# Hard Grade 12 Trig. Questions

1. Jan 11, 2012

### tahayassen

Hard Grade 12 Trig. Questions :(

1. The problem statement, all variables and given/known data

http://img831.imageshack.us/img831/8389/daumequation13263429629.png [Broken]

2. Relevant equations

No equations are required.

3. The attempt at a solution

http://img845.imageshack.us/img845/9519/34851996.png [Broken]

http://img692.imageshack.us/img692/1841/24101089m.png [Broken]

http://img694.imageshack.us/img694/6687/41337501.png [Broken]

http://img9.imageshack.us/img9/9702/91594220.png [Broken]

Last edited by a moderator: May 5, 2017
2. Jan 11, 2012

### Mentallic

Re: Hard Grade 12 Trig. Questions :(

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 cancelled out.

3. Jan 11, 2012

### chapstic

Re: Hard Grade 12 Trig. Questions :(

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

Last edited: Jan 11, 2012
4. Jan 11, 2012

### SammyS

Staff Emeritus
Re: Hard Grade 12 Trig. Questions :(

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.

Last edited by a moderator: May 5, 2017
5. Jan 11, 2012

### Curious3141

Re: Hard Grade 12 Trig. Questions :(

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.

Last edited by a moderator: May 5, 2017
6. Jan 11, 2012

### tahayassen

Re: Hard Grade 12 Trig. Questions :(

You guys make these problems seem really easy...

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

7. Jan 12, 2012

Re: Hard Grade 12 Trig. Questions :(

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

8. Jan 12, 2012

### tahayassen

Re: Hard Grade 12 Trig. Questions :(

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

9. Jan 12, 2012

### Curious3141

Re: Hard Grade 12 Trig. Questions :(

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

10. Jan 12, 2012

### tahayassen

Re: Hard Grade 12 Trig. Questions :(

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

11. Jan 12, 2012

### Mentallic

Re: Hard Grade 12 Trig. Questions :(

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

12. Jan 12, 2012

### tahayassen

Re: Hard Grade 12 Trig. Questions :(

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 [Broken]

Last edited by a moderator: May 5, 2017
13. Jan 12, 2012

### Curious3141

Re: Hard Grade 12 Trig. Questions :(

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).

14. Jan 12, 2012

### Curious3141

Re: Hard Grade 12 Trig. Questions :(

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.

Last edited by a moderator: May 5, 2017
15. Jan 12, 2012

### Mentallic

Re: Hard Grade 12 Trig. Questions :(

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 learnt it yet.

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

16. Jan 12, 2012

### tahayassen

Re: Hard Grade 12 Trig. Questions :(

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.

17. Jan 12, 2012

### tahayassen

Re: Hard Grade 12 Trig. Questions :(

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.

18. Jan 12, 2012

### Mentallic

Re: Hard Grade 12 Trig. Questions :(

What do you mean the maximum for the sin function?

What does sin(2x) equal?

19. Jan 12, 2012

### Mentallic

Re: Hard Grade 12 Trig. Questions :(

There you go! That's much better

20. Jan 13, 2012

### tahayassen

Re: Hard Grade 12 Trig. Questions :(

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?

Last edited: Jan 13, 2012