Question

How do you verify the identity $\cos (\dfrac{\pi }{2} + x) = - \sin x$?

Answer 1

This question is based on trigonometric identities. In this question we have to prove this
trigonometric identity$\cos (\dfrac{\pi }{2} + x) = - \sin x$. To prove this we need to know the
trigonometric identity$\cos (A + B) = \cos A\cos B - \sin A\sin B$. To prove this identity we will apply

trigonometric identity and put the corresponding values of $\sin \dfrac{\pi }{2}$and $\cos \dfrac{\pi }{2}$to desired results. To solve this question knowing the basic trigonometric identities is must.

Complete step by step solution:
Let us try to solve this question in which we are asked to prove that 0$\cos (\dfrac{\pi }{2} + x) = - \sin x$.
To prove this identity we use the following trigonometric identity $\cos (A + B) = \cos A\cos B - \sin A\sin B$ to expand$\cos (\dfrac{\pi }{2} + x)$.
Since we know the value of sine function and cosine function at $\dfrac{\pi }{2}$. Putting the values of $\sin \dfrac{\pi }{2}$ and $\cos \dfrac{\pi }{2}$ in the formula we get the required result. Let’s formally prove the trigonometric identity.
To prove: $\cos (\dfrac{\pi }{2} + x) = - \sin x$
Proof:
$\cos (\dfrac{\pi }{2} + x) = \cos \dfrac{\pi }{2}\cos x - \sin \dfrac{\pi }{2}\sin x$where
$A = \dfrac{\pi }{2}$and $B = x$
Now, putting the values of $\sin \dfrac{\pi }{2}$and $\cos \dfrac{\pi }{2}$in the above formula, we get
$\cos (\dfrac{\pi }{2} + x) = 0 \cdot \cos x - 1 \cdot \sin x$
Because $\sin \dfrac{\pi }{2} = 1$and$\cos \dfrac{\pi }{2} = 0$.
$\cos (\dfrac{\pi }{2} + x) = - \sin x$which is our required result.
Hence we prove that the$\cos (\dfrac{\pi }{2} + x) = - \sin x$.

Note: Sine and cosine trigonometric functions have phase differences of $\dfrac{\pi }{2}$. Sine and cosecant functions have positive values in quadrant $1$ and $2$. Similarly, cosine function and secant functions have positive values in quadrant $1$ and $2$. Similarly, cosine function and secant function have positive values in quadrant $1$ and $4$. Similarly, tangent function and cotangent function have positive values in quadrant $1$ and $3$. For trigonometric questions knowing trigonometric identities and the trigonometric functions values at common angle values.