Question

if the period of the function f(x)= sin^4x -cos^4x is mπ/n where m and n are co prime then find the value of (mn)^2​

Answer 1

A composite number can be written as a product

of prime numbers in

Answer 2

Answer:

The value of (mn)^2​=1.

Step-by-step explanation:

Given: Period of $f(x)= sin^4x -cos^4x$ is mπ/n ; m & n are coprime.

To find: the value of $(mn)^{2}$.

$f(x)= sin^4x -cos^4x$

      $=( sin^2x -cos^2x)( sin^2x+cos^2x)$

We know that, $cos^2x-sin^2x =cos2x$ and $sin^2x +cos^2x =1$.

$f(x)=-cos2x$

Now we know that period of cosx is $2\pi$ so the period of cos2x is $\pi$.

=> Period of $f(x)=-cos2x$ is $\pi$.

Now compare mπ/n with $\pi$ we have $m=1,n=1$

=> $(mn)^{2} =1^{2}$

              =1

Hence, if the period of the function f(x)= sin^4x -cos^4x is mπ/n where m and n are co prime then the value of (mn)^2​=1.

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