How to Find the Area Between Two Graphs: A Trigonometric Approach

[Ocasta]

Homework Statement

Find the area of the region enclosed between y=4sin(x) and y=2cos(x) from x=0 to x=0.8pi.

Homework Equations

$\int^{0.8\pi}_0 dx$

$g(x) = 4\sin(x)$

$f(x) = 2\cos(x)$

The Attempt at a Solution

This problem needs to be split up into two parts.

$\int_0 ^n [f(x) - g(x)] dx + \int_n^{0.8\pi} [g(x) - f(x)] dx$

My major problem is finding n.

I set:

$f(n) = g(n) \rightarrow 4\sin(n) = 2 \cos(n) \rightarrow 2\sin(n) = cos(n) \rightarrow 2 = \frac{cos(n)}{sin(n)} \rightarrow 2 = \cot(n)$

I'm having trouble finding that point n. I've worked out that it's near

$\frac{15\pi}{96} = n$


and with that n I have:
$\int_0^n [2 \cos(x) - 4 \sin(x)] dx + \int_n^{0.8\pi} [2 \cos(x) - 4\sin(x)] dx$

$-2 \sin(x) - (-4) \cos(x)]_0^\frac{15\pi}{96} + [-2 \sin(x) - (-4) \cos(x)]_\frac{15\pi}{96}^{0.8\pi}$

 

[junaid314159]

Looks like you did a very good job. You got cot(n) = 2. If you take the reciprocal of both sides, you will see that it is the same as tan(n) = 1/2. If you are allowed to solve the problem numerically, then you can find arctan(0.5) on your calculator and evaluate the two integrals using the decimal value. If you need an exact answer, then you can try substituting arctan(1/2) as it is and evaluate the sin and cos of arctan(1/2) using your knowledge of trig.

Junaid Mansuri