andycampbell1
May3-11, 05:34 AM
1. The problem statement, all variables and given/known data
Hi
I am doing a maths past paper and I would like some help to understand where part of the solution came from. The part I do not understand is the bit at the end which says tan\alpha = 5\pi/4. It would be great if someone could help me understand where this came from.
The question is
y = sin t - cos t in the form A cos(\omegat+\alpha)
2. Relevant equations
3. The attempt at a solution
The solution is
y=sin t -cos t = A cos(\omegat+\alpha)
= A cos t cos \alpha - A sin t sin \alpha
A cos \alpha = -1
- A sin \alpha = 1
Squaring the above equations and adding side by side implies that A = \sqrt{2}
Dividing the above equations side by side gives
tan \alpha= 1
tan is positive in the first and third quadrant. We choose the third quadrant since from the above equation sin and cos are negative.
Hence \alpha = \pitan-1 = 5\pi/4
y= sin t - cos t = \sqrt{2}cos(t+ 5\pi/4)
What I do not understand is where did 5\pi come from? I have checked my notes and it does not mention anything about where pi comes from.
Hi
I am doing a maths past paper and I would like some help to understand where part of the solution came from. The part I do not understand is the bit at the end which says tan\alpha = 5\pi/4. It would be great if someone could help me understand where this came from.
The question is
y = sin t - cos t in the form A cos(\omegat+\alpha)
2. Relevant equations
3. The attempt at a solution
The solution is
y=sin t -cos t = A cos(\omegat+\alpha)
= A cos t cos \alpha - A sin t sin \alpha
A cos \alpha = -1
- A sin \alpha = 1
Squaring the above equations and adding side by side implies that A = \sqrt{2}
Dividing the above equations side by side gives
tan \alpha= 1
tan is positive in the first and third quadrant. We choose the third quadrant since from the above equation sin and cos are negative.
Hence \alpha = \pitan-1 = 5\pi/4
y= sin t - cos t = \sqrt{2}cos(t+ 5\pi/4)
What I do not understand is where did 5\pi come from? I have checked my notes and it does not mention anything about where pi comes from.