How to get $A$ and $B$ from $A\csc 10^\circ+B=$ $\sin 10^\circ+\cos 60^\circ+\cos 40^\circ+\sin 70^\circ+\sin 90^\circ$?

Question

The problem is as follows:

Find $A+B$ from:

$A\csc 10^\circ+B=\sin 10^\circ+\cos 60^\circ+\cos 40^\circ+\sin 70^\circ+\sin 90^\circ$

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{5}\\ 2.&\textrm{2}\\ 3.&\textrm{1}\\ 4.&\textrm{4}\\ \end{array}$

What I did to attempt solving this problem was as follows: Please note that what itis written below follows the right side of the equation so it can be equated properly.

$\sin 10^\circ+\cos 60^\circ+\cos 40^\circ+\sin 70^\circ+\sin 90^\circ$

The curious thing which I noticed is that I could arrange them in the following way:

$\sin 10^\circ + \sin 30^\circ + \sin 50^\circ + \sin 70^\circ + \sin 90^\circ$

But that's it. Its just the part where I got stuck. Is there any trick here?. Can someone help me here?.

Since this problem was found in a section belonging to prosthaphaeresis topic on my workbook. Thus what could it be done here?.

What it confuses me the most is how to get to appear the cosecant of $10^\circ$. So all and all can someone help me with the appropiate method for this?.

Answer 1

When angles are in arithmetic progression, the formula for sum of their sines (cosines) is given by

$$\sin A + \sin (A+D) + \sin (A+2D) + \ldots + \sin (A+(n-1)D) =$$ $$ \frac{\sin nD/2}{\sin D/2}\, \sin \left( \frac{2A+(n-1)D}{2} \right)$$

This should remind you of an usual arithmetic series sum $$A+(A+D)+(A+2D)+\ldots+(A+(n-1)D)=$$ $$n \times \frac{\text{first term + last term}}{2}$$

In a memorable way, $$\sum_{k=0}^{n-1} \sin(A+kD)= \frac{\sin n\tfrac{D}{2}}{\sin \tfrac{D}{2}}\, \sin \left( \frac{\text{first angle + last angle}}{2} \right)$$ $$\sum_{k=0}^{n-1} \cos(A+kD)= \frac{\sin n\tfrac{D}{2}}{\sin \tfrac{D}{2}}\, \cos \left( \frac{\text{first angle + last angle}}{2} \right)$$ Please note, in sum of cosines, only the second factor changes.

Now to solve the problem, we first use $\cos \theta = \sin (90-\theta)$ to get the angles in AP. $$\sin 10^\circ+\cos 60^\circ+\cos 40^\circ+\sin 70^\circ+\sin 90^\circ$$ $$=\cos 80^\circ+\cos 60^\circ+\cos 40^\circ+\cos 20^\circ+1$$ $$=1+\frac{\sin 4\cdot \tfrac{20^\circ}{2}}{\sin \tfrac{20^\circ}{2}}\, \cos \left( \frac{20^\circ+80^\circ}{2} \right)=1+\frac{\sin 40^\circ \cos 50^\circ}{\sin 10^\circ}$$ $$=1+\frac{2\sin^2 40^\circ}{2\sin 10^\circ}=1+\frac{1-\cos 80^\circ}{2\sin 10^\circ}=\frac{1+\sin 10^\circ}{2\sin 10^\circ}$$ $$=\frac{1}{2}+\frac{1}{2}\csc 10^\circ$$