# The order of $C=\sum_{k=0}^{90}(90-k)\cos(k ^{\circ}), S=\sum_{k=0}^{90}(90-k)\sin(k ^{\circ})$, and $T=\sum_{k=0}^{90}(90-k)\tan(k ^{\circ})$.

This is multiple choice questions, where using calculators is not allowed. Candidates have, on average, $$2$$ minutes $$30$$ seconds to solve it. MY ATTEMPT:

$$k ^{\circ} = (\text{positive constant} \times k)$$ radians. That positive constant is $$\pi/180 ^{\circ}$$.

So, for simplicity, we can replace $$k ^{\circ}$$ by $$k$$ since the sign of inequality will not be changed when we multiply by a positive constant.

$$C \approx 90 \int_0^{\pi/2}\cos(x)dx-\int_0^{\pi/2}x \cos(x)dx=90(1)-(\frac{\pi}{2}-1) \approx 89.4$$

$$S \approx 90 \int_0^{\pi/2}\sin(x)dx-\int_0^{\pi/2}x \sin(x)dx=90(1)-(1) = 89$$

From here, $$S. Option D is the only valid one. That is; $$S.

Just to check, I used Excel, I found that D is the correct one. However, I am not sure if my approach was okay or not.

Your help to solve this problem in an easier and faster way would be appreciated. THANKS!

As $$k$$ increases, $$\cos k^\circ$$ and $$90-k$$ both decrease and $$\sin k^\circ$$ increases.
Using the fact that $$\cos k^\circ = \sin (90^\circ - k^\circ)$$ and the Rearrangement inequality, $$C>S$$ follows immediately.
For $$T$$, $$\tan 90^\circ$$ is not defined. Excluding the case $$k=90$$, $$S < T$$ follows from $$\cos x \le 1$$. The inequality $$T>C$$ will require a bit more work.