The value of sin-1cos29-5 is a-b- where a - b are coprime- Then the value of b2-a2 is

Sin^ 1cos nbsp;29π5 can be written as nbsp; π2 cos^ 1cos(6π π5)=π2 π5=3π10 Comparing with nbsp;aπb, we have a=3 and b=10 So nbsp;b^2 a^2=100 9=91 ...

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Standard XIII

Mathematics

Properties Derived from Trigonometric Identities

The value of ...

Question

The value of ${sin}^{- 1} cos \frac{29 π}{5}$ is $\frac{a π}{b}$ , where $a & b$ are coprime. Then the value of $b^{2} - a^{2}$ is

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Solution

${sin}^{- 1} cos \frac{29 π}{5}$ can be written as
$\frac{π}{2} - {cos}^{- 1} cos (6 π - \frac{π}{5}) = \frac{π}{2} - \frac{π}{5} = \frac{3 π}{10}$
Comparing with $\frac{a π}{b}$ , we have $a = 3$ and $b = 10$
So $b^{2} - a^{2} = 100 - 9 = 91$

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Properties Derived from Trigonometric Identities

Standard XIII Mathematics

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The value of sin−1cos29π5 is aπb, where a & b are coprime. Then the value of b2−a2 is

sin−1cos29π5 can be written as π2−cos−1cos(6π−π5)=π2−π5=3π10 Comparing with aπb, we have a=3 and b=10 So b2−a2=100−9=91

The value of ${sin}^{- 1} cos \frac{29 π}{5}$ is $\frac{a π}{b}$ , where $a & b$ are coprime. Then the value of $b^{2} - a^{2}$ is

${sin}^{- 1} cos \frac{29 π}{5}$ can be written as
$\frac{π}{2} - {cos}^{- 1} cos (6 π - \frac{π}{5}) = \frac{π}{2} - \frac{π}{5} = \frac{3 π}{10}$
Comparing with $\frac{a π}{b}$ , we have $a = 3$ and $b = 10$
So $b^{2} - a^{2} = 100 - 9 = 91$