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Mar 1 2014 07:09pm
if cos2x = -1/49, find tanx for -3pi/2 < x < -pi

tips for any helpful replies, thanks :D
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Mar 1 2014 07:13pm
Quote (togepi @ Mar 1 2014 08:09pm)
if cos2x = -1/49, find tanx for -3pi/2 < x < -pi

tips for any helpful replies, thanks :D


u could find the angle.. using cos inverse and dividing 2. and plug the angle u got into tan and see if its in between those ranges

This post was edited by FamilyGuyViewer on Mar 1 2014 07:14pm
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Mar 2 2014 04:01am
cos(2x) = 2.cos(x)² - 1 = 1 - 2.sin(x)²

Given that cos(2x) = -1/49, plug in this value into the 2 formulas and solve for cos(x) and for sin(x), using the fact that -3pi/2 < x < -pi to help choosing between negative or positive square roots.

Then, form the quotient sin(x) / cos(x).
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Mar 2 2014 01:07pm
Quote (FamilyGuyViewer @ Mar 1 2014 06:13pm)
u could find the angle.. using cos inverse and dividing 2. and plug the angle u got  into tan and see if its in between those ranges


Quote (feanur @ Mar 2 2014 03:01am)
cos(2x) = 2.cos(x)² - 1 = 1 - 2.sin(x)²

Given that cos(2x) = -1/49, plug in this value into the 2 formulas and solve for cos(x) and for sin(x), using the fact that -3pi/2 < x < -pi to help choosing between negative or positive square roots.

Then, form the quotient sin(x) / cos(x).


ty both, that helped
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