- #1

Vital

- 108

- 4

## Homework Statement

Hello!

I am doing exercises on sinusoid functions from the beginning of Trigonometry.

I hoped I understood the topic, but it seems not quite, because I don't get the results authors show as examples for one of possible answers, as there can be a few answers to the same exercise. I will be grateful for your help and explanation on what I am doing wrong.

Here is the task:

Show that the function is a sinusoid by rewriting it in the forms C(x) = A cos(ωx + φ) + B and S(x) = A sin(ωx + φ) + B for ω > 0.

## Homework Equations

I will show one example, and, maybe, later add more, if it is necessary.

f(x) = 3√3 sin(3x) - 3cos(3x)

Here is the answer from the book, and below I post how I tried to solve the task:

answer from the book f(x) = 3√3 sin(3x) - 3cos(3x) = 6 sin( 3x + (11π/6) ) = 6cos( 3x + (4π/3) )

## The Attempt at a Solution

My solution:

f(x) = 3√3 sin(3x) - 3cos(3x)

(1) rewrite the expression in the sinusoid form:

f(x) = A sin(3x) cos(φ) - A cos(3x) sin(φ)

Clearly, B = 0, ω = 3.

(2) Find coefficients:

3√3 = A cos(φ)

3 = A sin(φ)

Given Pythagorean identity we have:

cos

^{2}+ sin

^{2}= 1

Multiply both sides by A

^{2}:

A

^{2}cos

^{2}+ A

^{2}sin

^{2}= A

^{2}

3√3

^{2}+ 3

^{2}= 36

Choose A = 6 (choosing positive 6)

(3) Find φ (works both ways - take cos or sin):

3√3 = A cos(φ)

3√3 = 6 cos(φ)

φ = π/6

(4) Solution:

f(x) = 3√3 sin(3x) - 3cos(3x) in sin form S(x) = A sin(ωx + φ) + B

f(x) = 6 sin(3x + π/6); the book gives this: f(x) = 6 sin( 3x + (11π/6) ) which is not the same:

if we add π/6, both cos and sin have positive values at this angle, while at +11π/6 sin is negative. If, however, it were -11π/6, then we would have ended at the same place as if we added π/6 with both cos and sin positive.

I don't see my mistakes in computations, and why my answer is wrong; and I don't see how authors achieved +11π/6 as the value of φ.

To find the same formula for cos I used cofunction identities:

sin(θ) = cos (π/2 - θ)

In my case θ = 3x + π/6, so cos ( π/2 - 3x - π/6) = cos ( - (3x - π/3) ) = cos (3x - π/3) which is not the same as cos( 3x + (4π/3) ) for same reasons stated above in discussion on sin. -π/3 has a positive cos value (quadrant IV), while +4π/3 has a negative cos value (quadrant III).

Thank you very much!