Question #75094

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Question #75094

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Answer 1 · 2017-04-16T23:39:46

$csc (2 arctan (\frac{3}{4})) = \frac{25}{24}$ $csc(2arctan(3/4))=25/24$

Explanation:

Because $sin$ $sin$ and $csc$ $csc$ are reciprocals:

$csc (2 arctan (\frac{3}{4})) = \frac{1}{sin (2 arctan (\frac{3}{4}))}$ $csc(2arctan(3/4))=1/sin(2arctan(3/4))$

Using the double angle identity $sin (2 θ) = 2 sin (θ) cos (θ)$ $sin(2theta)=2sin(theta)cos(theta)$ :

$= \frac{1}{2 sin (arctan (\frac{3}{4})) cos (arctan (\frac{3}{4}))}$ $=1/(2color(blue)(sin(arctan(3/4)))color(red)(cos(arctan(3/4)))$

We can find the values of $sin (arctan (\frac{3}{4}))$ $sin(arctan(3/4))$ and $cos (arctan (\frac{3}{4}))$ $cos(arctan(3/4))$ using a similar method.

Note that when $θ = arctan (\frac{3}{4})$ $theta=arctan(3/4)$ , then $tan (θ) = \frac{3}{4}$ $tan(theta)=3/4$ . That is, where $θ$ $theta$ is an angle in a right triangle, the side opposite $θ$ $theta$ is $3$ $3$ and the leg adjacent to $θ$ $theta$ is $4$ $4$ . The Pythagorean theorem tells us that the hypotenuse is $5$ $5$ .

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We then see that:

$sin (arctan (\frac{3}{4})) = sin (θ) = \frac{opposite}{hypotenuse} = \frac{3}{5}$ $color(blue)(sin(arctan(3/4)))=sin(theta)="opposite"/"hypotenuse"=3/5$

$cos (arctan (\frac{3}{4})) = cos (θ) = \frac{adjacent}{hypotenuse} = \frac{4}{5}$ $color(red)(cos(arctan(3/4)))=cos(theta)="adjacent"/"hypotenuse"=4/5$

So the original expression is:

$= \frac{1}{2 (\frac{3}{5}) (\frac{4}{5})}$ $=1/(2color(blue)((3/5))color(red)((4/5)))$

$= \frac{25}{24}$ $=25/24$

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Question #75094

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