can anyone prove this?2cos^2(45°-A) = 1 +sin2A
Answers
Answer:
Step-by-step explanation:
Consider the problem,
sin
2
A+sin
2
B+sin
2
c=2+2cosA.cosB.cosc
We can write sin
2
A as,
sin
2
A=
2
1−cos(2A)
Therefore,
LHS=
2
1−cos(2A)
+
2
1−cos(2B)
+
2
1−cos(2C)
=
2
3
−(cos(2A)+cos(2B)+cos(2C))
=
2
1
(3−(2cos(A+B)cos(A−B)+cos(2C)))
C=180−(A+B)
cos(C)=cos(180−(A+B))
cos(C)=−cos(A+B)
Therefore,
=
2
3
−(−2cos(C)cos(A−B)+cos(2C))
cos(2C)=2cos
2
(C)−1
And,
=
2
1
(3−(−2cosC)cos(A−B)+2cos
2
(C)−1)
=
2
1
(4−(2cos(C)cos(C)−cos(A−B)))
=
2
1
(4−2cos(C)(−cos(A+B)−cos(A−B)))
=
2
1
(4+2cos(C)(cos(A+B)cos(A−B)))
=
2
1
(4+2cos(C)×2cos(A)cos(B))
=2+2cos(A)cos(B)cos(C)
Therefore, If A+B+C=180, sin
2
A+sin
2
B+sin
2
c=2+2cosA.cosB.cosc