Consider a simple pendulum oscillating in a vertical plane, with θ...

Let $T$ be the tension in the string, $m$ be the mass of the bob, $v$ be its velocity, and $L$ be the length of the pendulum.

The forces acting on the bob in the radial direction are the tension $T$ directed towards the pivot and the component of weight $mg\cos\theta$ directed away from the pivot.

Using Newton's second law for circular motion, the net radial force provides the necessary centripetal acceleration:

$T - mg\cos\theta = \dfrac{mv^2}{L}$

$T = mg\cos\theta + \dfrac{mv^2}{L}$

For the tension to be equal to $mg\cos\theta$, the centripetal term must be zero:

$\dfrac{mv^2}{L} = 0 \Rightarrow v = 0$

The velocity of the pendulum bob is zero only at the extreme positions of its oscillation.

Therefore, the tension in the string is equal to $mg\cos\theta$ at the extreme positions.