Trig: put 2sinx-cosx in Asin(x+a) form, find max value,....

sqleung

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Hello. I'm in urgent need of help for this question in my trigonometry assignment. I have done some questions but this one completely stumped me. I need exact answers and I need them in radian form as well. There are three parts to this question:

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Given f(x) = 2sin x - cos x

1: Write f(x) in the form Asin(x + a)

So far, for this one, I expanded Asin(x + a) to form:
Asinxcosa + Acosxsina

From here, I managed to make out that:
Acos a = 2
Asin a = -1

So I get sin a / cos a = -1/2
That means tan a = -1/2

Now I'm stuck. I need my answers in EXACT form (eg. using irrational numbers). I also need my answers in radian form so I'm not exactly sure how to go about this.

2: Find the maximum value of 2sin x - cos x and the corresponding values of x.

This one I cannot do because I haven't managed to do the first one. Also, I'm rather confused on the wording of this problem.

3: Find the least positive value of p such that f(x) = f(x + p)

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Thankyou. Of course, any help will be appreciated. You don't have to do them all. I just need someone to "shed some light" for me because I'm really stuck.
 
sqleung said:
Given f(x) = 2sin x - cos x

1: Write f(x) in the form Asin(x + a)
I agree with your equalities. You got as far as tan(a) = -1/2. Use this to draw a right triangle, and find the length of the hypotenuse. This will allow you to determine that A = sqrt[5]. :wink:

As for the angle measure, this is, as you note, not one of the "nice" angle measures. Are you sure that you have to have the measure itself, rather than just using something like "arctan(-1/2)"...? Unless I'm reading Wikipedia wrong, I think you're kind of stuck with the inverse tangent. Sorry! :oops:

Eliz.
 
If you are required to provide exact values, simply provide them.

\(\displaystyle 2\sin(x)-cos(x)=\sqrt{5}\sin\left(x+atan\left(\frac{1}{2}\right)\right)\)

Done. Why does there need to be more.

On the other one, f(x) = f(x+p). By itself, I agree this may sound evasive or unclear, but if you read the definition of the "period" of a function, you should find very similar language. This should clear up your question. In case the definition is hiding, the question is asking for the period of the new function, which should be the same as the period of the old function.
 
I know this question has been posted 5 days ago but I'm struggling on it again. Sorry for troubling you guys but you've all been very helpful but this time, it's really urgent... :D

2: Find the maximum value of 2sin x - cos x and the corresponding values of x.

Okay, using my calculator, I managed to get square root 5 as the maximum value. However, I'm stuck on getting the corresponding values of x. Is there an equation I have to solve? Could you please provide some sort of starting point (or show me how to do it)?

3: Find the least positive value of p such that f(x) = f(x + p)

For this one, I really don't know what to do. I tried a couple of things. First, I tried:

2sinx - cosx = f(x + p)
= 2sin(x + p) - cos (x + p)
= 2(sinxcosp + cosxsinp) - (cosxcosp - sinxsinp)
= 2sinx - cosx

Perhaps I could incorporate my method above with something? I also found the period of the equation to be 2pi. However, I'm not too certain that this is the actual answer (there's got to be more to it than just 2pi)...:?

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Thankyou, all help is appreciated. Sorry for all the trouble. :(
 
2) The maximum value of the sine function always is associated with the argument at pi/2.

x + atan(1/2) = pi/2 -- Solve for x.

3) Part of the idea is to know when you are done and to release prejudices. If you prove it, and you still don't believe it, you need to fix your thinking, not your proof.

f(x) = f(x+p) where p is the minimum possible value for this property is the definition of "period". You cannot find the period and not find p. You cannot find p and not find the period. They are the same thing.

When faced with this, y = a*sin(b(x-c)), what is the period of this function? If b = 1, what is the period of this function?

What say you?
 
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