If A+B+C = π then prove that cos 2A + cos 2B - cos 2C =1-4sinAsinBcosC
twisha , 7 Years ago
Grade 12th pass
Trigonometry> If A+B+C = π then prove that cos 2A + cos...
1 Answers
Arun
Last Activity: 7 Years ago
cos2A + cos2B - cos2C
= cos2A - 2sin[( 2B + 2C)/2]sin[(2B - 2C)/2]
= cos2A - 2sin( B + C )sin(B - C)
We have : A + B + C = 180° => B + C = 180° - A
= cos2A - 2sin( 180° - A )sin(B - C)
= cos2A - 2( sin180°cosA - sinAcos180° )sin( B - C )
= cos2A + 2sinAsin(B - C)
= 1 - 2sin²A + 2sinAsin(B - C)
= 1 - 2sinA[ sinA - sin(B - C) ]
= 1 - 2sinA( 2cos[(A + B - C)/2]sin[(A - B + C)/2] )
= 1 - 4sinA.cos( A + B - C )/2.sin(A - B + C ) /2
We have : A + B + C = 180° => A + B = 180° - C
And : A + C = 180° - B
= 1 - 4sinA.cos[( 180° - C - C )/2].sin[( 180° - B - B )/2]
= 1 - 4sinA.cos( -2C )/2.sin( -2B)/2
= 1 - 4sinA.cos(-C).sin(-B)
= 1 - 4sinA.sinB.cosC
= cos2A - 2sin[( 2B + 2C)/2]sin[(2B - 2C)/2]
= cos2A - 2sin( B + C )sin(B - C)
We have : A + B + C = 180° => B + C = 180° - A
= cos2A - 2sin( 180° - A )sin(B - C)
= cos2A - 2( sin180°cosA - sinAcos180° )sin( B - C )
= cos2A + 2sinAsin(B - C)
= 1 - 2sin²A + 2sinAsin(B - C)
= 1 - 2sinA[ sinA - sin(B - C) ]
= 1 - 2sinA( 2cos[(A + B - C)/2]sin[(A - B + C)/2] )
= 1 - 4sinA.cos( A + B - C )/2.sin(A - B + C ) /2
We have : A + B + C = 180° => A + B = 180° - C
And : A + C = 180° - B
= 1 - 4sinA.cos[( 180° - C - C )/2].sin[( 180° - B - B )/2]
= 1 - 4sinA.cos( -2C )/2.sin( -2B)/2
= 1 - 4sinA.cos(-C).sin(-B)
= 1 - 4sinA.sinB.cosC
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