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3 years ago

How to express cos 3x in terms of cosx

1 answer:

Elena-2011 [213]3 years ago

5 0

$\cos3x=\cos(x+2x)=\cos x\cos2x-\sin x\sin2x$
$=\cos x(\cos^2x-\sin^2x)-2\sin^2x\cos x$
$=\cos^3x-\cos x\sin^2x-2\sin^2x\cos x$
$=\cos^3x-3\cos x\sin^2x$
$=\cos^3x-3\cos x(1-\cos^2x)$
$=\cos^3x+3\cos^3x-3\cos x$
$=4\cos^3x-3\cos x$

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Answer:

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Step-by-step explanation:

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2 years ago

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jeyben [28]

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$\displaystyle2\pi\int_R^{R+2r}2xy\,\mathrm dx$

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So the volume is

$\displaystyle4\pi\int_R^{R+2r}x\sqrt{r^2-(x-(R+r))^2}\,\mathrm dx$

Substitute

$x-(R+r)=r\sin t\implies\mathrm dx=r\cos t\,\mathrm dt$

and the integral becomes

$\displaystyle4\pi r^2\int_{-\pi/2}^{\pi/2}(R+r+r\sin t)\cos^2t\,\mathrm dt$

Notice that $\sin t\cos^2t$ is an odd function, so the integral over $\left[-\frac\pi2,\frac\pi2\right]$ is 0. This leaves us with

$\displaystyle4\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}\cos^2t\,\mathrm dt$

Write

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3 years ago

Which logarithmic equation is equivalent to this exponential equation? <br><br> 12=5^x

Brut [27]

Given:

The equation is:

$12=5^x$

To find:

The logarithmic equation that is equivalent to the given exponential equation.

Solution:

According to the property of logarithm:

$a=b^x\Leftrightarrow \log_b a=x$

We have,

$12=5^x$

Here, $a=12,b=5$ . By using the above property of logarithm, we get

$\log_{5}12=x$

$x=\log_{5}12$

Therefore, the correct option is C.

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3 years ago