Question

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?

Answer 1

`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.

Final Answer