period of the function $y = \sin^2\frac{\sqrt2x+3}{6\pi}$

Question

Today I came across a question

The fundamental period of the function $y = \sin^2\frac{\sqrt2x+3}{6\pi}$ is $\lambda \pi^2$ then the value of $\frac{\lambda}{\sqrt2}$ is ___

I tried to equate the angle to $\pi$. Then I got $x=\frac{6\pi^2-3}{\sqrt2}$, which is not in the form of $\lambda\pi^2$. So even on further simplifying I am not getting the correct answer, which is 3. How to solve it?

Equating the angle to $\pi$ doesn't work when there is a constant inside the parentheses. — Matt Samuel, Feb 28 '16 at 15:54
@MattSamuel Yes...this is working...thanks..can you give me a reference for how this works? — manshu, Feb 28 '16 at 15:59
The 3 is a phase shift, it doesn't affect the period. I would think this would be in your textbook. — Matt Samuel, Feb 28 '16 at 16:00

score 1 · Accepted Answer · answered Feb 28 '16 at 15:58

1

HINT:

As $\cos2A=1-2\sin^2A$

the required period will be that of $$\cos\dfrac{\sqrt2x+3}{3\pi}=f(x)$$

Now for $f(x+T)=f(x),$

using Prosthaphaeresis Formulas

we need $\sin\dfrac{\sqrt2T}{6\pi}=0\iff\dfrac{\sqrt2T}{6\pi}=n\pi$ where $n$ is any integer

answered Feb 28 '16 at 15:58

lab bhattacharjee

274,582

1

The same can also be reached by using http://math.stackexchange.com/questions/175143/prove-sinab-sina-b-sin2a-sin2b – lab bhattacharjee Feb 28 '16 at 16:08

period of the function $y = \sin^2\frac{\sqrt2x+3}{6\pi}$

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