Trig Quadratic Identity & Volume Between Two Curves (1 Viewer)

j1mmy_ · Mar 31, 2013

How do you do these questions,

1) Noting that $2cos^{2}x\equiv 1+cos2x$ , prove that $8cos^{4}x\equiv 3+4cos2x+cos4x$

2) Sketch on the same diagram, the curves $y=cos x, y=cos^2x, for (0\leq x\leq \frac{\pi}{2})$ . Find the area enclosed between these curves and the volume generated when this area is rotated about the x axis.

Here's the diagram for Question 2:

TheMathStudent · Mar 31, 2013

1. $\begin{align*}4cos^4(x)&=1+2cos(2x)+4cos^2(2x)\\&=1+2cos(2x)+2(2cos^2(2x))\\&=1+2cos(2x)+2(1+cos(4x)) \end{align*}$
expand and simplify

2. All you need to note is that $cos^2(x)=\frac{1+cos(2x)}{2}$ now it should be easier to sketch

j1mmy_ · Mar 31, 2013

TheMathStudent said:
1. $\begin{align*}4cos^4(x)&=1+2cos(2x)+4cos^2(2x)\\&=1+2cos(2x)+2(2cos^2(2x))\\&=1+2cos(2x)+2(1+cos(4x)) \end{align*}$
expand and simplify

2. All you need to note is that $cos^2(x)=\frac{1+cos(2x)}{2}$ now it should be easier to sketch

Could you help me on the bit with finding the volume, that's the one I'm stuck on as shown in title.

nightweaver066 · Mar 31, 2013

$$The $ y=cos^2x $ graph is below the $ y=cosx$

$V= \pi \int ^{\frac{\pi}{2}} _0 (cosx)^2 - (cos^2x)^2 \; dx$

$= \pi \int ^{\frac{\pi}{2}} _0 cos^2x - cos^4x \; dx$

$Now just use what you've proven before and integrate$

j1mmy_ · Mar 31, 2013

nightweaver066 said:
$$The $ y=cos^2x $ graph is below the $ y=cosx$

$V= \pi \int ^{\frac{\pi}{2}} _0 (cosx)^2 - (cos^2x)^2 \; dx$

$= \pi \int ^{\frac{\pi}{2}} _0 cos^2x - cos^4x \; dx$
$Now just use what you've proven before and integrate$

Cool, thanks!

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Trig Quadratic Identity & Volume Between Two Curves (1 Viewer)

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