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1 year ago

Using radians find the amplitude in period of each function then graph. Question 9

Mathematics

1 answer:

sleet_krkn [62]1 year ago

6 0

we have the function

$y=\frac{1}{2}sin4\theta$

Remember that

The Amplitude is the height from the center line to the peak (or to the trough). Or we can measure the height from highest to lowest points and divide that by 2

In the general form of the equation of the sine

$y=Asin(B(x+c))+D$

The amplitude is the factor A

In this problem

The amplitude is 1/2

using a graphing tool

Note that the amplitude is the height from the center line to the peak (the center line is the x-axis, so the distance is 0.5.

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2 years ago

Evaluate the integral following

alina1380 [7]

Answer:

$\displaystyle{4\tan x + \sin 2x - 6x + C}$

Step-by-step explanation:

We are given the integral of:

$\displaystyle{\int 4(\sec x - \cos x)^2 \, dx}$

First, we can use a property to separate a constant out of integrand:

$\displaystyle{4 \int (\sec x - \cos x)^2 \, dx}$

Next, expand the expression (integrand):

$\displaystyle{4 \int \sec^2 x - 2\sec x \cos x + \cos^2 x \, dx}$

Since $\displaystyle{\sec x = \dfrac{1}{\cos x}}$ then it can be simplified to:

$\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2\dfrac{1}{\cos x} \cos x + \cos^2 x \, dx}\\\\\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \cos^2 x \, dx}$

Recall the formula:

$\displaystyle{\int \dfrac{1}{\cos ^2 x} \, dx = \int \sec ^2 x \, dx = \tan x + C}\\\\\displaystyle{\int A \, dx = Ax + C \ \ \tt{(A \ and \ C \ are \ constant.)}$

For $\displaystyle{\cos ^2 x}$ , we need to convert to another identity since the integrand does not have a default or specific integration formula. We know that:

$\displaystyle{2\cos^2 x -1 = \cos2x}$

We can solve for $\displaystyle{\cos ^2x}$ which is:

$\displaystyle{2\cos^2 x = \cos2x+1}\\\\\displaystyle{\cos^2x = \dfrac{\cos 2x +1}{2}}$

Therefore, we can write new integral as:

$\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \dfrac{\cos2x +1}{2} \, dx}$

Evaluate each integral, applying the integration formula:

$\displaystyle{\int \dfrac{1}{\cos^2x} \, dx = \boxed{\tan x + C}}\\\\\displaystyle{\int -2 \, dx = \boxed{-2x + C}}\\\\\displaystyle{\int \dfrac{\cos 2x +1}{2} \, dx = \dfrac{1}{2}\int \cos 2x +1 \, dx}\\\\\displaystyle{= \dfrac{1}{2}\left(\dfrac{1}{2}\sin 2x + x\right) + C}\\\\\displaystyle{= \boxed{\dfrac{1}{4}\sin 2x + \dfrac{1}{2}x + C}}$