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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 as a cosine with a phase displacement."
"Draw an appropriate sketch on your paper and use it to write as a cosine with a phase displacement."
"Use the composite argument property to write as a linear combination of cos and sin ."
My response: (gave up)
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
...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."
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.
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 and give the same degree measure for angle A."
Same thing as above. What are the equivalent fractions for 0.96 and 0.28?"Show that both and give the same degree measure for angle B."
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.
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: . .
Let be in this right triangle:
Code:* * * 5 * * * * 3 * * * θ * * - - - - - - - - * 4
Substitute into : .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
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:
Turns out it's exactly the same as yours