The motion of a simple pendulum is given by $\theta = A\cos\left(\sqrt{\frac g l}t\right)$ Find the tension in the string of this pendulum; assume that $\theta \ll 1$
the vertical component of the force due to gravity must be equal to the tension hence: $$T = F\cos\theta$$ by small-angle approximation $$\cos\theta = 1 -\frac{\theta^2}{2}$$ $$T = mg\left(1 -\frac{\theta^2}{2}\right)$$ $$T = mg\left(1-\frac{A^2\cos^2\left(\sqrt{\frac g l}t\right)}{2}\right)$$ by trigonometry: $$T = mg\left({1-\frac{A^2}{2}+\frac{A^2}{2}\sin^2\left(\sqrt{\frac g l}t\right)}\right)$$
which is not the desired answer, what did I get wrong? true answer : $$mg\left(1-\frac{1}{2}A^2+3\frac{A^2}{2}\sin^2\left(\sqrt{\frac g l}t\right)\right)$$
Update: I forgot to include the centripetal force in the equation hence:
$$T = Fcos\theta + F_c$$ $$F_c = \frac{mv^2} {r} = \omega^2 r$$ $$\omega^2 = (g/l)A^2\sin^2\left(\sqrt{\frac g l}t\right)$$ $$r = A$$
but this adding this to our previous answer does not reach the desired form
Your work is almost fine, note that for the centripetal part we have that $r=l$ then
$$F_c=m\omega^2l = mgA^2\sin^2\left(\sqrt{\frac g l}t\right)$$
and then
$$T = F\cos\theta + F_c=mg\left({1-\frac{A^2}{2}+\frac{A^2}{2}\sin^2\left(\sqrt{\frac g l}t\right)}\right)+mgA^2\sin^2\left(\sqrt{\frac g l}t\right)=$$
$$=mg\left(1-\frac{1}{2}A^2+3\frac{A^2}{2}\sin^2\left(\sqrt{\frac g l}t\right)\right)$$