proof for sin^(-1)x-sin^(-1)y
Answers
Answer:
LHS = RHS
Step-by-step explanation:
Let: sin
−1
(x)=A,sin
−1
(y)=B,sin
−1
(z)=C
On taking sin both sides in each of them we get:
⇒x=sinA,y=sinB,z=sinC ........ [As sin(sin
−1
(x))=x]
Given that: sinx+siny+sinz=π
⇒A+B+C=π
A+B=π−C ..... (1)
To prove: x
1−x
2
+y
1−y
2
+z
1−z
2
=2xyz
LHS:
=sinA
1−sin
2
A
+sinB
1−sin
2
B
+sinC
1−sin
2
C
As we know, {
1−sin
2
A
=cosA} , LHS simplifies to:
=sinAcosA+sinBcosB+sinCcosC
=
2
2sinAsinA+2sinBcosB+2sinCcosC
... Multiply and Divide by 2
As we know, 2sinAcosA=sin(2A) so LHS:
=
2
sin2A+sin2B+2sinCcosC
=
2
2sin(A+B)cos(A−B)+2sinCcosC
[Identity:sinx+siny=2sin(
2
x+y
)cos(
2
x−y
)]
=
2
2sin(π−C)cos(A−B)+2sinCcosC
..... From (1)
=
2
2sinC(cos(A−B)+cos(π−(A+B)))
[Identity:sin(π−C)=sinC]
2
2sinC(cos(A−B)−cos((A+B)))
[Identity:cos(π−x)=−cosx]
2
2sinC(2sin(A)sin(B))
[Identity:cosy−cosx=2sin(
2
x+y
)sin(
2
x−y
)]
=2sinAsinBsinC
=2xyz
LHS = RHS
Hence proved