OHHHH NOOOOO!!!!
I sure hope you don't think that
cos(A+B) = cosA + cosB !!!!!!
and what is with this 2cos theta = sin theta??
You mean like 2cos60º = sin60º
or 2(1/2) = .866 ??
We are dealing with a phase shift here, namely a shift of cos(theta) by pi/2 to the left.
So visualize cos(theta) moved 90º to the left, would you not have the standard sine curve reflected in the x-axis?
So it would be -sin(theta) which is choice D
or you can do it by using the proper expansion of cos(A+B) which is
cosAcosB - SinAsinB
cos (theta + pi/2)
= cos(theta)cos(pi/2) - sin(theta)sin(pi/2)
but cos(pi/2) = 1 and sin(pi/2) = 0
which leaves
cos(theta)cos(pi/2) - sin(theta)sin(pi/2)
= cos(theta)*0 - sin(theta)*1
= -sin(theta)
Which expression equivalent to: cos(theta + pi/2).
A)cos theta
B)-cos theta
C)sin theta
D)-sin theta
my book doesnt give examples of this but my crack at it would be C b/c distributive property? and 2cos theta = sin theta
3 answers
Good job Reiny.
Guido
Guido
Didn't know I was being graded, lol!
On the other hand, just give me a raise.
On the other hand, just give me a raise.