Question 889064: Given tan theta = 8/5 and pi < theta < 3pi/2. Find the exact value of sin theta/2, cos theta/2, tan theta/2.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i'll use x instead of theta since it's easier to type.
your problem statement becomes:
Given tan x = 8/5 and pi < x < 3pi/2. Find the exact value of sin x/2, cos x/2, tan x/2.
pi < x < 3pi/2.
in degrees, that means 180 < x < 270.
that puts x in the third quadrant.
tan(x) in the third quadrant is positive, but x and y in the third quadrant are negative, so tan(x) in the third quadrant is equal to -8/-5.
the value of x in the first quadrant is arctan(8/5) = 57.99461679 degrees.
the value of x in the third quarter is 180 + 57.99461679 degrees which is equal to 237.99461679 degrees.
you can test if this angle is correct by taking the tangent of it.
you will get tan(x) = 1.6 which is equivalent to 8/5, or more correctly, -8/-5.
half of 237.99461679 degrees is 118.9973084 degrees is equal to x/2.
118.9973084 degrees is in the second quadrant.
use your calculator to get:
sin(x/2) = .8746424812...
cos(x/2) = -.4847685324...
tan(x/2) = -1.804247642...
if my assumptions are correct, these are the values we should get when we use the half angle formulas.
the hypotenuse of the triangle containing the reference angle of x is equal to sqrt(8^2+5^2) = sqrt(64+25) = sqrt(89)
so our triangle has:
an opposite side from angle x of -8
an adjacent side from angle x of -5
a hypotenuse of sqrt(89)
we get:
sin(x) = -8/sqrt(89) = -.847998304...
cos(x) = -5/sqrt(89) = -.52999894...
tan(x) = -8/-5 = 1.6
the half angle identify formulas are:
sin(x/2) = +/- sqrt((1-cos(x))/2)
cos(x/2) =+/- sqrt((1+cos(x))/2)
tan(x/2) = +/- sqrt((1-cos(x))/(1+cos(x)))
it's safe to say that, if x is in the third quadrant, than x/2 is in the second quadrant.
a quick test reveals this to be true.
180/2 = 90 which is the lowest value that x/2 will take and is on the border between the first quadrant and the second quadrant.
270/2 = 135 which is in the middle of the second quadrant.
if x is in the third quadrant, x/2 has to be in the second quadrant.
now the formula for sin(x/2) is equal to +/- sqrt((1-cos(x))/2)
that becomes:
sin(x/2) = +/- .8746424812...
since sine is positive in the second quarter, then sin(x/2) = .8746424812.
this is the same as we got up above, so we are doing ok so far.
now the formula for cos(x/2) is equal to +/- sqrt((1+cos(x))/2)
that becomes:
cos(x/2) = +/- .4847685324...
since cosine is negative in the second quarter, then cos(x/2) = -.4847685324...
since this is the same as we got earlier, this is ok as well.
now the formula for tan(x/2) is equal to +/- sqrt((1-cos(x))/(1+cos(x))).
since tan is negative in the second quarter, we'll be looking for - sqrt... and not plus.
that becomes:
tan(x/2) = -1.804247642...
once again, this agrees with what we got above, so the assumption is that we did this correctly.
your solutions are:
sin(x/2) = .8746424812...
cos(x/2) = -.4847685324...
tan(x/2) = -1.804247642...
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