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# Simplify the following expression: cos(acos(2/x)-asec(x))

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Simplify the following expression: cos(acos(2/x)-asec(x))

Guest Dec 5, 2014

#1
+92806
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$$\\cos(acos(2/x)-asec(x))\\\\ =cos(acos(2/x)-acos(1/x)) \\\\ Let\;\;\theta_1=acos(2/x)\qquad and \qquad \theta_2=acos(1/x)\\\\ =cos(\theta_1-\theta_2) \\\\ = cos(\theta_1)cos(\theta_2)+sin(\theta_1)sin(\theta_2)\\\\$$

etc

Now you have the cos values

To get the sine values I would draw a right angled tiangle for each of them that is theta1 and theta 2

I'd mark the adj and the hypotenuse according to the given ratios and then I'd find the opposite sides using phythagoras' theorem.

You can read the sin values off the triangles.

If you understand all this then you can finish it yourself.

Think about it if you need more help post again and I can explain better/more.

Melody  Dec 5, 2014
#1
+92806
+5

$$\\cos(acos(2/x)-asec(x))\\\\ =cos(acos(2/x)-acos(1/x)) \\\\ Let\;\;\theta_1=acos(2/x)\qquad and \qquad \theta_2=acos(1/x)\\\\ =cos(\theta_1-\theta_2) \\\\ = cos(\theta_1)cos(\theta_2)+sin(\theta_1)sin(\theta_2)\\\\$$

etc

Now you have the cos values

To get the sine values I would draw a right angled tiangle for each of them that is theta1 and theta 2

I'd mark the adj and the hypotenuse according to the given ratios and then I'd find the opposite sides using phythagoras' theorem.

You can read the sin values off the triangles.

If you understand all this then you can finish it yourself.

Think about it if you need more help post again and I can explain better/more.

Melody  Dec 5, 2014