Question

If A+B+C = π , prove that :-
cos²A + cos²B - cos²C = 1 - 2sinA sinB cosC.

Answer 1

Hey there!!

A + B + C = π
It means these are the angles of a triangle as we know that sum of angles of a triangle is 180°.

Now, Let's consider

2 sinA . sinB . cosC
= sinA (2 sinB.cosC)
= sinA (sin(180-A) + sin(B-C) )
= sinA (sinA + sin(B-C) )
= sin²A + sinA sin(B-C)
= sin²A + {2sinA sin(B-C) / 2}
= sin²A + {cos(A-B+C) - cos(A+B-C) / 2}
= sin²A + {cos(A+C-B) - cos(A+B-C) / 2}
= sin²A + {cos(180-B-B) - cos(180-C-C) / 2}
= sin²A + {cos(180-2B) - cos(180-2C) / 2}
= sin²A - {cos2B + cos2C / 2}
= ( 2sin²A - cos2B + cos2C ) / 2
= 1/2 { 2sin²A - cos2B + cos2C }
= 1/2 { 2sin²A - 1+2sin²B + 1-2sin²C }
=
1/2 { 2sin²A + 2sin²B - 2sin²C }
= sin²A + sin²B - sin²C
= sin²A + sin²B - sin²C = 2 sinA . sinB . cosC
= 1 - cos²A + 1-cos²B -1+cos²C =2sinA.sinB.cosC
= cos²A + cos²B - cos²C
= 1- 2sinA.sinB.cosC

Hence Proved

HOPE IT HELPS!!

Answer 2

Step-by-step explanation:

cosC = cos(180−(A+B))  

........ = −cos(A+B)  

........ = −(cosA cosB − sinA sinB)  

........ = sinA sinB − cosA cosB  

LHS = cos²A + cos²B + cos²C  

....... = cos²A + cos²B + (sinA sinB − cosA cosB)²  

....... = cos²A + cos²B + sin²A sin²B − 2 sinA sinB cosA cosB + cos²A cos²B  

....... = cos²A + cos²B + (1−cos²A) (1−cos²B) − 2 sinA sinB cosA cosB + cos²A cos²B  

....... = cos²A + cos²B + 1 − cos²A − cos²B + cos²A cos²B − 2 sinA sinB cosA cosB + cos²A cos²B  

....... = 1 − 2 sinA sinB cosA cosB + 2cos²A cos²B  

....... = 1 − 2 cosA cosB (sinA sinB − cosA cosB)  

....... = 1 − 2 cosA cosB cosC  

....... = RHS