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A. Increases, then decreases.

B. Decreases and then increases.

C. Decreases throughout.

D. Increases throughout.

E. Decreases, increases and then decreases again.

Answer

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Given trigonometric function is $4\cos \left( {\dfrac{\theta }{2}} \right)$.

The lower interval is $\dfrac{\pi }{4}$ and the upper interval is $\dfrac{{5\pi }}{4}$.

Now, we need to analyse this problem step by step. The angle value also gets divided by two in the function. Hence, we must not forget to divide the intervals also.

The basic $\cos \theta $ function is multiplied by four in the given problem. But it does not make any difference as if the function is increasing, by multiplying it with a constant value increases the amplitude only. Same in the case if the function is a decreasing function and it is multiplied with any constant value.

As all of us know that $\cos \theta $ is a decreasing function in the interval $\left( {0,\pi } \right)$.

So, we can get one clarity that when the value of $\theta $ increases in the range of $\left( {0,\pi } \right)$, the value of $\cos \theta $ will be decreasing.

But in the function, it is given the angle as $\left( {\dfrac{\theta }{2}} \right)$. So, let us divide the interval boundaries also with two.

So, we get,

The lower boundary is \[\dfrac{\pi }{8}\].

The upper boundary is \[\dfrac{{5\pi }}{8}\].

Now, both values are within the range of the interval $\left( {0,\pi } \right)$.

Hence the function is decreasing in that interval and multiplication of the function with constant does not make any difference, we can say that

As $\theta $ increases from $\dfrac{\pi }{4}$ to $\dfrac{{5\pi }}{4}$, the value of $4\cos \left( {\dfrac{\theta }{2}} \right)$ decreases throughout.