How do you find the exact value of the five remaining trigonometric function given #cottheta=1/2# and the angle is in standard position in quadrant III?

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John D. Share
Feb 9, 2018

Answer:

#sintheta = -(2sqrt5)/5 " "" "" "costheta = -sqrt5/5#

#csctheta = -sqrt5/2" "" "" "color(white)"x"sectheta = -sqrt5#

#tantheta = 2" "" "" "" "" "color(white)".-"cottheta = 1/2#

Explanation:

First, let's find #tantheta#. We know that it will be #1/cottheta#, so:

#color(red)(tantheta) = 1/cottheta = 1/(1/2) = color(red)2#

Now, using the identities #tantheta = sintheta/costheta# and #sin^2theta + cos^2theta = 1#, we can find the cosine:

#tantheta = 2#

#sintheta/costheta = 2#

#sqrt(sin^2theta)/costheta = 2#

#sqrt(sin^2theta+cos^2theta - cos^2theta)/costheta = 2#

#sqrt(1 - cos^2theta) = 2costheta#

#1 - cos^2theta = 4cos^2theta#

#1 = 5cos^2theta#

#1/5 = cos^2theta#

#+-sqrt5/5 = costheta#

Now, since the angle is in Q3 where both coordinates are negative, we can say that #costheta# is negative, so:

#color(red)(costheta = -sqrt5/5#

To find the sine, remember the identity #sin^2theta+cos^2theta = 1#.

#sin^2theta + (-sqrt5/5)^2 = 1#

#sin^2theta + 1/5 = 1#

#sin^2theta = 4/5#

#sintheta = +- (2sqrt5)/5#

Again, both coordinates are negative, so sine must be negative.

#color(red)(sintheta = -(2sqrt5)/5#

Remember that #csctheta = 1/sintheta#, and #sectheta = 1/costheta#

#color(red)(csctheta) = 1/sintheta = 1/(-(2sqrt5)/5) = color(red)(-sqrt5/2#

#color(red)(sectheta) = 1/costheta = 1/(-sqrt5/5) = color(red)(-sqrt5#

The values of all five remaining trig function have been found at this point, and their solutions are highlighted above in red.

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