Question

Problem Set 1, section 1A, problem 6b: Can someone please explain how it is determined that Acos(y)=1 and Asin(y)=-1? I didn't see any of this covered in any of the independent study lecture videos, so I'm assuming it's review material (if it is, I haven't seen it in a long time). Thanks.

Answer 1

can you state the full problem?

Answer 2

Express the following in the form A*sin(x+c):

sinx-cosx

Answer 3

From the Sum Identity
http://www.mathportal.org/algebra/trigonometry/trigonometric-formula.php
we know that

sin (x + y) = sin(x) cos(y) + cos(x) sin(y)

we can get a similar form as the required form
(substitute c = y and multiply both sides with A
A*sin (x + c) = A*sin(x) cos(c) + A*cos(x) sin(c)

we need the form sin(x)-cos(x)
A*sin (x + c) = A*sin(x) cos(c) + A*cos(x) sin(c) = sin(x)-cos(x)
Hence


as you can see from the drawing
wee need to solve the system of equations
A*cos(c) = 1
A*sin(c) = -1

(which explains your original question!)
This is equaivalent to
cos(c) = 1/A
sin(c) = -1/A

then
from the first we can see that A = 1/cos(c) and substitute in in the second
then we get:
sin(c)/cos(c) = -1 or sin(c) =- cos(c), and this is at c = -pi/4 or 3*pi/4
now
cos(-pi/4) = 1/sqrt(2) so A = sqrt(2), which has to check with
sin((-pi/4) =-/sqrt(2) so A = sqrt(2) indeed
So we have the first solution: sqrt(2) * sin(x-pi/4) = sin(x)-cos(x)

or
cos(3pi/4) = -1/sqrt(2) so A = -sqrt(2), which has to check with
sin((3pi/4) =1/sqrt(2) so A =-sqrt(2) indeed
So the second solution is -sqrt(2) * sin(x+3pi/4) = sin(x)-cos(x)

This can be shown in a graph

Answer 4

@ERMiller

Answer 5

Thanks for your help. I still don't see why A*cos(c) = 1 and A*sin(c) = -1. It seems like you arrived at that first and went from there. I understand the process once we determine A*cos(c) = 1 and A*sin(c) = -1, I just don't get how that is determined.

Answer 6

This is the case because of the general property
sin (x + y) = sin(x) cos(y) + cos(x) sin(y)

Do you know how that property? (Have you ever seen how that was proven?)

Answer 7

Anyway, if you use that 'law', you can see that if sin(x) - cos(x) is of the form
A*(sin(x) cos(y) + cos(x) sin(y)), it means you is equal to A*sin(x+y)

sin(x) - cos(x) is only of that form if
A*cos(y) = 1
and
A*sin(y) = 1

if you backsubstitute it into
A*(sin(x) cos(y) + cos(x) sin(y)) = A*cos(y) sin(x) +A*sin(y)cos(x)
we get the needed
sin(x)-cos(x)

For
A*cos(y) = 1
and
A*sin(y) = 1
to be true
then
y = -pi/4 and A = sqrt(2)
or
y = 3*pi/4 and A = -sqrt(2)

From the graph you can see that that is the right solution
(whether you use y or c makes no difference)

Answer 8

Here is a nice proof of the identities
like sin (x + y) = sin(x) cos(y) + cos(x) sin(y)
http://www.youtube.com/watch?v=zcEMKv5yIYs

Answer 9

Oh, ok - I had never seen it proven despite being aware of the property. Thanks so much for taking all that time to explain it. I really appreciate it.

Answer 10

It's a good repetition for me too, I hadn't looked at the proof since secondary school andymore.

These and more trigonometry identities are used very often. It often helps to just think "Is there any one of those I can use?"

Answer 11

A*cos(y) = 1 and A*sin(y) = 1 just fits the bill