Verify identity list steps. Cot(t)(1-cos^2(t))=cos(t)sin(t)

1 answer:

4 0

Remember: We have to work from either the LHS or the RHS.
(Left hand side or the Right hand side)

You should already know this:

$\huge{Cot(t) = \frac{1}{tan(t)} = \frac{1}{\frac{sin(t)}{cos(t)}} = 1\div \frac{sin(t)}{cos(t)} = 1\times \frac{cos(t)}{sin(t)}=\boxed{\frac{cos(t)}{sin(t)}}$

You should also know this:

$sin^2(t) + cos^2(t) = 1\\\\\boxed{sin^2(t)} = 1 - cos^2(t)$

So plugging in both of those into our identity, we get:

$\frac{cos(t)}{sin(t)}\cdot sin^2(t) = cos(t)\cdot sin(t)$

Simplify the denominator on the LHS (Left Hand Side)

We get:

$cos(t) \cdot sin(t) = cos(t) \cdot sin(t)$

LHS = RHS

Therefore, identity is verified.

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How do you do this question?

Ksivusya [100]

Answer:

V = (About) 22.2, Graph = First graph/Graph in the attachment

Step-by-step explanation:

Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.

$\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}$

The plus 1 in '1 + 2/x' is shifting this graph up from where it is rotating, but the negative 1 is subtracting the area between the y-axis and the shaded region, so that when it's flipped around, it becomes a washer.

$V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\$

$\int _1^3\left(1+\frac{2}{y}\right)^2dy=4\ln \left(3\right)+\frac{14}{3}, \int _1^31dy=2\\\\=> \pi \left(4\ln \left(3\right)+\frac{14}{3}-2\right)\\=> \pi \left(4\ln \left(3\right)+\frac{8}{3}\right)$

Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.