Sum of Sines (to the fourth power)

Using the fact that $\sin^{n}(x) = \cos^{n}(90-x)$ for any real value of n, you can replace any $\sin^{4}(x)$ in the series where $x \geq 46$ with $\cos^{4}(90-x)$
This allows us to "pair" cosines with sines of the same input value, only the term $\sin^{4}(45)$ is not able to be paired.
so $\displaystyle\sum_{k=1}^{89} (\sin^{4}k) = \displaystyle\sum_{k=1}^{44}( \sin^{4}k +\cos^{4}k ) + \sin^{4}(45)$
Letting $x = \sin^{2}k$ and $y = \cos^{2}k$ , $\sin^{4}k +\cos^{4}k = x^{2}+y^{2} = (x+y)^{2} - 2xy = (\sin^{2}k +\cos^{2}k)^{2} -2 \sin^{2}k \cos^{2}k$
This can be simplified to $1-2 \sin^{2}k \cos^{2}k$ by the Pythagorean identity. you can also sub $\sin^{2}k \cos^{2}k = \frac{\sin^{2}k}{4}$
Now we have $\displaystyle\sum_{k=1}^{89} (\sin^{4}k) = \displaystyle\sum_{k=1}^{44}( \sin^{4}k +\cos^{4}k ) + \sin^{4}(45) = \displaystyle\sum_{k=1}^{44}(1- \frac{1}{2}\sin^{2}(2k)) +\sin^{4}(45)$
$\displaystyle\sum_{k=1}^{44}(1-\frac{1}{2} \sin^{2}(2k)) = \displaystyle\sum_{k=1}^{44}(1) - \frac{1}{2} \displaystyle\sum_{k=1}^{44}\sin^{2}(2k)$
Now notice that using $\sin^{2}(x) = \cos^{2}(90-x)$ all of the sines in the series may be paired with a cosine of the same input, giving:
$\displaystyle\sum_{k=1}^{44}\sin^{2}(2k) = \displaystyle\sum_{k=1}^{22}(\sin^{2}(2k)+\cos^{2}(2k)) = \displaystyle\sum_{k=1}^{22}(1) = 22$ Substituting and simplifying: $\displaystyle\sum_{k=1}^{44}(1) - \frac{1}{2} \displaystyle\sum_{k=1}^{44}\sin^{2}(2k) +\sin^{4}(45) = (44)-(\frac{22}{2})+(\frac{1}{4}) = 33.25$

$\begin{aligned} S & = \sum_{k=1}^{89} \sin^4 k^\circ \\ & = \sum_{k=1}^{44} \sin^4 k^\circ + \sin^4 45^\circ + \color{#3D99F6} \sum_{k=46}^{89} \sin^4 k^\circ & \small \color{#3D99F6} \text{Since }\sin (90^\circ - \theta) = \cos \theta \\ & = \sum_{k=1}^{44} \sin^4 k^\circ + \frac 14 + \color{#3D99F6} \sum_{k=1}^{44} \cos^4 k^\circ \\ & = \sum_{k=1}^{44} \left({\color{#3D99F6}\sin^4 k^\circ + \cos^4 k^\circ}\right) + \frac 14 & \small \color{#3D99F6} \text{As }(\sin^2 \theta+\cos^2 \theta)^2 = \sin^4 \theta + 2 \sin^2 \theta \cos^2 \theta + \cos^4 \theta \\ & = \sum_{k=1}^{44} \left({\color{#3D99F6}1-2\sin^2 k^\circ \cos^2 k^\circ}\right) + 0.25 & \small \color{#3D99F6} \text{Note that } \sin 2\theta = 2\sin \theta \cos \theta \\ & = 44.25 - \color{#3D99F6} \frac 12 \sum_{k=1}^{44} \sin^2 2k^\circ & \small \color{#3D99F6} \text{And } \sin^2 \theta = \frac 12(1 - \cos 2\theta) \\ & = 44.25 - \color{#3D99F6} \frac 14 \sum_{k=1}^{44} \left(1-\cos 4k^\circ \right) \\ & = 33.25 - \color{#3D99F6} \frac 14 \sum_{k=1}^{44} \cos 4k^\circ & \small \color{#3D99F6} \text{Using } \sum_{k=a}^b f(k) = \sum_{k=a}^b f(a+b-k) \\ & = 33.25 - \color{#3D99F6} \frac 1{\color{#D61F06}2} \sum_{k=1}^{44} \left(\cos 4k^\circ + \color{#D61F06} \cos (180^\circ - 4k^\circ) \right) & \small \color{#D61F06} \text{Note that } \cos (180^\circ - \theta) = - \cos \theta \\ & = 33.25 - \color{#3D99F6} \frac 12 \sum_{k=1}^{44} \left(\cos 4k^\circ \color{#D61F06} - \cos 4k^\circ \right) \\ & = \boxed{33.25} \end{aligned}$

First we use the double-angle identities to show $\begin{aligned} \sin^4 k &= \left(\frac{1}{2} - \frac{1}{2}\cos(2k)\right)^2 \\ &= \frac{1}{4} - \frac{1}{2}\cos(2k) + \frac{1}{4}\cos^2(2k) \\ &= \frac{1}{4} - \frac{1}{2}\cos(2k) + \frac{1}{4}\left(\frac{1}{2} + \frac{1}{2}\cos(4k)\right) \\ &= \frac{3}{8} - \frac{1}{2}\cos(2k) + \frac{1}{8}\cos(4k) \end{aligned}$

Then $\sum_{k=1}^{89} \sin^4k = \sum_{k=1}^{89} \left(\frac{3}{8} - \frac{1}{2}\cos(2k) + \frac{1}{8}\cos(4k)\right) = \frac{3}{8}(89) - \frac{1}{2}\sum_{k=1}^{89}\cos(2k) + \frac{1}{8}\sum_{k=1}^{89}\cos(4k)$ and to evaluate this, we can compute each of the two remaining sums separately as $\sum_{k=1}^{89}\cos(2k) = \cos(2(45)) + \sum_{k=1}^{44}\left(\cos(2k) + \cos(180-2k)\right) = 0 + \sum_{k=1}^{44} (0) = 0$ $\sum_{k=1}^{89}\cos(4k) = \left(\sum_{k=1}^{360/4}\cos(4k)\right) - \cos(360) = 0 - 1 = -1.$

Putting it all back into our equation for the sum, we have $\sum_{k=1}^{89} \sin^4k = \frac{3}{8}(89) - \frac{1}{2}(0) + \frac{1}{8}(-1) = \frac{133}{4} = \boxed{33.25}$