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Trigonometry 14 Online
OpenStudy (anonymous):

find cos x/2 if tan x = -3/4

jimthompson5910 (jim_thompson5910):

Let's ignore the negative sign in -3/4 for now So if tan x = 3/4, then we have this right triangle |dw:1399252244613:dw|

jimthompson5910 (jim_thompson5910):

this is because tan(angle) = opposite/adjacent

jimthompson5910 (jim_thompson5910):

what is the hypotenuse of the right triangle? |dw:1399252296698:dw|

OpenStudy (anonymous):

hypotenuse is 5

jimthompson5910 (jim_thompson5910):

good

jimthompson5910 (jim_thompson5910):

|dw:1399252387972:dw|

jimthompson5910 (jim_thompson5910):

using this triangle, what are sin(x) and cos(x) ?

OpenStudy (anonymous):

sin is 3/5 and cos is 4/5

jimthompson5910 (jim_thompson5910):

both are correct

jimthompson5910 (jim_thompson5910):

To find cos(x/2), we now use the formula \[\Large \cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1+\cos(x)}{2}}\] So plug in cos(x) = 4/5 and simplify to get... \[\Large \cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1+\cos(x)}{2}}\] \[\Large \cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1+\frac{4}{5}}{2}}\] \[\Large \cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{\frac{5}{5}+\frac{4}{5}}{2}}\] \[\Large \cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{\frac{5+4}{5}}{2}}\] \[\Large \cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{\frac{9}{5}}{2}}\] \[\Large \cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{\frac{9}{5}}{\frac{2}{1}}}\] \[\Large \cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{9}{5}*\frac{1}{2}}\] \[\Large \cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{9}{10}}\] \[\Large \cos\left(\frac{x}{2}\right) = \pm\frac{\sqrt{9}}{\sqrt{10}}\] \[\Large \cos\left(\frac{x}{2}\right) = \pm\frac{3}{\sqrt{10}}\] \[\Large \cos\left(\frac{x}{2}\right) = \pm\frac{3*\sqrt{10}}{\sqrt{10}*\sqrt{10}}\] \[\Large \cos\left(\frac{x}{2}\right) = \pm\frac{3\sqrt{10}}{10}\]

jimthompson5910 (jim_thompson5910):

If you don't have to rationalize the denominator, then you can stop at \[\Large \cos\left(\frac{x}{2}\right) = \pm\frac{3}{\sqrt{10}}\]

jimthompson5910 (jim_thompson5910):

Now the problem is this: tan(x) = -3/4 so x is in quadrant 2 or quadrant 4 (this is where tan(x) is negative) if x is in quadrant 2, then 90 < x < 180 (degrees) cut each part in half to get 45 < x/2 < 90 so x/2 is in quadrant 1 if x is in quadrant 2 ---------------------------------------- if x is in quadrant 4, then 270 < x < 360 (degrees) cut each part in half to get 135 < x/2 < 180 so x/2 is in quadrant 2 if x is in quadrant 4 ---------------------------------------- so x/2 is either in quadrant 1 or quadrant 2 that means cos(x/2) is either positive or negative we don't have enough info to say one way or the other

jimthompson5910 (jim_thompson5910):

so that's why we are forced to keep the plus/minus (unless more info comes along)

OpenStudy (anonymous):

cool, thank you

jimthompson5910 (jim_thompson5910):

you're welcome

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