Trig graphs help

[james hoffa]

I have three questions in my book about the composite argument property that I'm not entirely sure about


"On your paper, draw an appropriate sketch and use it to write [MATH]y=4cosx+3sinx[/MATH] as a cosine with a phase displacement."

My response:

y=5cos(x)-0.64350169

"Draw an appropriate sketch on your paper and use it to write [MATH]y=-3cosx-1sinx[/MATH] as a cosine with a phase displacement."

My response:

y=-3.1622777cos?-3.4633437

 

[Deleted member 4993]

I have three questions in my book about the composite argument property that I'm not entirely sure about


"On your paper, draw an appropriate sketch and use it to write [MATH]y=4cosx+3sinx[/MATH] as a cosine with a phase displacement."

y = 4 cos(x) + 3 sin(x)

let cos(?) = 0.8 and sin(?) = 0.6, then

y = 5 * cos(x - ?)

My response:

y=5cos(x)-0.64350169

"Draw an appropriate sketch on your paper and use it to write [MATH]y=-3cosx-1sinx[/MATH] as a cosine with a phase displacement."

My response:

y=-3.1622777cos?-3.4633437

follow the method shown above

 

[soroban]

Hello, james hoffa!

\(\displaystyle \text{Write }\:y \:=\:4\cos x\,+\,3\sin x\,\text{ as a cosine with a phase displacement.}\)

\(\displaystyle \text{My response: }\;y\:=\:5\cos(x -0.643501109)\) . . Correct!

Since you got the right answer, I assume you know the procedure.
. . But I'll review it anyway.


\(\displaystyle \text{We have: }\;y \;=\;4\cos x + 3\sin x\)

\(\displaystyle \text{Divide by 5: }\;\frac{y}{5} \;=\;\frac{4}{5}\cos x + \frac{3}{5}\sin x\;\;{\bf[1]}\)


\(\displaystyle \text{Let }\theta\text{ be an angle in this right triangle:}\)

                        *
                     *  *
            5     *   - *
               *        * 3
            *           *
         * @            *
      *  *  *  *  *  *  *
               4

\(\displaystyle \text{Hence: }\;\cos\theta = \frac{4}{5},\;\;\sin\theta = \frac{3}{5} \quad\hdots\;\text{ and: }\:\theta = \arccos\tfrac{4}{5}\)


\(\displaystyle \text{Substitute into }\bf{[1]}:\) . \(\displaystyle \frac{y}{5} \;=\;\cos\theta\cos x + \sin\theta\sin x\)

. . . .\(\displaystyle \text{And we have: }\;\frac{y}{5} \;=\;\cos(x - \theta) \quad\Rightarrow\quad y \;=\;5\cos(x - \theta)\)


\(\displaystyle \text{Therefore: }\;y \;=\;5\cos\left(x - \arccos \tfrac{4}{5}\right)\)

 

[BigGlenntheHeavy]

\(\displaystyle Another \ way, \ to \ wit: \ y \ = \ 4cos(x)+3sin(x), \ Let \ y \ = \ bcos(x-a)\)

\(\displaystyle Then \ bcos(x-a) \ = \ b[cos(x)cos(a)+sin(x)sin(a)] \ = \ bcos(a)cos(x)+bsin(a)sin(x)\)

\(\displaystyle Hence, \ 4 \ = \ bcos(a) \ and \ 3 \ = \ bsin(a)\)

\(\displaystyle Squaring \ and \ adding \ the \ above \ two \ equations, \ we \ get:\)

\(\displaystyle b^2[cos^2(a)+sin^2(a)] \ = \ 25, \ \implies \ b \ = \ 5, \ cos(a) \ = \ 4/5 \ and\)

\(\displaystyle sin(a) \ = \ 3/5 \ (first \ quadrant), \ a \ = \ arcsin(3/5) \ = \ arccos (4/5)\)

\(\displaystyle Hence, \ putting \ it \ all \ together, \ we \ get;\)

\(\displaystyle y \ = \ 4cos(x)+3sin(x) \ = \ bcos(x-a) \ = \ 5cos[x-arccos(4/5)]\)

 

[BigGlenntheHeavy]

\(\displaystyle Second \ one.\)

\(\displaystyle y \ = \ -3cos(x)-sin(x)\)

\(\displaystyle Let \ y \ = \ bcos(x+a) \ = \ bcos(a)cos(x)-bsin(a)sin(x)\)

\(\displaystyle Hence, \ bcos(a) \ = \ -3 \ and \ bsin(a) \ = \ 1\)

\(\displaystyle As \ before, \ squaring \ and \ adding \ gives \ b^2[cos^2(a)+sin^2(a)] \ = \ 10, \ \implies \ b \ = \ \sqrt{10}\)

\(\displaystyle Ergo, \ cos(a) \ = \ \frac{-3}{\sqrt{10}} \ and \ sin(\pi+a) \ = \ \frac{-1}{\sqrt{10}}, \ third \ quadrant\)

\(\displaystyle Hence, \ y \ = \ -3cos(x)-sin(x) \ = \ bcos(x+a) \ = \ \sqrt{10}cos\bigg[x+arccos\bigg(\frac{-3}{\sqrt{10}}\bigg)\bigg]\)