If A,B,C are the angles of a triangle,then the value of sin^2A+sin^2B+sin^2C-2cosA.cosB.cosC=?

2 Answers
P dilip_k
May 6, 2018

#sin^2A+sin^2B+sin^2C-2cosAcosBcosC#

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

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

#=1/2(3-(cos2A+cos2B+cos2C)-2cosAcosBcosC#
#=1/2(3-(2cos(A+B)cos(A-B)+2cos^2C-1)-2cosAcosBcosC#

#=1/2(3-2cos(pi-C)cos(A-B)-2cos^2C+1)-2cosAcosBcosC#

#=1/2(4+2cos(C)cos(A-B)-2cos^2C)-2cosAcosBcosC#

#=2+cos(C)cos(A-B)-cos^2C)-2cosAcosBcosC#

#=2+cos(C)(cos(A-B)-cosC)-2cosAcosBcosC#

#=2+cos(C)(cos(A-B)-cos(pi-(A+B))-2cosAcosBcosC#

#=2+cos(C)(cos(A-B)+cos(A+B))-2cosAcosBcosC#

#=2+cosC(2cosAcosB)-2cosAcosBcosC#

#=2#

Dean R.
May 6, 2018

# sin ^2A + sin ^2 B +sin^ 2 C − 2 cos A cos B cos C =2 #

Explanation:

For a triangle we know #A+B+C=180^circ.# Supplementary angles have the same sines and opposite cosines. We apply the sum angle formulas and grind it out, simplifying with #cos^2 theta+sin^2 theta=1.#

# sin ^2A + sin ^2 B +sin^ 2 C − 2 cos A cos B cos C #

# = sin ^2A + sin ^2 B +sin^ 2(180^circ - (A+B))− 2 cos A cos B cos (180^circ - (A+B)) #

# = sin ^2A + sin ^2 B +sin^ 2(A+B)+ 2 cos A cos B cos (A+B) #

# = sin ^2A + sin ^2 B +( sin Acos B + cos A sin B )^2+ 2 cos A cos B ( cos A cos B - sin A sin B) #

# = sin ^2A + sin ^2 B + sin^2 Acos^2 B + 2 cos A sin A cos B sin B + cos^2 A sin^2 B + 2 cos^2 A cos^2 B - 2 cos A sin A cos B sin B #

# = sin ^2A + sin ^2 B + sin^2 Acos^2 B + cos^2 A sin^2 B + 2 cos^2 A cos^2 B #

# = ( sin ^2 A + cos^2 A cos^2 B + cos^2 A sin^2 B ) + (sin^2 B + sin^2 A cos^2B + cos ^2 A cos^2 B )#

# = ( sin ^2 A + cos^2 A( cos^2 B + sin^2 B ) ) + (sin^2 B + cos^2 B(sin^2 A + cos ^2 A ) )#

# = ( sin ^2 A + cos^2 A ) + (sin^2 B + cos^2 B )#

# = 1 + 1 #

#= 2#

Related questions
Impact of this question
15910 views around the world
You can reuse this answer
Creative Commons License