Question

Express the following in the form : y = A sin (bx+c)

sin x - cos x

Answer 1

So you'll want to get rid of the cos x in your expression. Manipulating the identity cos^2 x + sin^2 x = 1,\[\cos x = \pm{\sqrt{1-\sin^2x}}\]
Therefore,\[\sin x - \cos x = \sin x - (\pm{\sqrt{1-\sin^2x}})\]
Next you'll want to employ double-angle formulas, as you can see from the form sin (bx+c).\[\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\]...Hmm. I think I'll post this and let others add their ideas.

Answer 2

I may be going in the wrong direction, but an idea I have is that the original problem will need to be manipulated using trig identities before double-angle formula can be utilized.

Answer 3

I thought perhaps to divide out sin x so that we are left with sin x (1- (cos x / sin x)), but I am not sure that is an identity, going back to my notes.

Answer 4

cos x/sin x = cot x

Answer 5

So you would end up with 1-cot x

Answer 6

\[\cos \theta = \sin \left( \pi/2-\theta \right)\]
that is probably a more useful identity than leaving myself with 1-cot x

Answer 7

And hey! sin(pi/2 - theta) is in (bx+c) format

Answer 8

\[\sin \theta - \cos \theta = \sin \theta - \sin \left( \pi/2-\theta \right)\]

Answer 9

Hum. Ho hum. You can't factor out the sine function...

Answer 10

Unless... OK, let's set sin(pi/2 - theta) =\[\sin(\pi/2)\cos(\theta)-\sin(\theta)\cos(\pi/2)\]\[\cos(\theta)*1-\sin(\theta)*0 = \cos(\theta)\]... = sin theta - cos theta.

Answer 11

Blah.

Answer 12

I feel like it is halfway there ... because I have two equations that I can manipulate into that format required and then it seems that the answer is Equation 1 - Equation 2:

\[eq 1. y= 1 \sin (1x+0)\]
\[eq 2. y = 1 \sin (-1x+\pi/2)\]

Answer 13

The black line is the sin x - cos x, and the red line is 1 sin (1x+0)... so there is an observable increase in amplitude, a observable change in phase to the right (i.e. -c/b is >0) and possibly no change in period, so b is likely 1... at least from the observation of the graphs.

Answer 14

My head hurts.

Answer 15

c appears to be -pi/4... through trial and error on the graphs...

Answer 16

yeah - You have been a trooper - I feel like progress was made, but it is still eluding me. I will have to sleep on it - thanks a bunch for your help.

Answer 17

trying to figure out how to do this with algebra. it is well known (more or less) that
\[\sin(x)+\cos(x)=\sqrt{2} \sin(x+\frac{\pi}{4})\] but i cannot remember the proof. could mimic it for yours. i'll try google

Answer 18

if someone has a good way to explain this that would be great, my method is guess and check-ish

Answer 19

i got \[\sqrt2\sin(x+ \frac{7\pi}{4}) \]

Answer 20

if you just want the answer the formula for linear combinations is here
http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Linear_combinations

Answer 21

oh wow thats extremely useful o.O i didnt know about that

Answer 22

maybe joemath has the proof of these formulas but i am sure they are derived in a standard text, which i seem not to have

Answer 23

but if you just want to use the formula you get
\[\sqrt{a^2+b^2}=\sqrt{2}\]

Answer 24

and
\[\sin^{-1}(-\frac{1}{\sqrt{2}})=-\frac{\pi}{4}\]

Answer 25

so answer is
\[\sqrt{2}\sin(x-\frac{\pi}{4})\]

Answer 26

Of no immediate help to me because I can't get there, Prof Jerison lists the answer as :
\[\sqrt{2}\sin(x-\pi/4)\]

So observationally, there is the phase shift of -pi/4 and some amplitude A greater than 1... with no change in period. I have to go, so will check in tonight. Great thinking on this combination.

Answer 27

formula gives exactly what you wrote

Answer 28

We have three inknowns here: A, b and c. Tangent had the right idea with
sin(u + v) in terms of the trig functioons of of u and of v. Here it's
A(sin(bx) cos c - cos(bx) sin c) = sin x - cos x for all x.
(A cos c) sin(bx) - (A sin c) cos x = sin x - cos x. This forces b = 1.
We now have A cos c sin x - A sin c cos x = sin x - cos x.
Set corresponding coefficients equal:
A cos c = 1
-A sin c = -1.
-Asin c/(Acos c) = -1/1
tan c = 1, so c = pi/4.
Squaring both sides of both equations and adding gives us
A^2 (cosx)^2 + b^2(sin x)^2 = 2 ++>a = sqrt(2).
[So sin x - cos x = sqrt(2)sin(x + pi/4).

Answer 29

\[\sqrt{2}\sin(x + \pi/4).\]

Answer 30

The only problem with that answer is when we check against the original equation it doesn't add up. c = -pi/4 will work
\[\sqrt{2}\sin(x - \pi/4)\]works.