PROVE THAT :COS 2A – COS 2B + COS 2C = -1 -4 COSA COSB COSC
PROVE THAT :
COS 2A – COS 2B + COS 2C = -1 -4 COSA COSB COSC
JYOTSNA , 7 Years ago
Grade 11
Trigonometry> PROVE THAT : COS 2A – COS 2B + COS 2C = -...
1 Answers
Arun
Last Activity: 7 Years ago
cosC = cos(180−(A+B))
........ = −cos(A+B)
........ = −(cosA cosB − sinA sinB)
........ = sinA sinB − cosA cosB
LHS = 2(cos²A + cos²B + cos²C) -3
....... =2 [cos²A + cos²B + (sinA sinB − cosA cosB)²] -3
....... = 2[cos²A + cos²B + sin²A sin²B − 2 sinA sinB cosA cosB + cos²A cos²B ] -3
....... = 2 [cos²A + cos²B + (1−cos²A) (1−cos²B) − 2 sinA sinB cosA cosB + cos²A cos²B] -3
....... = 2[cos²A + cos²B + 1 − cos²A − cos²B + cos²A cos²B − 2 sinA sinB cosA cosB + cos²A cos²B ] -3
....... = 2 − 4 sinA sinB cosA cosB + 4cos²A cos²B -3
....... = -1 − 4 cosA cosB (sinA sinB − cosA cosB)
....... = -1 − 4 cosA cosB cosC
....... = RHS
........ = −cos(A+B)
........ = −(cosA cosB − sinA sinB)
........ = sinA sinB − cosA cosB
LHS = 2(cos²A + cos²B + cos²C) -3
....... =2 [cos²A + cos²B + (sinA sinB − cosA cosB)²] -3
....... = 2[cos²A + cos²B + sin²A sin²B − 2 sinA sinB cosA cosB + cos²A cos²B ] -3
....... = 2 [cos²A + cos²B + (1−cos²A) (1−cos²B) − 2 sinA sinB cosA cosB + cos²A cos²B] -3
....... = 2[cos²A + cos²B + 1 − cos²A − cos²B + cos²A cos²B − 2 sinA sinB cosA cosB + cos²A cos²B ] -3
....... = 2 − 4 sinA sinB cosA cosB + 4cos²A cos²B -3
....... = -1 − 4 cosA cosB (sinA sinB − cosA cosB)
....... = -1 − 4 cosA cosB cosC
....... = RHS
Regards
Arun (askIITians forum expert)
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