trig identity

I have the following problem:
y = -sqrt(12)*sin t - 2*cos t

and I'm putting it into the
following form:

y = C*sin(t + a)
Where a is my phase shift.

C = sqrt([-sqrt(12)]^2 + (-2)^2)
C = 4

cos a = -sqrt(12)/4 = -sqrt(3)/2
sin a = -2/4 = -1/2

This would mean our reference angle
lies in quadrant three. We also
know that the adjacent side is
-sqrt(3) and that the opposite side
is -1. This tells us that we have a
30:60:90 triangle as our reference
triangle. This would tell us then
that the angle is 180 + 30 or
210 degrees or 7pi/6 radians.

Which gives us y = 4*sin(t + 7pi/6)

But my book says y = 4*sin(t + 4pi/3)

This would be valid only if our
adjacent side was 1 unit
and out opposite side was sqrt(3)
units, producing a 60 degree angle
or pi/3 radians.

Am I missing something or is
my book simply mistaken?

--
conrad

.

Relevant Pages

  • Re: Putnam Exam 2004 -- [*SPOILERS*]
    ... the area and side lengths of a triangle satisfy ... > then that will always be opposite an acute angle. ...
    (sci.math)
  • Re: trig identity
    ... This would mean our reference angle ... -sqrtand that the opposite side ... 30:60:90 triangle as our reference ...
    (sci.math)
  • Re: Calculate Angle and Length in Triangle
    ... triangle, where Angle1 is opposite Side1 etc ... If angles A & B are not internal angles opposite sides a & b adapt as ... PS the above formula validates Dave's "would be" triangles. ... The angle provided is not the angle of a corner. ...
    (microsoft.public.excel.programming)
  • Re: Most efficient way to learn/review/remember math?
    ... At this point I draw an equilateral triangle. ... Each angle is 2pi/3. ... Side length 1 is opposite the right angle. ... Sin = Opposite over Hypotenuse. ...
    (sci.math)
  • Re: irreducible polynomial in integrand
    ... AB and BC of a reqular pentagon. ... triangle whose apex angle is 108 degrees and whose other angles are ... degrees and apex angle 108 degrees. ... has base angles 36degrees and apex angle 108 degrees. ...
    (sci.math)