Question

Using the trigonometric identity for sine of sum of angles, without expanding each term separately, prove that
$sin\left( {{40}^{\circ }}+\theta \right)\cos \left( {{10}^{\circ }}+\theta \right)-\cos \left( {{40}^{\circ }}+\theta \right)\sin \left( {{10}^{\circ }}+\theta \right)=\dfrac{1}{2}$.

Answer 1

Hint: In this given question, we can use the transformation formula for the sine of the difference of two angles that is, $\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)$ in order to reach to our answer. We can replace A and B with $\left( {{40}^{\circ }}+\theta \right)$ and $\left( {{10}^{\circ }}+\theta \right)$ respectively and simplify the so formed equation and then put the required value of trigonometric ratios and we will arrive at our answer.

Complete step-by-step answer:
In the given question, we are asked to prove the following equation:
$sin\left( {{40}^{\circ }}+\theta \right)\cos \left( {{10}^{\circ }}+\theta \right)-\cos \left( {{40}^{\circ }}+\theta \right)\sin \left( {{10}^{\circ }}+\theta \right)=\dfrac{1}{2}$.
Here, in this solution, we are going to make use of the following transformation formula for the sine of the difference of two angles, that is:
$\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)..............(1.1)$, in the process of solving the equation with replaced values of A and B as $\left( {{40}^{\circ }}+\theta \right)$ and $\left( {{10}^{\circ }}+\theta \right)$ respectively.
The process of proving that the Left Hand Side (LHS) is equal to the Right Hand Side (RHS) is as follows:
$LHS=sin\left( {{40}^{\circ }}+\theta \right)\cos \left( {{10}^{\circ }}+\theta \right)-\cos \left( {{40}^{\circ }}+\theta \right)\sin \left( {{10}^{\circ }}+\theta \right)$
$=\sin \left( {{40}^{\circ }}+\theta -{{10}^{\circ }}-\theta \right)$
$=\sin \left( {{60}^{\circ }} \right)$
$=\dfrac{1}{2}$
$=RHS$
Hence, we have proved the required condition is that the Left Hand Side (LHS) is equal to the Right Hand Side (RHS).
Therefore, we come to a conclusion that $sin\left( {{40}^{\circ }}+\theta \right)\cos \left( {{10}^{\circ }}+\theta \right)-\cos \left( {{40}^{\circ }}+\theta \right)\sin \left( {{10}^{\circ }}+\theta \right)=\dfrac{1}{2}$.

Note: Here, we are using the transformation formula for the sine of the difference of two angles, that is:
$\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)$.
But in some other questions we may also have to use the transformation formula for the sine of the sum of two angles, that is:
$\sin A\cos B+\cos A\sin B=\sin \left( A+B \right)$.