What is the value of 5 cos⁻¹(1/2)+7sin⁻¹(−1/2)

The correct option is π/2.

To find the value of the expression $5{\cos }^{-1}\left(\frac{1}{2}\right)+7{\sin }^{-1}\left(-\frac{1}{2}\right)$, we evaluate each inverse trigonometric term individually using their principal value branches.

First, let us find the value of ${\cos }^{-1}\left(\frac{1}{2}\right)$:
The principal value branch of ${\cos }^{-1}\left(x\right)$ is $\left[0,\pi \right]$.
Since $\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}$ and $\frac{\pi }{3}\in \left[0,\pi \right]$, we have:
${\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$

Second, let us find the value of ${\sin }^{-1}\left(-\frac{1}{2}\right)$:
The principal value branch of ${\sin }^{-1}\left(x\right)$ is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.
Using the property ${\sin }^{-1}(-x)=-{\sin }^{-1}\left(x\right)$, we get:
${\sin }^{-1}\left(-\frac{1}{2}\right)=-{\sin }^{-1}\left(\frac{1}{2}\right)$
Since $\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$, it follows that:
${\sin }^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi }{6}$

Now, substitute these principal values back into the original expression:
$5{\cos }^{-1}\left(\frac{1}{2}\right)+7{\sin }^{-1}\left(-\frac{1}{2}\right)=5\left(\frac{\pi }{3}\right)+7\left(-\frac{\pi }{6}\right)$

Simplify the terms by finding a common denominator (which is 6):
$=\frac{5\pi }{3}-\frac{7\pi }{6}$
$=\frac{10\pi }{6}-\frac{7\pi }{6}$
$=\frac{10\pi -7\pi }{6}$
$=\frac{3\pi }{6}$
$=\frac{\pi }{2}$

Thus, the final value of the expression is $\frac{\pi }{2}$.