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3 years ago

Verify the identity: 2cosx(sin3x-sinx) = sin4x

1 answer:

8 0

$\bf \qquad \qquad \textit{triple-angle identity}\\\\
sin(3\theta)=3sin(\theta)-4sin^3(\theta)\\\\
\left. \qquad \qquad \right.\textit{quad-angle identity}\\\\
sin(4\theta)=
\begin{cases}
8sin(\theta)cos^3(\theta)=4sin(\theta)cos(\theta)\\\\
\boxed{4sin(\theta)cos(\theta)-8sin^3(\theta)cos(\theta)}
\end{cases}\\\\
-----------------------------\\\\
2cos(x)[sin(3x)-sin(x)]=sin(4x)\\\\
-----------------------------\\\\$

$\bf 2cos(x)[sin(3x)-sin(x)]\implies 2cos(x)[3sin(x)-4sin^3(x)-sin(x)]
\\\\\\
2cos(x)[2sin(x)-4sin^3(x)]\implies \boxed{4sin(x)cos(x)-8sin^3(x)cos(x)}$

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Evaluate the integral e^xy w region d xy=1, xy=4, x/y=1, x/y=2

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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

$\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv$
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3 years ago

Determine which equations below, when combined with the equation 3x-4y=2, would form a system with no solutions. Choose all that

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3 years ago

Read 2 more answers

Mr. Gutierrez had $100 to purchase candy for his students that completed their work. He

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Answer:

$21.50

Step-by-step explanation:

Mr. Gutierrez had $100 to purchase candy for his students that completed their work.

He bought 8 bags of jolly ranchers

Each bag of jolly ranchers cost $3.25.

Hence, the cost of 8 bags of jolly ranchers = 8 × $3.25

= $26

He also bought 25 bags of assorted chocolate and each bag of assorted chocolate cost $2.10.

Hence, the cost of 25 bags of assorted chocolates = 25 × $2.10

= $52.5

Therefore, the amount of money Mr. Gutierrez has left over after these purchases is calculated as:

Total amount - Sum of ( Cost of 8 bags of jolly ranchers + 25 bags of assorted chocolates)

= $100 - ( $26 + $52.5)

= $100 - $78.50

= $21.50

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3 years ago

A bag of marbles contains 3 green marbles, 7 red marbles, 2 white

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A bread recipe calls for 2 cups of wheat flour and 3 1/4 cups of white flour. How much flour does the recipe call for altogether

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Answer:

Step-by-step explanation:

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