Hard Grade 12 Trig. Questions

[tahayassen]

Hard Grade 12 Trig. Questions :(

Homework Statement

http://img831.imageshack.us/img831/8389/daumequation13263429629.png

Homework Equations

No equations are required.

The Attempt at a Solution

http://img845.imageshack.us/img845/9519/34851996.png http://img692.imageshack.us/img692/1841/24101089m.png http://img694.imageshack.us/img694/6687/41337501.png http://img9.imageshack.us/img9/9702/91594220.png

 

[Mentallic]

1) Expand the RHS using sin(9x)=sin(5x+4x), then see where you can go from there.

2) $cos(\pi/6+\pi/4)=cos(10\pi/24)$ and you do know the double angle formulae.

3) sin(2x) = ?

4) There are common factors in the numerator and denominator that can be canceled out.

 

[chapstic]

This attempt to help is warned with its been quite some time since I've done trig such as this.

#1 should be easily solved using product of sums identity.

sin$\alpha$*cos$\beta$ = $\frac{1}{2}$[sin($\alpha$ + $\beta$) + sin($\alpha$ - $\beta$)]

#2 this one seems tricky, but some form of the sum or difference formula

 

[SammyS]

Homework Statement

http://img831.imageshack.us/img831/8389/daumequation13263429629.png

The Attempt at a Solution

http://img845.imageshack.us/img845/9519/34851996.png

You really shouldn't put so many problems in one thread. ... 2 at most.

For #1, you're almost there.

Next step:

$=\{2\sin(4x)\cos(4x)\}\sin(x)+2\cos^2(4x)\sin(x)-\sin(x)$​

Factor sin(x) out of the last two terms.

You have double angle forms for sine & for cosine .

Then it's just recognizing angle addition.

 

[Curious3141]

Homework Statement

http://img831.imageshack.us/img831/8389/daumequation13263429629.png

Homework Equations

No equations are required.

The Attempt at a Solution

http://img845.imageshack.us/img845/9519/34851996.png http://img692.imageshack.us/img692/1841/24101089m.png http://img694.imageshack.us/img694/6687/41337501.png http://img9.imageshack.us/img9/9702/91594220.png

For #3, you'd better make sure you've got the question exactly right, because there seem to be no real solutions for the equation you gave.

For #2, another way to do it is to see it as $\cos(\frac{\pi}{3} - \frac{\pi}{8})$, then use the angle sum and half-angle formulae to resolve it.

For #1 (it's like a countdown ), SammyS has already given you the killer hint.

 

[tahayassen]

You guys make these problems seem really easy...

It's almost like you guys do these for breakfast. xD

 

[Bread18]

Is the 3rd question right? Because sinxcosx =/= 1

 

[tahayassen]

Yes, the third question is copied correctly. I'll ask my teacher.

 

[Curious3141]

Yes, the third question is copied correctly. I'll ask my teacher.

Well, if you've learned a bit of calculus, try to find the minimum value of $f(x) = \cot x + \tan x$.

 

[tahayassen]

Well, if you've learned a bit of calculus, try to find the minimum value of $f(x) = \cot x + \tan x$.

I don't know Calculus, so I used a graphing calculator. It's negative infinity, right?

 

[Mentallic]

1) I always find these kinds of questions to be easier if you go from simple to more complicated, rather than the other way around (which doesn't apply in many other cases) because using trig sum formulae and such are always easier to expand than contract or factorize.

$$\sin(9x)$$
$$=\sin(5x+4x)$$
$$=\sin(5x)\cos(4x)+\cos(5x)\sin(4x)$$

Now the LHS is

$$2\sin(5x)\cos(4x)-\sin(x)$$

so cancelling out one of the $\sin(5x)\cos(4x)$ leaves us with the task of showing

$$\sin(5x)\cos(4x)-\sin(x)=\cos(5x)\sin(4x)$$

And if we take the $\sin(x)$ over to one side, we can see why

$$\sin(5x)\cos(4x)-\cos(5x)\sin(4x)=\sin(x)$$

Because of the difference formula for sines.


2) whichever approach you want to take, whether it be mine or curious' just depends on what you think you'll find easier. I personally find mine simpler (hence why I chose to give that as advice) but that doesn't mean you will

3) yes, there aren't any real solutions.

I don't know Calculus, so I used a graphing calculator. It's negative infinity, right?

Yes it is, but I think curious was meant to add "in the interval $[0,\pi]$"

 

[tahayassen]

Yes it is, but I think curious was meant to add "in the interval $[0,\pi]$"

Isn't it still negative infinity? At that interval, I see parabolas with a range of negative infinity to negative two, and parabolas with a range of two to positive infinity.

http://img339.imageshack.us/img339/2084/19143388.png

 

[Curious3141]

I don't know Calculus, so I used a graphing calculator. It's negative infinity, right?

As Mentallic said, I should've been more precise. OK, I meant the min. value of $f(x) = |\cot x + \tan x|$ for $x \in (0,2\pi)$.

