2[1+2cos((2A+2B)/2)cos((2A-2B)/2)+cos2(pi-(A+B))]#

#=1/2[1+2cos(A+B)cos(A-B)+cos(2pi-2(A+B))]#

#=1/2[1+2cos(A+B)cos(A-B)+2cos^2(A+B)-1]#

#=1/2[2cos(A+B){cos(A-B)+cos(A+B)}]#

#=cos(pi-C)*2cosAcosB=-2cosAcosBcosC#

Answer 2 · 2018-07-15T15:44:08

Below.

Let we take the values of #A+B+C=pi#

#C= pi-(A+B)#----------(i)

Multiplying LHS by #1/2# we get,

#1/2(2 cos^2A + 2 cos^2B - 2 sin^2 C)#

#=># #1/2(1+ cos 2A+1+cos 2B - {1-cos2C})#

#=># #1/2( 1+ cos 2A+1+cos 2B - 1+ cos2C)#

#=># #1/2(1+ cos 2A+cos 2B+cos2C)#

Using the formula of #cos{2A+2B} # we get,

#=># # 1/2 (1+ 2cos[{2A+2B}/2] cos [{2A-2B}/2]+ cos2 C)#

Putting the value of equation (i) in #cos 2C # we get,

#=># # 1/2 (1+ 2cos[{2A+2B}/2] cos [{2A-2B}/2]+ cos2[pi-{A+B}])#

Resolving this equation further,

#=># #1/2( 1+ 2cos[A+B]cos[A-B]+2cos^2(A+B)-1)#

#=># #(cos[pi-C]*2cosAcosB) #

This gives the final output as,

#=># #-2cosA*cosB*cosC#