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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Greatness on March 19, 2011, 08:00:54 pm

Title: Swarles' questions
Post by: Greatness on March 19, 2011, 08:00:54 pm: 1) If sin(pi/10) = (sqrt(5)-1)/4, then what is the exact value of sec(pi/5)
I didnt really know how to do this but i got an annswer of (sqrt(5)-1)....

2) simplify: ((sin(3x)/sin(x)) + ((cos(3x)/cos(x))
Again not sure how to do this, i tried using compound angle formulas and cancel stuff out but not sure if my answers are right

Thanks :)
Title: Re: Swarles' questions
Post by: kamil9876 on March 19, 2011, 08:23:59 pm: 1) you can use the double angle formula to find $cos(\pi/5)=cos(2\pi/10)$ and then use sec=1/cos.

2) what makes you unsure?
Title: Re: Swarles' questions
Post by: Greatness on March 19, 2011, 08:32:31 pm: 1) First i found cos(pi/10) by using the trig identity then used double angle formula to solve for cos(pi/5) then took the reciprocal of that.

2) I'm unsure because the working is quite long but i changed the initial eqn to this: ((sin(2x+x)/sin(x)) + ((cos(2x+x)/cos(x)) and did some cancelling and got an answer of sin(x)cos(x) + 2cos(2x)
i dont know if that will simplfify any further.
Title: Re: Swarles' questions
Post by: luffy on March 20, 2011, 03:39:23 pm: Quote from: swarley on March 19, 2011, 08:32:31 pm
1) First i found cos(pi/10) by using the trig identity then used double angle formula to solve for cos(pi/5) then took the reciprocal of that.

2) I'm unsure because the working is quite long but i changed the initial eqn to this: ((sin(2x+x)/sin(x)) + ((cos(2x+x)/cos(x)) and did some cancelling and got an answer of sin(x)cos(x) + 2cos(2x)
i dont know if that will simplfify any further.

Below is how I would do it: (Apologies if I made any errors)

1) There is probably a MUCH easier way to do this. ( I stress the word MUCH)

$\sin \frac{\pi}{10} = \frac{\sqrt{5} - 1}{4}$

$\sin^2 \frac{\pi}{10} + \cos^2 \frac{\pi}{10} = 1$

$\cos^2 \frac{\pi}{10} = 1 - \sin^2 \frac{\pi}{10}$

$\therefore \cos^2 \frac{\pi}{10} = 1 - (\frac{\sqrt{5} - 1}{4})^2$

$\therefore \cos^2 \frac{\pi}{10} = 1 - \frac{6 - 2\sqrt{5}}{16}$

$\therefore \cos^2 \frac{\pi}{10} = 1 - \frac{3 - \sqrt{5}}{8}$

$\therefore \cos^2 \frac{\pi}{10} = \frac{5 + \sqrt{5}}{8}$

Now, recall that $\cos 2x = 2\cos^2 x - 1$

$\therefore \cos \frac{\pi}{5} = 2\cos^2 \frac{\pi}{10} - 1$

$\therefore \cos \frac{\pi}{5} = 2(\frac{5 + \sqrt{5}}{8}) - 1$

$\therefore \cos \frac{\pi}{5} = \frac{5 + \sqrt{5}}{4}- 1$

$\therefore \cos \frac{\pi}{5} = \frac{1 + \sqrt{5}}{4}$

$\therefore \sec \frac{\pi}{5} = \frac{1}{\cos \frac{\pi}{5}} = \frac{4}{1 + \sqrt{5}}$

2) $\frac{\sin 2x+x}{\sin x} + \frac{\cos 2x+x}{\cos x}$

$=\frac{\sin 2x\cos x + \cos 2x\sin x}{\sin x} + \frac{\cos 2x\cos x - \sin 2x\sin x}{\cos x}$

Now recall that $\sin 2x = 2\sin x\cos x$ and $\cos 2x = 2\cos^2 x - 1$

$\frac{(2\sin x\cos x)\cos x + (2\cos^2 x - 1)\sin x}{\sin x} + \frac{(2\cos^2 x - 1)\cos x - (2\sin x\cos x)\sin x}{\cos x}$

$=\frac{2\sin x\cos^2 x + 2\sin x\cos^2 x - \sin x}{\sin x} + \frac{2\cos^3 x - \cos x - 2\sin^2 x\cos x}{\cos x}$

$= 4\cos^2 x - 1 + 2\cos^2 x - 1 - 2\sin^2 x$

$=6\cos^2 x - 2\sin^2 x - 2$

$= 8\cos^2 x - 4$

I skipped a couple of steps, but the latex was quite long (in my defense) :D
Title: Re: Swarles' questions
Post by: Greatness on March 20, 2011, 03:52:48 pm: What did you do to get from the 2nd to 3rd step??? Sry for being noob :P And thanks!
Title: Re: Swarles' questions
Post by: luffy on March 20, 2011, 03:55:21 pm: Quote from: swarley on March 20, 2011, 03:52:48 pm
What did you do to get from the 2nd to 3rd step??? Sry for being noob :P And thanks!

For which question?

EDIT: After further inspection, I realized your probably talking about question 2 :P
Pretty much, I used the rules of $\sin 2x = 2\sin x\cos x$ and $\cos 2x = 2\cos^2 x - 1$
Title: Re: Swarles' questions
Post by: Greatness on March 20, 2011, 03:59:59 pm: Oh you did Q1 as well!! For Q2 :)
Title: Re: Swarles' questions
Post by: luffy on March 20, 2011, 04:03:11 pm: Wait, I'll fix my original solution to make it easier to understand.
Title: Re: Swarles' questions
Post by: Greatness on March 20, 2011, 04:54:07 pm: Thank you!!!
Title: Re: Swarles' questions
Post by: Greatness on March 22, 2011, 04:41:00 pm: I found out today that there is much quicker way to doing Q2.
Take the common denominator first, which gives you a compund angle, which then you can simplify. :)
Title: Re: Swarles' questions
Post by: luffy on March 22, 2011, 06:04:53 pm: Quote from: swarley on March 22, 2011, 04:41:00 pm
I found out today that there is much quicker way to doing Q2.
Take the common denominator first, which gives you a compund angle, which then you can simplify. :)

Yeah.... Come to think of it, that sounds like an easier way of doing it. If you are comfortable with compound angle/double angle formulas, it won't take too long to do it either way. Nonetheless, simplifying denominators is what you should do in future cases to save those vital few seconds :D
Title: Re: Swarles' questions
Post by: Greatness on September 04, 2011, 07:34:15 pm: A tad confused with this kinematics question. Not quite sure where to start for a) i was thinking a=v*dv/dx but something went wrong lol A hint to get me started would be awesome!
Thanks!
Title: Re: Swarles' questions
Post by: tony3272 on September 04, 2011, 07:49:32 pm: A hint would be that we're given a value for the Constant Acceleration :P
Title: Re: Swarles' questions
Post by: Greatness on September 04, 2011, 07:55:26 pm: XD ah man so lame. So a=-9.8, s=-44.1, u=0 and v=?: v=+/- 29.4 but v>=0 so v=29.4m/s?