Spin in two different direction

i tried this using spherical coordinates, it turned out to be too tedious. i decided to use cartesian coordinate because of a nice relation i saw: $2\cos(\theta/2) \mid\theta \phi +\rangle =2 \cos^2(\theta/2) \mid \uparrow \rangle+2\cos(\theta/2)\sin(\theta/2) e^{i\phi}\mid \downarrow\rangle\\=(1+cos(\theta))\mid \uparrow \rangle+\sin(\theta)(\cos(\phi)+i\sin(\phi))\mid \downarrow\rangle= (1+z)\mid \uparrow \rangle+(x+iy)\mid \downarrow\rangle$ so the inner product is: $4\cos(\theta_1/2)\cos(\theta_2/2) \langle\theta_2 \phi_2 +\mid\theta_1 \phi_1 +\rangle=(1+z_1)(1+z_2)+(x_1+iy_1)(x_2-iy_2)$ squaring the norm of both sides and using half angle identities $4(1+z_1)(1+z_2)\mid\langle\theta_2 \phi_2 +\mid\theta_1 \phi_1 +\rangle\mid^2 = [(1+z_1)(1+z_2)+(x_1+iy_1)(x_2-iy_2)][(1+z_1)(1+z_2)+(x_1-iy_1)(x_2+iy_2)]$ the RHS uses the fact $|z|^2 = z \overline{z}$ . instead of fully expanding we can use $(a+b)(a+\overline{b})= a^2 +a(b+\overline{b})+|b|^2 = a^2+2a Re(b) +|b|^2$ with a.b been obvious substitutes. we have $4(1+z_1)(1+z_2)\mid\langle\theta_2 \phi_2 +\mid\theta_1 \phi_1 +\rangle\mid^2\\= (1+z_1)^2(1+z_2)^2+2(1+z_1)(1+z_2) Re((x_1+iy_1)(x_2-iy_2)) + (x_1^2+y_1^2)(x_2^2+y_2^2) \\ =(1+z_1)^2(1+z_2)^2+2(1+z_1)(1+z_2)(x_1x_2+y_1 y_2) + (1-z_1^2)(1-z_2^2)\\= (1+z_1)(1+z_2)[(1+z_1)(1+z_2)+2(x_1x_2+y_1 y_2) +(1-z_1)(1-z_2)]$ where the normality of the vector was used to find $x^2+y^2+z^2=1$ . cancelling the common factor we have $4\mid\langle\theta_2 \phi_2 +\mid\theta_1 \phi_1 +\rangle\mid^2= (1+z_1)(1+z_2)+2(x_1x_2+y_1 y_2) +(1-z_1)(1-z_2)\\= 1+z_1+z_2+z_1z_2+2x_1x_2+2y_1y_2 +1-z_1-z_2+z_1+z_2= 2(1+x_1x_2+y_1y_2+z_1z_2) \\ |<\theta_2 \phi_2+|\theta_1 \phi_1 +>|^2= \dfrac{1+x_1x_2+y_1y_2+z_1z_2}{2} = \dfrac{1+n_1 .n_2}{2}$ the cosine of the angle between the two unit vectors( $\pi/3$ ) is the dot product of the two unit vectors.. hence this turns into $\mid\langle\theta_2 \phi_2 +\mid\theta_1 \phi_1 +\rangle\mid^2 = \dfrac{1+\cos(\pi/3)}{2} = \cos^2(\pi/6)= \boxed{\dfrac{3}{4}}$