# Math Help - Composite argument property and linear combination of cosx/sinx

1. ## Composite argument property and linear combination of cosx/sinx

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 $y=4cosx+3sinx$ 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 $y=-3cosx-1sinx$ as a cosine with a phase displacement."

My response:

$y=-3.1622777cosθ-3.4633437$

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

My response: (gave up)

Precalculus Resources - Keymath.com

3. I have not looked at the first or second problem. But regarding the third:
"Use the composite argument property to write $y=7cos(x-20°)$ as a linear combination of cos and sin ."

My response: (gave up)
Use the cosine of a difference formula:
\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}

"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 ... 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."

5. Originally Posted by wiseguy

"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:
\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 $cos^{-1} 0.6$ and $sin^{-1} 0.8$ 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 $cos^{-1} 0.96$ and $sin^{-1} 0.28$ give the same degree measure for angle B."
Same thing as above. What are the equivalent fractions for 0.96 and 0.28?

6. 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 :/

7. 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.

8. Originally Posted by eumyang
...
What are the equivalent fractions for 0.96 and 0.28?
0.96=3/5 and 0.28=4/5, correct me if I'm wrong.

9. Originally Posted by wiseguy
0.96=3/5 and 0.28=4/5, correct me if I'm wrong.
No. 0.96 = 24/25 and 0.28 = 7/25.

10. 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
. . $y\:=\:4\cos x+3\sin x$ .as a cosine with a phase displacement.

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

Let $\theta$ be in this right triangle:

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

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

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

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

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

11. EDIT: thanks allot for the help on this topic. It's really great to get help from the experts over the net.

**
Thanks for the help!

I ended up figuring it out by putting the original equation in my calculator, taking the first maximum value, then using the y value as my new amplitude and the x value as the phase displacement. I ended up with this equation:

y=5cos(x-0.64350423)

Turns out it's exactly the same as yours

Originally Posted by Soroban
Hello, wiseguy!

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

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

Let $\theta$ be in this right triangle:

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

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

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

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

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