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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Gloamglozer on April 07, 2010, 01:41:45 am

Title: Compound Angle & Vector Questions
Post by: Gloamglozer on April 07, 2010, 01:41:45 am: 1. Simplify $tan(y) - tan(x+y)$

2. Simplify $tan(\frac{\pi}{2} - y)$

Isn't $tan(\frac{\pi}{2})$ undefined?

3. If $sin(x)=\frac{4}{5}$ for $x \in [0,\frac{\pi}{2}]$ and $tan(y)=\frac{5}{12}$ for $y \in [\pi,\frac{3\pi}{2}]$ , find the exact values of $sin(x+y)$

4. Let i be a unit vector in the east direction and j be a unit vector in the north direction. A runner heads off in a direction which is $\frac{\pi}{3}$ west of north. Find a unit vector in this direction.

I've already got the answer as $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$ but I was just wondering if that is the correct answer and if it is possible to do it using a unit circle.
Title: Re: Compound Angle & Vector Questions
Post by: TrueTears on April 07, 2010, 01:53:07 am: 1. $\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$

Thus we have $\tan(y) - \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$

Common denominator we got:

$\frac{(1-\tan(x)\tan(y))\tan(y)-\tan(x)-\tan(y)}{1-\tan(x)\tan(y)}$

$\frac{\tan(y)-\tan(x)\tan^2(y)-\tan(x)-\tan(y)}{1-\tan(x)\tan(y)}$

$\frac{-\tan(x)(\tan^2(y)+1)}{1-\tan(x)\tan(y)}$

$\frac{-\tan(x)\sec^2(x)}{1-\tan(x)\tan(y)}$

2.

$\tan\left(\frac{\pi}{2}-y\right) = \cot(y)$ [Remember this complementary angle formula!]

3.

$\sin(x+y) = \sin(x)\cos(y) + \sin(y)\cos(x)$

Since $\sin^2(x)+\cos^2(x) = 1$

We have $\cos(x) = \sqrt{1-\sin^2(x)} = \sqrt{1-\frac{16}{25}} = \frac{3}{5}$ [We take positive answer because x is in the first quadrant]

We have $\frac{\sin(y)}{\cos(y)} = \frac{5}{12}$

Thus $\frac{\sin^2(y)}{\cos^2(y)} = \frac{25}{144}$

But $\cos^2(y) = 1-\sin^2(y)$

So $\frac{\sin^2(y)}{1-\sin^2(y)} = \frac{25}{144}$

Solving for $\sin(y)$ we get $\sin(y) = -\frac{5}{13}$ Since y is in 3rd quadrant.

Also we know that $\sin^2(y) = 1-\cos^2(y)$ so $\frac{1-\cos^2(y) }{\cos^2(y)} = \frac{25}{144}$ Solving for $\cos(y)$ we get $\cos(y) = -\frac{12}{13}$ again its negative because y is in the 3rd quadrant.

Thus we have:

$\sin(x+y) = \frac{4}{5}\left(-\frac{12}{13}\right) + -\frac{5}{13}\left(\frac{3}{5}\right) = \text{something}$
Title: Re: Compound Angle & Vector Questions
Post by: TrueTears on April 07, 2010, 02:07:57 am: For 4) If it is $\frac{\pi}{3}$ west of north, then it makes an angle of $\frac{\pi}{6}$ with the negative x axis which means it makes an angle of $\frac{5\pi}{6}$ with the positive x axis.

A unit vector in this direction is given by its directional cosines, ie, $<\cos\left(\frac{5\pi}{6}\right), \sin\left(\frac{5\pi}{6}\right)> = <-\frac{\sqrt{3}}{2}, \frac{1}{2}>$

So yes you are right.

Ahh, haven't done maths in sooo long, so rusty at it now :knuppel2: not that i was any good to begin with anyway
Title: Re: Compound Angle & Vector Questions
Post by: Martoman on April 07, 2010, 02:14:04 am: 2) $tan(\frac{\pi}{2} -(\frac{\pi}{2} -y)) = cot(y)$
To answer your question, $tan(\frac{\pi}{2})$ is definately undefined. You just have to ask yourself what this question was asking. If indeed y = $\pi$ or some multiple of it, then yes, it would be undefined. HOWEVER for any other value in the circle this is defined. If y = $\frac{2\pi}{5}$ It would be valid, and so on.

3) $sin(x+y) = sin(x)cos(y) + cos(x)sin(y)$

Those we know sin(x). We have to work out the others given the info.

For the $sin(x)$ case, draw a right angled triangle, with 4 as the opposite side, 5 as the hypotenuse. Then, you have a 3,4,5 triangle. Since it is in the 1st quadrant, cos(x),tan(x) anything you pick will be positive. Since we are dealing with the x variable, we want to find cos(x) so we can work out $sin(x+y)$

From the triangle $cos(x) = adj/hyp = 3/5$

Simiarly with the "y" variable, we want to know cos(y) and sin(y) but are given tan(y). Note it is in the third quadrant, hence sin(y), cos(y) are going to be negative.

