# Thread: Trig Question - Urgent

1. ## Trig Question - Urgent

Sorry for bothering you again but this is rather (very?) urgent. I have one question that I want to ask and if you could help me, it would be greatly appreciated. Also, would you be able to do them in radian measure and give the exact answers (eg. without decimals).:

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Given f(x) = 2sinx - cosx

Write in the form of Asin(x + a)
I've managed to do this one with the answer being:
square root 5 (x + arctan -0.5)

Find the maximum value of 2sinx - cosx and the corresponding values of x

Okay, I've found that the maximum value is square root 5 (using my trusty calculator) but I'm quite stuck on finding the corresponding values of x. Could you perhaps provide a starting point of some sort or guide me through this one?

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

Now this is the one I need urgent help on. I have absolutely no idea how to go about this one. I assume the p would mean the period. Do I simply find the period of square root 5 (x + arctan -0.5)?

I tried some working out but am uncertain about it:

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

?!

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Thankyou VERY much guys. All help is definitely appreciated.

2. $\sqrt5\sin(x+\tan^{-1}\frac{-1}{2})$

The max of the sine function occurs when argument = Pi

so setting the argument of the function equal to pi will give the max

$\pi=x+\tan^{-1}\frac{-1}{2}$

3. I'm not too sure I understand what you mean.

I got the maximum value of my equation as square root 5 and I figured this out by graphing the equation onto my calculator and using a feature which helps me find the maximum point of the graph which was square root 5.

Sorry for the trouble.

4. Given f(x) = 2sinx - cosx

Write in the form of Asin(x + a)
I've managed to do this one with the answer being:
square root 5 (x + arctan -0.5)
This should be $\sqrt {5} \sin (x+ \arctan (-.5))
$
. You probably knew that already but I thought I'd point it out because you had it written without the sin twice in your post.

The max of the sine function occurs when argument = Pi
I think TheEmptySet meant $\frac {\pi}{2}$

Do I simply find the period of square root 5 (x + arctan -0.5)?
Yes.

5. So, for:

Find the maximum value of 2sinx - cosx and the corresponding values of x

I was wondering, where did you get the pi/2 from and how does it help me find the corresponding values of x? Is there some equation I have to eventually solve?

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

I found the period of square root 5 sin(x + arctan -0.5)

But however, that doesn't really seem like the answer. The least positive value of p? I think there's more to it than that...

6. Originally Posted by sqleung
So, for:

Find the maximum value of 2sinx - cosx and the corresponding values of x

I was wondering, where did you get the pi/2 from and how does it help me find the corresponding values of x? Is there some equation I have to eventually solve?

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

I found the period of square root 5 sin(x + arctan -0.5)

But however, that doesn't really seem like the answer. The least positive value of p? I think there's more to it than that...
No there's not. It's that simple.

The maximum value of $\sqrt {5} \sin (x+ \arctan (-0.5))$ is clearly $\sqrt{5}$. You get this value when the sine function reaches its maximum value of 1. This happens when what you're 'sining' is equal to pi/2. So the maximum value of $\sqrt{5}$ occurs when $x + \arctan(-0.5) = \frac{\pi}{2}$.

7. So I get:

x = (pi/2) - (arctan -0.5)

And for all values of x, it'll be:

x = (pi/2) - (arctan -0.5) + 2kpi where k is an integer?

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I'm still not understanding fully how you got pi/2 though. So care to shed some light?

Sorry for being such a pain.

Thankyou.

8. Originally Posted by sqleung
So I get:

x = (pi/2) - (arctan -0.5)

And for all values of x, it'll be:

x = (pi/2) - (arctan -0.5) + 2kpi where k is an integer?

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I'm still not understanding fully how you got pi/2 though. So care to shed some light?

Sorry for being such a pain.

Thankyou.