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

Question

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

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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, $S<C$. Option D is the only valid one. That is; $S<C<T$.

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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.

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

Answer 1

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.