s i n ( 1 2 c o s − 1 4 5 ) = Options: A . &nbsp 1 √ 10 B . &nbsp − 1 √ 10 C . &nbsp 1 10 D . &nbsp − 1 10

Answer: Option A : A Let 0 ≤ x ≤ π So, c o s − 1 4 5 = x ⇒ c o s x = 4 5 ....(i) Now s i n ( 1 2 c o s − 1 4 5 ) = s i n ( x 2 ) ....(ii) From (i), c o s x = 4 5 ⇒ 1 − 2 s i n 2 x 2 = 4 5 ⇒ 2 s i n 2 x 2 = 1 − 4 5 = 1 5 ⇒ s i n x 2 = √ 1 10 .

sin(12cos−145)=

Discussion Forum : Inverse Trigonometric Functions

Question -

$s i n (\frac{1}{2} c o s^{- 1} \frac{4}{5}) =$

Options:

A .

\frac{1}{\sqrt{10}}

B .

- \frac{1}{\sqrt{10}}

C .

\frac{1}{10}

D .

- \frac{1}{10}

Answer: Option A
:
A
Let

0 \leq x \leq π

So,

c o s^{- 1} \frac{4}{5} = x \Rightarrow c o s x = \frac{4}{5}

....(i)
Now

s i n (\frac{1}{2} c o s^{- 1} \frac{4}{5}) = s i n (\frac{x}{2})

....(ii)
From (i),

c o s x = \frac{4}{5} \Rightarrow 1 - 2 s i n^{2} \frac{x}{2} = \frac{4}{5}

\Rightarrow 2 s i n^{2} \frac{x}{2} = 1 - \frac{4}{5} = \frac{1}{5} \Rightarrow s i n \frac{x}{2} = \sqrt{\frac{1}{10}}

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