Question

Prove the identity. $\sin \left( {\pi + x} \right) = - \sin x$. Have to show the statements and the rules?

Answer 1

Here, we will use the basic identities of the trigonometric functions on the LHS of the given equation of the trigonometric function for the simplification of the equation. Then by solving the simplified equation by substituting the values, we will get the required property.

Complete step by step solution:
Given property is $\sin \left( {\pi + x} \right) = - \sin x$.
First, we will take the LHS of the given equation. Therefore, we get
$LHS = \sin \left( {\pi + x} \right)$
Now we will apply the basic properties of the trigonometric sine function i.e. $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$. Therefore, by applying this property we get
$\Rightarrow LHS = \sin \pi \cos x + \cos \pi \sin x$
We know that the value of $\sin \pi = 0$ and $\cos \pi = - 1$. Therefore, by putting these values in the equation we get
$\Rightarrow LHS = 0 \times \cos x + \left( { - 1} \right) \times \sin x$
Multiplying the terms, we get
$\Rightarrow LHS = 0 - \sin x$
$\Rightarrow LHS = - \sin x$
So, we can clearly see that it is equal to the RHS of the given equation. Therefore, we get
$\Rightarrow \sin \left( {\pi + x} \right) = - \sin x$
Hence, proved.

Note: In this type of questions we can apply the basic trigonometric properties to simplify and solve the equation. We need to keep in mind the different properties of the trigonometric function and also which quadrant function is positive or negative. We know that in the first quadrant all the functions i.e. sin, cos, tan, cot, sec, cosec are positive. In the second quadrant, only the sin and cosec functions are positive and all the other functions are negative. In the third quadrant, only tan and cot function is positive. Similarly in the fourth quadrant, only the cos and sec function is positive.