Question

What is the value of $\dfrac{{\sin 150^\circ + \sin 210^\circ + \sin 810^\circ }}{{\sin 810^\circ }}$?

Answer 1

In this question, we are given a trigonometric equation and we have to find its value. The trigonometric ratios given to us in the question are more than $90^\circ$ and we are told the values of these ratios up to $90^\circ$ only. So, at first, we will split these ratios. For example: we can write these ratios as $\sin 150^\circ = \sin \left( {180 - 30} \right)^\circ$. So, the first step is to split all the given ratios in this manner. Then, find their values and put them in the given equation. Solve them and the resultant answer will be the required answer.

Complete step-by-step solution:
We are given a trigonometric equation $\dfrac{{\sin 150^\circ + \sin 210^\circ + \sin 810^\circ }}{{\sin 810^\circ }}$ and we have to find its value.
First, let us expand them as we have only been taught the values of these ratios up to $90^\circ$.
$\Rightarrow \dfrac{{\sin \left( {180 - 30} \right)^\circ + \sin \left( {180 + 30} \right)^\circ + \sin \left( {2 \times 360 + 90} \right)^\circ }}{{\sin \left( {2 \times 360 + 90} \right)^\circ }}$
Now, we know that $\sin \left( {180 + x} \right)^\circ = - \sin x^\circ$, $\sin \left( {180 - x} \right)^\circ = \sin x^\circ$. Using these, we will simplify the above equation,
$\Rightarrow \dfrac{{\sin 30^\circ - \sin 30^\circ + \sin 90^\circ }}{{\sin 90^\circ }}$
On simplifying, we will get –
$\Rightarrow \dfrac{{\sin 90^\circ }}{{\sin 90^\circ }}$

Hence, The value of $\dfrac{{\sin 150^\circ + \sin 210^\circ + \sin 810^\circ }}{{\sin 810^\circ }}$ $= 1$.

Note: 1) There are always 2 methods to expand a trigonometric ratio. One method has been used above in this question. The other can be explained below.
We wrote $\sin \left( {150} \right)^\circ$ as $\sin \left( {180 - 30} \right)^\circ$. It can also be written as -
$\sin \left( {150} \right)^\circ = \sin \left( {90 + 60} \right)^\circ$
Similarly, $\sin 210^\circ = \sin \left( {270 - 60} \right)^\circ$. In this case, sin changes to cos because they are expanded using $90^\circ$ and $270^\circ$.
2) While expanding $\sin 270^\circ$, we put a sign of subtraction because $\sin 270^\circ$ will be in the $3^{rd}$ quadrant, and in the $3^{rd}$ quadrant, sin is negative.
3) $\sin 810^\circ$ was expanded in the following way:
At first, we look for a multiple of 360, closest to 810. This is because it will complete the whole rounds at the quadrants. In this case, it is 720.
Then, we add something to 720 to make it 810. In this case, it will be 90.
Hence, our $\sin 810^\circ$ has now become $\sin \left( {2 \times 360 + 90} \right)^\circ$.