Thread: 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 as a cosine with a phase displacement."

My response:



"Draw an appropriate sketch on your paper and use it to write as a cosine with a phase displacement."

My response:



"Use the composite argument property to write as a linear combination of cos and sin ."

My response: (gave up)

This site might help you
Precalculus Resources - Keymath.com

I have not looked at the first or second problem. But regarding the third:

"Use the composite argument property to write as a linear combination of cos and sin ."

My response: (gave up)

Use the cosine of a difference formula:

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 ... 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."

Originally Posted by wiseguy

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:

There's no need to take the arccosine, because the problem didn't ask for an angle.

"Show that and 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 and give the same degree measure for angle B."

Same thing as above. What are the equivalent fractions for 0.96 and 0.28?

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 :/

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.

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.

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.

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
. . .as a cosine with a phase displacement.

Divide by 5: . .[1]


Let be in this right triangle:

Code:

                        *
                     *  *
             5    *     *
               *        * 3
            *           *
         * θ            *
      * - - - - - - - - *
                4

Then: .


Substitute into [1]: .

. . . . . . . . . . . . . . .

. . . . . . . . . . . . . . .

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: . .[1]


Let be in this right triangle:

Code:

                        *
                     *  *
             5    *     *
               *        * 3
            *           *
         * θ            *
      * - - - - - - - - *
                4

Then: .


Substitute into [1]: .

. . . . . . . . . . . . . . .

. . . . . . . . . . . . . . .