That's 2. It should be obvious therefore that the left hand side of the equation you were given can never be equal to 1, not for real values of x anyway.

You will immediately see this when trying to arrive at a solution whether you use your method (which is perfectly valid) and you finish with: $\sin{x}\cos{x} = 1 \Rightarrow \frac{1}{2}\sin{2x} = 1 \Rightarrow \sin{2x} = 2$ which has no real solution for x, or if you used the other common method of letting $\tan x = m$, then setting up a quadratic in m, only to find that this has a negative discriminant (so again, no real solutions).

 

[Curious3141]

Isn't it still negative infinity? At that interval, I see parabolas with a range of negative infinity to negative two, and parabolas with a range of two to positive infinity.

http://img339.imageshack.us/img339/2084/19143388.png

Please see my edit. The absolute value of the LHS of the equation will always be $\geq 2$, so the expression can never be equal to 1. You see that "dead space" in the graph between -2 and 2 on the vertical axis? So the curve never "touches" 1 and your equation has no real solutions.

 

[Mentallic]

Isn't it still negative infinity?]

Ugh I meant $[0,\pi/2]$ :yuck:

If you arrive at a question that doesn't have any real solutions, at least simplify the expression to get it down to a form that you can easily argue why it has no real solutions because you can't just say because it looks like it on my graphing calculator, or start using calculus when you haven't learned it yet.

From $\sin(x)\cos(x)=1$ you only have one step to go.

 

[tahayassen]

Solve sin(x)cos(x) when x = pi/4 (the maximum for the sin function)

sin(pi/4)*cos(pi/4)
=1/2

Therefore, the maximum for sinxcosx is 0.5.

 

[tahayassen]

Actually, my brain just hatched an idea.

sin(x)cos(x)
=0.5*2*sin(x)cos(x)
=0.5*sin(2x)

Therefore, the maximum is 0.5.

 

[Mentallic]

Solve sin(x)cos(x) when x = pi/4 (the maximum for the sin function)

What do you mean the maximum for the sin function?

What does sin(2x) equal?

 

[Mentallic]

Actually, my brain just hatched an idea.

sin(x)cos(x)
=0.5*2*sin(x)cos(x)
=0.5*sin(2x)

Therefore, the maximum is 0.5.

There you go! That's much better

 

[tahayassen]

I just realized a mistake I made.

When I go from line 3 to 4 in question 3 in my original post, aren't I getting rid of a restriction?

Any insight?

Also, I asked my teacher about question 3. He said that if he thinks that the answer provides non-real roots, then he wants the non-real roots. How would I go about finding those?

 

[Mentallic]

I just realized a mistake I made.

When I go from line 3 to 4 in question 3 in my original post, aren't I getting rid of a restriction?

Any insight?

Yes, that $\sin(x)\cos(x)\neq 0$ but we've already shown that the solutions to this equation are complex, which aren't zero. As a note, you should always check that your solutions satisfy the original equation. It's easy to forget that the initial equation had divisions, or you squared the values along the way and your solution set might not satisfy the original solution set. Just be weary of this.

Also, I asked my teacher about question 3. He said that if he thinks that the answer provides non-real roots, then he wants the non-real roots. How would I go about finding those?

Have you been taught how to find these complex numbers? It's just as a heads up so we know how much detail to provide. It could be long and demanding, or it could be short and sweet.

 

[Curious3141]

I just realized a mistake I made.

When I go from line 3 to 4 in question 3 in my original post, aren't I getting rid of a restriction?

What do you mean by a "restriction"?

Also, I asked my teacher about question 3. He said that if he thinks that the answer provides non-real roots, then he wants the non-real roots. How would I go about finding those?

I would ask your teacher if he was kidding.

But if he's not, are you familiar with these, for starters?

$\sin ix = i\sinh x$

$\cos ix = \cosh x$

Because the easiest way to get there is to use those, and to use those, you need to know hyperbolic trig and inverse hyperbolic trig in addition to understanding complex numbers quite well. It's not going to be easy to make that "leap", if the current level of the questions you're being set is any indication.

At any rate, I get x to be: $\frac{\pi}{4} + i\ln(2+\sqrt{3})$ or $\frac{5\pi}{4} + i\ln(2+\sqrt{3})$ . That'll give you at least *some* of the solutions with the real part within the required range of [0, 2∏]. I'm pretty sure I'm missing some solutions. Complex trig can be very irritating that way. :tongue:

If you want to know how to get there, you really have to demonstrate at least basic knowledge of hyperbolic trig and complex numbers, as I mentioned.

