Question
Question: How do you verify the identity \( \sin \left( \dfrac{\pi }{2}+x \right)=\cos x \) ?...
How do you verify the identity sin(2π+x)=cosx ?
Solution
Hint : We explain the process of finding values for associated angles. We find the rotation and the position of the angle for (2π+x) . 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(2π+x) .
For general form of sin(y) , we need to convert the value of y into the closest multiple of 2π and add or subtract a certain value α from that multiple of 2π to make it equal to y .
Let’s assume y=k×2π+α , k∈Z . Here we took the addition of α . We also need to remember that ∣α∣≤2π .
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 from y .
For the given angle (2π+x) , we can express as (2π+x)=1×2π+x .
The value of k is odd which means the trigonometrical ratio changes to cos(x) .
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(2π+x)=sin(1×2π+x)=cos(x) .
Thus, the identity sin(2π+x)=cosx is verified.
So, the correct answer is “ cosx ”.
Note : We need to remember that the easiest way to avoid the change of ratio thing is to form the multiple of π instead of 2π . It makes the multiplied number always even. In that case we don’t have to change the ratio. If x=k×π+α=2k×2π+α . Value of 2k is always even.