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Question: How do you verify the identity \( \sin \left( \dfrac{\pi }{2}+x \right)=\cos x \) ?...

How do you verify the identity sin(π2+x)=cosx\sin \left( \dfrac{\pi }{2}+x \right)=\cos x ?

Explanation

Solution

Hint : We explain the process of finding values for associated angles. We find the rotation and the position of the angle for (π2+x)\left( \dfrac{\pi }{2}+x \right) . We explain the changes that are required for that angle. Depending on those things we find the solution.

Complete step-by-step answer :
We need to find the ratio value for sin(π2+x)\sin \left( \dfrac{\pi }{2}+x \right) .
For general form of sin(y)\sin \left( y \right) , we need to convert the value of yy into the closest multiple of π2\dfrac{\pi }{2} and add or subtract a certain value α\alpha from that multiple of π2\dfrac{\pi }{2} to make it equal to yy .
Let’s assume y=k×π2+αy=k\times \dfrac{\pi }{2}+\alpha , kZk\in \mathbb{Z} . Here we took the addition of α\alpha . We also need to remember that απ2\left| \alpha \right|\le \dfrac{\pi }{2} .
Now we take the value of k. If it’s even then keep the ratio as sin and if it’s odd then the ratio changes to cot ratio from cos.
Then we find the position of the given angle as quadrant value measured in counter clockwise movement from the origin and the positive side of X-axis.
If the angle falls in the first or second quadrant then the sign remains positive but if it falls in the third or fourth quadrant then the sign becomes negative.
Depending on the sign and ratio change the final angle becomes α\alpha from yy .
For the given angle (π2+x)\left( \dfrac{\pi }{2}+x \right) , we can express as (π2+x)=1×π2+x\left( \dfrac{\pi }{2}+x \right)=1\times \dfrac{\pi }{2}+x .
The value of k is odd which means the trigonometrical ratio changes to cos(x)\cos \left( x \right) .
The position of the angle is in the second quadrant. The angle completes the half of half-circle 1 times.
Therefore, the sign remains positive.
The final form becomes
sin(π2+x)=sin(1×π2+x)=cos(x)\sin \left( \dfrac{\pi }{2}+x \right)=\sin \left( 1\times \dfrac{\pi }{2}+x \right)=\cos \left( x \right) .
Thus, the identity sin(π2+x)=cosx\sin \left( \dfrac{\pi }{2}+x \right)=\cos x is verified.
So, the correct answer is “ cosx\cos x ”.

Note : We need to remember that the easiest way to avoid the change of ratio thing is to form the multiple of π\pi instead of π2\dfrac{\pi }{2} . It makes the multiplied number always even. In that case we don’t have to change the ratio. If x=k×π+α=2k×π2+αx=k\times \pi +\alpha =2k\times \dfrac{\pi }{2}+\alpha . Value of 2k2k is always even.