 

[Mentallic]

At any rate, I get x to be: $\frac{(2k+1)\pi}{4} + i\ln(2+\sqrt{3})$

Tssk tssk, I expected more out of you! Clearly the answer should be

$$x\approx \frac{(2k+1)\frac{3141}{1000}}{4}+i\ln(2+\sqrt{3})$$

:tongue:

 

[Curious3141]

Tssk tssk, I expected more out of you! Clearly the answer should be

$$x\approx \frac{(2k+1)\frac{3141}{1000}}{4}+i\ln(2+\sqrt{3})$$

:tongue:

LOL, look the TS might think you were being serious! :rofl: Let's leave the "Almost ∏" jokes to our "secret forum" .

 

[Mentallic]

LOL, look the TS might think you were being serious! :rofl: Let's leave the "Almost ∏" jokes to our "secret forum" .

Oh wow, now I guess it's not so secret anymore, eh? Haha nah but seriously I never even considered that there would be secret areas to the forum until the day I was upgraded. It was a shock really!

And I agree with Curious here. It's much more likely that the question is flawed as opposed to expecting you to solve it, unless of course your year 12 is much different to what mine was.
If you don't know hyperbolic trig, just ask your teacher to solve it for you

 

[Curious3141]

Oh wow, now I guess it's not so secret anymore, eh? Haha nah but seriously I never even considered that there would be secret areas to the forum until the day I was upgraded. It was a shock really!

Well, I kind of figured there would be one for the Mentors and Admins (as I've been a mod on another forum that also had "secret" areas for staff), but didn't expect there'd be one here for HHs. Who knows how many other secrets lurk?

 

[Mentallic]

Well, I kind of figured there would be one for the Mentors and Admins (as I've been a mod on another forum that also had "secret" areas for staff), but didn't expect there'd be one here for HHs. Who knows how many other secrets lurk?

Are you trying to get me curious? :tongue:

By the way, tahayassen, sorry about spamming your help thread. We're still here for you!

 

[Curious3141]

Are you trying to get me curious? :tongue:

Are you trying to make me mental? :tongue2:

By the way, tahayassen, sorry about spamming your help thread. We're still here for you!

But I don't think he's here for us, not any more. That little green light next to his username has been off for a while. :zzz: I think we've scared him off.

Seriously, we'd better stop spamming in H/W threads. I'm stopping right...NOW.

 

[tahayassen]

Have you been taught how to find these complex numbers? It's just as a heads up so we know how much detail to provide. It could be long and demanding, or it could be short and sweet.

Nope. D:

I would ask your teacher if he was kidding.

But if he's not, are you familiar with these, for starters?

$\sin ix = i\sinh x$

$\cos ix = \cosh x$

Because the easiest way to get there is to use those, and to use those, you need to know hyperbolic trig and inverse hyperbolic trig in addition to understanding complex numbers quite well. It's not going to be easy to make that "leap", if the current level of the questions you're being set is any indication.

At any rate, I get x to be: $\frac{\pi}{4} + i\ln(2+\sqrt{3})$ or $\frac{5\pi}{4} + i\ln(2+\sqrt{3})$ . That'll give you at least *some* of the solutions with the real part within the required range of [0, 2∏]. I'm pretty sure I'm missing some solutions. Complex trig can be very irritating that way. :tongue:

If you want to know how to get there, you really have to demonstrate at least basic knowledge of hyperbolic trig and complex numbers, as I mentioned.

Nope, I'm not familiar with either.

Tssk tssk, I expected more out of you! Clearly the answer should be

$$x\approx \frac{(2k+1)\frac{3141}{1000}}{4}+i\ln(2+\sqrt{3})$$

:tongue:

I wish I could understand the joke.

Oh wow, now I guess it's not so secret anymore, eh? Haha nah but seriously I never even considered that there would be secret areas to the forum until the day I was upgraded. It was a shock really!

And I agree with Curious here. It's much more likely that the question is flawed as opposed to expecting you to solve it, unless of course your year 12 is much different to what mine was.
If you don't know hyperbolic trig, just ask your teacher to solve it for you

Well, I can't ask any teacher in my entire school for help, because apparently, these questions are going to make up our final exam.

They've given these and 12 other questions to the students for them to work on two weeks prior to the exam. They've said that you can use any resource available. The final exam is going to consist of 8 of these questions.

 

[SammyS]

Tssk tssk, I expected more out of you! Clearly the answer should be

$$x\approx \frac{(2k+1)\frac{3141}{1000}}{4}+i\ln(2+\sqrt{3})$$

:tongue:

...

I wish I could understand the joke.

The joke has to do with Curious3141's user name, the four digits at the end come from π.

 

[3.14159253589]

Homework Statement

http://img831.imageshack.us/img831/8389/daumequation13263429629.png

Homework Equations

No equations are required.

The Attempt at a Solution

http://img845.imageshack.us/img845/9519/34851996.png


http://img692.imageshack.us/img692/1841/24101089m.png


http://img694.imageshack.us/img694/6687/41337501.png


http://img9.imageshack.us/img9/9702/91594220.png

If sin(x)cos(x)=1, then x is not a real number. So are you looking for real numbers? If you are not, then you can use exponential forms of sin and cos.