# Rewrite the given function as a sinusoid - why so different

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1. Apr 29, 2017

### Vital

1. The problem statement, all variables and given/known data
Hello!
I am at the inverse trigonometric functions section, and the following exercise asks to rewrite the given function as a sinusoid of a form S(x) = A sin(ωx + φ).

I thought I have understood the approach to solving such tasks, and it went pretty smoothly, until I hit the two functions below. Please, take a look at them, and help me to understand why in the first case we set φ = π + arcsin(some_value) and in the other we set φ = arcsin(-some_value), if in both cases initially we have arcsin(-some_value), namely the some_value is negative in both cases, so in both cases sin(φ) has negative values. I give answers to these questions below each function, but I am not sure those are correct answers.

2. Relevant equations
Here are two functions, my solutions and answers from the book. Please, help me to understand the difference.

3. The attempt at a solution
First function:

f(x) = -cos(x) - 2√2 sin(x)

-1 = A sin(φ) and - 2√2 = A cos(φ)

cos2(φ) + sin2(φ) = 1
multiply both sides by A2
A2 cos2(φ) + A2 sin2(φ) = A2

(-1) 2 + (- 2√2)2 = A2
A = 3 (taking the positive answer from the square root)

Then, -1 = A sin(φ) => sin(φ) = -⅓ => arcsin(-⅓) = φ

f(x) = 3 sin(x + π + arcsin (⅓) ) = 3 sin(x + 3.4814)

Before moving to the next example, I would like to note that I understand that if we find arcsin(⅓) we get a reference angle with sin value is Quadrant I or Quadrant II, where sin is positive; and as far as in the given function we have a negative value of sin(φ), then by adding arcsin(⅓) to π we get to the desired Quadrant III, where sin(φ) has negative value. Then, do I understand correctly that the answer:

f(x) = 3 sin(x + -arcsin (-⅓) ) = 3 sin(x - 0.3398) is also correct, but gives us the angle in Quadrant IV, instead of Quadrant III as in the previous answer?

Second function:
f(x) = 2sin(x) - cos(x)
2 = A cos(φ) and -1 = A sin(φ)
A2 cos2(φ) + A2 sin2(φ) = A2
A = √5
φ = arcsin(-1/√5) = arcsin(-√5/5)
f(x) = √5 sin( x + arcsin(-√5/5)) = √5 sin( x - 0.4636)

And here we could also give another answer:
f(x) = √5 sin( x + π + arcsin(√5/5)) = √5 sin( x + 3.6052), correct?

Thank you very much!

2. Apr 29, 2017

This item comes up quite often in optics and interference and other places and the function takes the form $F(t)=Acos(\omega t)+Bsin(\omega t)$. By factoring out $\sqrt{A^2+B^2}$, the result is $F(t)=\sqrt{A^2+B^2}(cos(\phi)cos(\omega t)+sin(\phi)sin(\omega t))=\sqrt{A^2+B^2}cos(\omega t-\phi)$ where $\phi=arctan(\frac{B}{A})$. Since $sin(\theta+\pi/2)=cos(\theta)$, we can write $F(t)=\sqrt{A^2+B^2}sin(\omega t-\phi+\pi/2)$.

3. May 2, 2017

### Vital

Thank you very much, but you are not answering my questions. Even though your answer helped me to learn something new, it doesn't help me to understand what is going on with the exercises. I have precise questions about the difference in the answer I get and the answer given in the book.

4. May 2, 2017

### FactChecker

When you decided to use the positive square root for A=3, you picked one of two possibilities. The negative root would have given the book answer. When you have a choice like that, it does not mean that both choices are equally correct. It just means that you should consider both possibilities and see which one (or both) work out. You need to check your final answer. Your final answer with A=3 does not match the original function. You should have also considered A=-3 and you would have gotten a match to the original function.

EDIT: Sorry, I forgot that A is used other places (where it should be positive?). So I will need to think about this some more.

Last edited: May 2, 2017
5. May 2, 2017

@FactChecker gave a good response=even my method isn't foolproof if you simply use the formula $\phi=arctan(B/A)$. If both $A$ and $B$ are negative (as it was for this problem), then $\phi$ lies in the 3rd quadrant instead of the first=(you add $\pi$ to the angle in the first quadrant that has $arctan(B/A)$).

Last edited: May 2, 2017
6. May 2, 2017

Scratch that=I need to check my answer. Editing... Yes, I get the result also that $f(x)=3sin(x+3.4814)$.

Last edited: May 2, 2017
7. May 2, 2017

### Ray Vickson

You need both $\sin(\phi) = -1/3$ and $\cos(\phi) = -2\sqrt{2}/3$, so $\phi$ must be in the third quadrant. After that, you can get a unique positive value $\phi = \pi + \arctan(\sqrt{2}/4)$. You made an error when you claimed that $3 \sin(x - \arcsin(-1/3))$ is also correct: it is not, and neither is $3 \sin(x+\arcsin(-1/3)).$

Last edited: May 2, 2017