Question 567944: To solve for all angles (theta) for which sin(theta) = -cos(theta), do I use the inverse function or the reciprocal?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! To solve for all angles (theta) for which sin(theta) = -cos(theta), do I use the inverse function or the reciprocal?
--------------
Do it like this:
sin = -cos
sin^2 = cos^2 = 1 - sin^2
2sin^2 = 1
sin^2 = 1/2
theta = pi/4, 3pi/4, 5pi/4, 7pi/4 + 2n*pi, n = 0,1,2,3...
Then eliminate Q1 & Q3 where the signs are the same.
----
--> theta = 3pi/4 + n*pi, n = 0,1,2,3...
or theta = 135 + n*180 degs, n = 0,1,2,3...
====================
A simpler approach:
sin = -cos
Divide by cos
tan = -1
|
|
|
