Suppose we are given

θ = sin
^{
–1
}
x

**
**

**
**

**
We need to find sin
**
**
θ
**
**
, cos
**
**
θ
**
**
and tan
**
**
θ
**

Now,

θ = sin
^{
–1
}
x

So,

sin θ = x

sin θ = x/1

And we know that

sin θ = (Side opposite to angle θ)/Hypotenuse

x/1 = AB/AC

So, AB = x, AC = 1

Now, in right triangle Δ ABC

By Pythagoras Theorem

AC
^{
2
}
= AB
^{
2
}
+ BC
^{
2
}

1
^{
2
}
= x
^{
2
}
+ BC
^{
2
}

1 – x
^{
2
}
= BC
^{
2
}

BC = √(1-x
^{
2
}
)

##
**
Value of sin
**
**
θ
**
**
, cos
**
**
θ
**
**
, tan
**
**
θ
**
**
when
**
**
θ
**
**
is given as inverse
**

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**

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