If sin^2 a + sin^2 b + sin^2 c + sin^2 d = 0 , then minimum of osa + Cosb + Cosc + Cosd is (A) 4 (C) - 6 (D) - 4 (B) 0
Answers
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
Answer:
A
Step-by-step explanation:
You just need to solve it well ,use
sin2=1-Cos2.
using this identity you must get 4