As you know, it is generally easier to resolve forces into components at right angles to each other. Obvious possibilities (in a two dimensional set up) are horizontal and vertical or along the string and a right angles to it.

Suppose that the simple pendulum has been displaced through angle $\theta$ and has just been released. In that case its velocity is zero and there will be no centripetal acceleration, that is no acceleration in the string direction.

In that case it makes sense to resolve forces along the string and at right angles to it. Along the string, because there is no acceleration, we have $$mg \cos \theta =T$$ [If the pendulum is moving, the body has a centripetal acceleration and the equation becomes $T - mg \cos \theta = \frac {mv^2}{\ell}$, but for many purposes we can neglect this centripetal term.]

At right angles to the string we have $$mg \sin \theta =ma$$ in which $a$ is the acceleration of the pendulum bob.

As you know, it is generally easier to resolve forces into components at right angles to each other. Obvious possibilities (in this two dimensional set up) are horizontal and vertical or along the string and a right angles to it.

Suppose that the simple pendulum has been displaced through angle $\theta$ and has just been released. In that case its velocity is zero and there will be no centripetal acceleration, that is no acceleration in the string direction.

In that case it makes sense to resolve forces along the string and at right angles to it. Along the string, because there is no acceleration, we have $$mg \cos \theta =T$$ [If the pendulum is moving, the body has a centripetal acceleration and the equation becomes $T - mg \cos \theta = \frac {mv^2}{\ell}$, but for many purposes we can neglect this centripetal term.]

At right angles to the string we have $$mg \sin \theta =ma$$ in which $a$ is the 'tangential' acceleration of the pendulum bob.