Often when dealing in
trigonometry, you have an
algebraic equation such as:
sinΘ = √3 / 2
Where you want to find
Θ.
Obviously:
Θ = sin
^{1}(√2 / 2)
Θ = sin
^{1}(0.70710678)
Θ = 45
^{o} or π/4
^{c} depending on your
unit of angular measure
For now I will deal in
radians (
superscript c) not
degrees (superscript o), but I will leave the information for degrees in for those of you who prefer them.:
Suppose we had:
sinΘ = √2 / 2
for 0
> Θ
> 2π
We are presented with a problem, how do we find the other values of Θ for which sinΘ = 0?
Sine, Cosine and Tangent are repeating waves! All we need to know is the
period, which is 2π
^{c} (360
^{o}) for sine and cosine, and π
^{c} (180
^{o}) for tangent and the shapes of the
graphs. There is an excellent node in
Trigonometry with the graphs.
A simplistic version for sine follows:
 _
 / \

 \_/

The
horizontal axis shows 2π, so we know that for sinΘ > 0 Θ = sin
^{1}Θ
and there will be
two values of Θ, both before π
^{c}
There are similar rules for all the trigonometric functions, but there is also an easy way to remember them:
pi/2
sin \  / all
\  /
\  /
\  /
\  /
\  /
\  /
pi / 4 rad _\  /
/ \  /_ pi / 4 rad
/ \  / \
pi \/ 0









tan  cos
3pi/2
The angles marked by circle sections are equal to that found by the inverse trigonometric function, in this case sin
^{1}. They drawn from the closest horizontal in the appropriate
quadrant. In "All" all results are positive, so if you have a
positive sine cos or tan, you draw a line in "All". The other quadrants work the same for specific functions.
This graph is between 0 and 2π
^{c} (360
^{o}), with each quadrant equal to π/2
^{c} (90
^{o}). To find the value of a line you've drawn, work out the angle from the starting position, (marked as 0) here. This is a
simple matter of
arithmatic. Say we have π/4 as our result (even though it would be exactly between the horizontal and the vertical, it is hard to draw), then we have found:
sin π/4 = sin ( π  π/4 ) = √3 / 2
Which is true!
Now, what if we have something nastier, like:
tanΘ = 0.5
Well, tan
^{1}0.5 = 0.463647609
^{c}
Remember, tan is
NEGATIVE here
pi/2
sin \  all
\ 
\ 
\ 
\ 
\ 
\ 
_\ 
0.463rad / \ 
/ \ 
pi \ 0
\ 
 \ / 0.463rad
 \ /
 \/
 \
 \
 \
 \
 \
tan  \ cos
3pi/2
so, from the starting position again, we see that
tan(π  0.463) = 0.5
tan(2π  0.463) = 0.5
So once again, we have our results.
A final exercise!
cos
^{2}Θ = 0.5
(cosΘ)
^{2} = 0.5
cosΘ = √(0.5)
Θ = cos
^{1}( √0.5 )
Θ = cos
^{1}(
±0.7071067812 )
Plus or Minus! This means, we have to draw lines in
all 4 quadrants.
pi/2
sin \  / all
\  /
\  /
\  /
\  /
\  /
\  /
pi / 4 rad _\  /
/ \  /_ pi / 4 rad
/ \  / \
pi \/ 0
 /\ 
pi / 4 rad \/  \ / pi / 4 rad
/  \ _/
/  \
/  \
/  \
/  \
/  \
/  \
tan /  \ cos
3pi/4
so:
cos
^{2}(π / 4) = 0.5
cos
^{2}((&pi  (π / 4)) = 0.5
cos
^{2}((&pi + (π / 4)) = 0.5
cos
^{2}((2&pi  (π / 4)) = 0.5
One final thing I haven't addressed, how do you remember the order of the quadrants? Either use the graphs to work it out, memorise them, or use a
mnemonic, such as:
The one I use:
All Stations To Crew
for the less politically correct:
All Sailors Take Cock
SamuraiSatan says The preferred mnemonic device used in my precalculus class is All students take calculus