SOLVED Composite argument property and linear combination of cosx/sinx

Jul 2010
101
0
I have three questions in my book about the composite argument property that I'm not entirely sure about (Thinking)


"On your paper, draw an appropriate sketch and use it to write \(\displaystyle y=4cosx+3sinx\) as a cosine with a phase displacement."

My response:

\(\displaystyle y=5cos(x)-0.64350169
\)

"Draw an appropriate sketch on your paper and use it to write \(\displaystyle y=-3cosx-1sinx\) as a cosine with a phase displacement."

My response:

\(\displaystyle y=-3.1622777cosθ-3.4633437\)

"Use the composite argument property to write \(\displaystyle y=7cos(x-20°)\) as a linear combination of cos and sin ."

My response: (gave up)
 
Jan 2010
278
138
I have not looked at the first or second problem. But regarding the third:
"Use the composite argument property to write \(\displaystyle y=7cos(x-20°)\) as a linear combination of cos and sin ."

My response: (gave up)
Use the cosine of a difference formula:
\(\displaystyle \begin{aligned}
y &= 7 cos (x - 20°) \\
&= 7[(cos x)(cos 20°) + (sin x)(sin 20°)] \\
&= 7(cos x)(cos 20°) + 7(sin x)(sin 20°) \\
&= (7cos 20°)cos x + (7sin 20°)sin x \\
&\approx 6.578cos x + 2.394sin x \\
\end{aligned}\)
 
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Jul 2010
101
0
Two more roadblocks...

I have two more questions about this section that are a little shady -

"Suppose that cos A = 0.6 and sin A = 0.8. Suppose that cos B = 0.96 and sin B = 0.28. Use the composite argument property to find cos (A + B)."

I responded with
cos (A − B) = cos A cos B + sin A sin B
cos(A-B)=0.6*0.96+0.8*0.28
cps(A-B)=0.576+0.224
cos(A+B)=0.129024

arccos0.129024=A+B
A+B=1.441411637

...I'm pretty sure this is correct, but a double check would be greatly appreciated :D... so far so good, until-

"Show that cos -10.6 and sin-10.8 give the same degree measure for angle A."

How am I suppose to use the "composite argument property" to answer this question? (If I'm suppose to use it, it does not say so but it is in the same section...) All I did was punch both of these in my calculator to prove that they have the same degree measure. Surely I cannot put this as an answer, my instructor will think I'm being a smart ---. Here's another just like it-

"Show that both cos-1 0.96 and sin -1 0.28 give the same degree measure for angle B."
(Surprised)
 
Last edited:
Jan 2010
278
138
I have two more questions about this section that are a little shady -

"Suppose that cos A = 0.6 and sin A = 0.8. Suppose that cos B = 0.96 and sin B = 0.28. Use the composite argument property to find cos (A + B)."

I responded with
cos (A − B) = cos A cos B + sin A sin B
cos(A-B)=0.6*0.96+0.8*0.28
cps(A-B)=0.576+0.224
cos(A+B)=0.129024

arccos0.129024=A+B
A+B=1.441411637
Be careful with your signs. Does the original problem say to find cos (A+B) or cos (A-B)? Because in your work you started with cos (A-B). Also, you multiplied in the final step instead of added.

If the problem is to find cos (A+B), then:
\(\displaystyle \begin{aligned}
cos\,(A + B) &= cos\,A\,cos\,B - sin\,A\,sin\,B \\
&= (0.6)(0.96) - (0.8)(0.28) \\
&= 0.352
\end{aligned}\)
There's no need to take the arccosine, because the problem didn't ask for an angle.

"Show that \(\displaystyle cos^{-1} 0.6\) and \(\displaystyle sin^{-1} 0.8\) give the same degree measure for angle A."
I don't think you need to use the sum/difference identities. Maybe draw a triangle? 0.6 = 3/5 and 0.8 = 4/5. You'll get a 3-4-5 triangle, and you can see that angle A is the same.

"Show that both \(\displaystyle cos^{-1} 0.96\) and \(\displaystyle sin^{-1} 0.28\) give the same degree measure for angle B."
Same thing as above. What are the equivalent fractions for 0.96 and 0.28?
 
Last edited:
Jul 2010
101
0
Yes, I double checked it, the problem does say cos(A+B). I copied the first line from a formula in my text book and wasn't being careful :/
 
Jul 2010
101
0
I'm pretty sure the two questions about showing that arccos0.6 and arcsin0.8 give the same degree measure are simply just hitting them into the calculator. I don't see any other way to do this.

Edit: The triangle idea is a good one. I didn't want to just put calculator answers for my teacher.
 

Soroban

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Hello, wiseguy!

I believe this is what they expected for the first one . . .


On your paper, draw an appropriate sketch ? and use it to write
. . \(\displaystyle y\:=\:4\cos x+3\sin x\) .as a cosine with a phase displacement.

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


Let \(\displaystyle \theta\) be in this right triangle:

Code:
                        *
                     *  *
             5    *     *
               *        * 3
            *           *
         * θ            *
      * - - - - - - - - *
                4

Then: .\(\displaystyle \cos\theta = \dfrac{4}{5},\;\sin\theta = \dfrac{3}{5}\)


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

. . . . . . . . . . . . . . .\(\displaystyle \dfrac{y}{5} \;=\;\cos(x - \theta)\)

. . . . . . . . . . . . . . .\(\displaystyle y \;=\;5\cos\left(x - \arcsin\frac{3}{5}\right)\)