Trig graphs help

james hoffa

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Jul 22, 2010
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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
 
james hoffa said:
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
 
Hello, james hoffa!

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

My response:   y=5cos(x0.643501109)\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.


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

Divide by 5:   y5  =  45cosx+35sinx    [1]\displaystyle \text{Divide by 5: }\;\frac{y}{5} \;=\;\frac{4}{5}\cos x + \frac{3}{5}\sin x\;\;{\bf[1]}


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


Code:
                        *
                     *  *
            5     *   - *
               *        * 3
            *           *
         * @            *
      *  *  *  *  *  *  *
               4

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


Substitute into [1]:\displaystyle \text{Substitute into }\bf{[1]}: . y5  =  cosθcosx+sinθsinx\displaystyle \frac{y}{5} \;=\;\cos\theta\cos x + \sin\theta\sin x

. . . .And we have:   y5  =  cos(xθ)y  =  5cos(xθ)\displaystyle \text{And we have: }\;\frac{y}{5} \;=\;\cos(x - \theta) \quad\Rightarrow\quad y \;=\;5\cos(x - \theta)


Therefore:   y  =  5cos(xarccos45)\displaystyle \text{Therefore: }\;y \;=\;5\cos\left(x - \arccos \tfrac{4}{5}\right)

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

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

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

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

b2[cos2(a)+sin2(a)] = 25,      b = 5, cos(a) = 4/5 and\displaystyle b^2[cos^2(a)+sin^2(a)] \ = \ 25, \ \implies \ b \ = \ 5, \ cos(a) \ = \ 4/5 \ and

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

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

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

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

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

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

As before, squaring and adding gives b2[cos2(a)+sin2(a)] = 10,      b = 10\displaystyle As \ before, \ squaring \ and \ adding \ gives \ b^2[cos^2(a)+sin^2(a)] \ = \ 10, \ \implies \ b \ = \ \sqrt{10}

Ergo, cos(a) = 310 and sin(π+a) = 110, third quadrant\displaystyle Ergo, \ cos(a) \ = \ \frac{-3}{\sqrt{10}} \ and \ sin(\pi+a) \ = \ \frac{-1}{\sqrt{10}}, \ third \ quadrant

Hence, y = 3cos(x)sin(x) = bcos(x+a) = 10cos[x+arccos(310)]\displaystyle Hence, \ y \ = \ -3cos(x)-sin(x) \ = \ bcos(x+a) \ = \ \sqrt{10}cos\bigg[x+arccos\bigg(\frac{-3}{\sqrt{10}}\bigg)\bigg]
 
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