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

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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,

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