We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

#1**+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

#1**+5 **

Best Answer

$$\\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