What is the value of sin−11π3 ?

Let the given be, sin−11π3 .sin−11π3=−sin11π3 ...{∵sin(-θ)=-sinθ}Substitute 11π=12π−1π in the above expression, we getsin−11π3=−sin12π−π3=−sin12π3−π3=−sin4π−π3 We know that, sin(A−B)=sinAcosB−cosAsinB .-sin4π−π3=-sin4πcosπ3−cos4πsinπ3 Then, substituting the values sin4π=0,cos4π=1,sin π 3 = 3 2 ,cos π 3 = 1 2 in the above expression, we get-sin4π−π3=-0×12−1×32 =−-32 Then, we get,sin−11π3=−−32 =32

What is the value of sin−11π3 ?

Let the given be, sin−11π3 .sin−11π3=−sin11π3 ...{∵sin(-θ)=-sinθ}Substitute 11π=12π−1π in the above expression, we getsin−11π3=−sin12π−π3=−sin12π3−π3=−sin4π−π3 We know that, sin(A−B)=sinAcosB−cosAsinB .-sin4π−π3=-sin4πcosπ3−cos4πsinπ3 Then, substituting the values sin4π=0,cos4π=1,sin π 3 = 3 2 ,cos π 3 = 1 2 in the above expression, we get-sin4π−π3=-0×12−1×32 =−-32 Then, we get,sin−11π3=−−32 =32

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What is the value of $\sin (- \frac{11 π}{3})$ ?

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\frac{\sqrt{3}}{2}

1

- 1

0

answer is C.

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Detailed Solution

Let the given be, $\sin (- \frac{11 π}{3})$ .
$\sin (- \frac{11 π}{3}) = - \sin (\frac{11 π}{3})$ $. . . {∵ \sin (- θ) = - sinθ}$
Substitute $11 π = 12 π - 1 π$ in the above expression, we get
$\begin{matrix} \sin (- \frac{11 π}{3}) = - \sin (\frac{12 π - π}{3}) \end{matrix}$

$\begin{matrix} = - \sin (\frac{12 π}{3} - \frac{π}{3}) \\ = - \sin (4 π - \frac{π}{3}) \end{matrix}$
We know that, $\sin (A - B) = \sin A \cos B - \cos A \sin B$ .
$- \sin (4 π - \frac{π}{3}) = - (\sin 4 πcos \frac{π}{3} - \cos 4 πsin \frac{π}{3})$
Then, substituting the values $\sin 4 π = 0, \cos 4 π = 1, \sin \frac{π}{3} = \frac{\sqrt{3}}{2}, \cos \frac{π}{3} = \frac{1}{2}$ in the above expression, we get
$\begin{matrix} - \sin (4 π - \frac{π}{3}) = - (0 \times \frac{1}{2} - 1 \times \frac{\sqrt{3}}{2}) \\ = - (- \frac{\sqrt{3}}{2}) \end{matrix}$
Then, we get,
$\begin{matrix} \sin (- \frac{11 π}{3}) = - (- \frac{\sqrt{3}}{2}) \\ = \frac{\sqrt{3}}{2} \end{matrix}$