Simplify (1 − cos x)(1 + cos x). 1 cos2 x sin2 x tan2 x

1 answer:

3 0

Answer

The expression is equivalent to sin^{2}x

The first step is to multiply the two expressions between parentheses :
(II)
There is a trigonometric identity that states :

Working with this expression :
⇒
(I)
Using the equation (I) in (II) :
⇒

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Make a change of coordinates:

$u(x,y)=xy$
$v(x,y)=\dfrac xy$

The Jacobian for this transformation is

$\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}$

and has a determinant of

$\det\mathbf J=-\dfrac{2x}y$

Note that we need to use the Jacobian in the other direction; that is, we've computed

$\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}$

but we need the Jacobian determinant for the reverse transformation (from $(x,y)$ to $(u,v)$ . To do this, notice that

$\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}$

we need to take the reciprocal of the Jacobian above.

The integral then changes to

$\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv$
$=\displaystyle\frac12\int_{v=}^{v=}\int_{u=}^{u=}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2$

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ExtremeBDS [4]

Alright here it is:

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I need help to these two questions please

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

If you're asking me grams specifically, and not kilograms, subtract 68 from 75 to get your answer. If it's only kilograms subtract 324 from 405. If it's both of them added together subtract 399 from 473, you're welcome!

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3 years ago

What's the correct answer for this?

11Alexandr11 [23.1K]

$\bf \qquad \qquad \textit{inverse proportional variation}\\\\
\textit{\underline{y} varies inversely with \underline{x}}\qquad \qquad y=\cfrac{k}{x}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\$

$\bf \textit{\underline{N} is inversely proportional to \underline{A}}\qquad N=\cfrac{k}{A}
\\\\\\
\textit{we also know that }
\begin{cases}
A=3\\
N=16
\end{cases}\implies 16=\cfrac{k}{3}\implies 16\cdot 3=k
\\\\\\
48=k\qquad thus\qquad \boxed{N=\cfrac{48}{A}}\\\\
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