Author Topic: Swarles' questions (Read 1551 times) Tweet Share

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

kamil9876 · « **Reply #1 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?

Greatness · « **Reply #2 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.

luffy · « **Reply #3 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)

Greatness · « **Reply #4 on:** March 20, 2011, 03:52:48 pm »

What did you do to get from the 2nd to 3rd step??? Sry for being noob

And thanks!

luffy · « **Reply #5 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 And thanks!

For which question?

EDIT: After further inspection, I realized your probably talking about question 2

Pretty much, I used the rules of $\sin 2x = 2\sin x\cos x$ and $\cos 2x = 2\cos^2 x - 1$

Greatness · « **Reply #6 on:** March 20, 2011, 03:59:59 pm »

Oh you did Q1 as well!! For Q2

luffy · « **Reply #7 on:** March 20, 2011, 04:03:11 pm »

Wait, I'll fix my original solution to make it easier to understand.

Greatness · « **Reply #8 on:** March 20, 2011, 04:54:07 pm »

Thank you!!!

Greatness · « **Reply #9 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.

luffy · « **Reply #10 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

Greatness · « **Reply #11 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!

tony3272 · « **Reply #12 on:** September 04, 2011, 07:49:32 pm »

A hint would be that we're given a value for the Constant Acceleration

Greatness · « **Reply #13 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?

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