Question

The period of \[{\sin ^p}x + {\cos ^p}x\]is \[\dfrac{\pi }{2}\]if is
(A). Even except 2
(B). Odd except 3
(C). Any natural number
(D). Any integer

Answer 1

Hint:
The best method to find the period of trigonometric functions is by substitution method. Substitute the values of p as 1,2,4 and so on and keep finding the period. You will soon find a pattern in the period and you will find the answer to the question.

Complete step by step solution:
Let \[f(x) = {\sin ^p}x + {\cos ^p}x\]
Step 1: Find the period of \[f(x)\] for \[p = 1\]
Let \[p = 1\]
\[f(x) = \sin x + \cos x\]
Period of \[f(x)\] is \[LCM(2\pi ,2\pi ) = 2\pi \]
So, options B, C, D do not hold true.
Step 2: Find the period of \[f(x)\] for \[p = 2\]
Now, if \[p = 2\]
\[f(x) = {\sin ^2}x + {\cos ^2}x = 1\]
Hence, period of \[f(x)\] is not \[\dfrac{\pi }{2}\] for \[p = 2\]
Step 3: Find the period of \[f(x)\] for \[p = 4\]
Also, if \[p = 4\]
\[\begin{array}{l}
f(x) = {\sin ^4}x + {\cos ^4}x\\
f(x) = 1 - \dfrac{1}{4}{\sin ^2}2x
\end{array}\]
Now the period of \[{\sin ^n}x\]is \[\pi \] when n is even.
So, period of \[{\sin ^2}2x\]is \[\dfrac{\pi }{2}\]
So, period of \[f(x)\] is \[\dfrac{\pi }{2}\] for even value of \[p\] except 2
Hence, the correct answer is (A).

Note:
The period is defined as the length of a function's cycle. Trig functions are cyclical, and when you graph them, you'll see the ups and downs of the graph and you'll see that these ups and downs keep repeating at regular intervals. Label your A, B, C, and D values and then Calculate your period. To solve such questions easily and effectively, learn the periods of common functions and use them in the problem to find periods. Also learn how multiplying or dividing the variables of functions affect the period of the function. Keep in mind that adding two or more trigonometric functions yield different periods based on the periods of individual functions. Plot graphs if required to understand how the periods work