So drawing a right angled triangle again, tan(y) implies the opposite side = 5, adjacent side = 12, so we have a 5,12,13 triangle.

Thus sin(y) = - $\frac{5}{13}$
$cos(y) = -\frac{12}{13}$

As per the $sin(x+y)$ equation we plug and chug our answer in to obtain the solution.

4) you are correct. Create a right angled triangle in the second quadrant which has 30 degrees in it, hypotenuse 1.
Title: Re: Compound Angle & Vector Questions
Post by: Martoman on April 07, 2010, 02:21:24 am: @truetears.

Not trying to rain on your parade, but since it is pythagorean triads, it is much easier to view it as a triangle. Heck, in any case it is. Its exactly what you have done, but I find for most people who need help with this sorta thing find it easier to view it this way, its *visually* intuitive to a greater or lesser extent than algebra.
Just saying. :angel:
Title: Re: Compound Angle & Vector Questions
Post by: TrueTears on April 07, 2010, 02:24:10 am: yeh i know, i purposely did the algebra becoz i realised my brain needed some mathematical exercise after a few month... so i didnt wanna take the easy path
Title: Re: Compound Angle & Vector Questions
Post by: Gloamglozer on April 08, 2010, 04:11:10 pm: Thank you TT and Martoman.

Quote from: TrueTears on April 07, 2010, 02:07:57 am
For 4) If it is $\frac{\pi}{3}$ west of north, then it makes an angle of $\frac{\pi}{6}$ with the negative x axis which means it makes an angle of $\frac{5\pi}{6}$ with the positive x axis.

A unit vector in this direction is given by its directional cosines, ie, $<\cos\left(\frac{5\pi}{6}\right), \sin\left(\frac{5\pi}{6}\right)> = <-\frac{\sqrt{3}}{2}, \frac{1}{2}>$

So yes you are right.

Ahh, haven't done maths in sooo long, so rusty at it now :knuppel2: not that i was any good to begin with anyway

You are a Maths god and we all know it. :P

With that vector question, do we still use the directional cosines if it is in another quadrant (cos and then sin)?
Title: Re: Compound Angle & Vector Questions
Post by: TrueTears on April 08, 2010, 06:16:32 pm: Quote from: Gloamglozer on April 08, 2010, 04:11:10 pm
Thank you TT and Martoman.

Quote from: TrueTears on April 07, 2010, 02:07:57 am
For 4) If it is $\frac{\pi}{3}$ west of north, then it makes an angle of $\frac{\pi}{6}$ with the negative x axis which means it makes an angle of $\frac{5\pi}{6}$ with the positive x axis.

A unit vector in this direction is given by its directional cosines, ie, $<\cos\left(\frac{5\pi}{6}\right), \sin\left(\frac{5\pi}{6}\right)> = <-\frac{\sqrt{3}}{2}, \frac{1}{2}>$

So yes you are right.

Ahh, haven't done maths in sooo long, so rusty at it now :knuppel2: not that i was any good to begin with anyway

You are a Maths god and we all know it. :P

With that vector question, do we still use the directional cosines if it is in another quadrant (cos and then sin)?
yeah just make sure its the angle made with the positive x axis :)
Title: Re: Compound Angle & Vector Questions
Post by: Gloamglozer on April 09, 2010, 12:05:38 am: Quote from: TrueTears on April 08, 2010, 06:16:32 pm
Quote from: Gloamglozer on April 08, 2010, 04:11:10 pm
Thank you TT and Martoman.

Quote from: TrueTears on April 07, 2010, 02:07:57 am
For 4) If it is $\frac{\pi}{3}$ west of north, then it makes an angle of $\frac{\pi}{6}$ with the negative x axis which means it makes an angle of $\frac{5\pi}{6}$ with the positive x axis.

A unit vector in this direction is given by its directional cosines, ie, $<\cos\left(\frac{5\pi}{6}\right), \sin\left(\frac{5\pi}{6}\right)> = <-\frac{\sqrt{3}}{2}, \frac{1}{2}>$

So yes you are right.

Ahh, haven't done maths in sooo long, so rusty at it now :knuppel2: not that i was any good to begin with anyway

You are a Maths god and we all know it. :P

With that vector question, do we still use the directional cosines if it is in another quadrant (cos and then sin)?
yeah just make sure its the angle made with the positive x axis :)

That question was in $R^2$ , just say it was in $R^3$ and k represented just straight up from the ground. And just say the question had a runner running up a hill for 6 km. Does that mean that k will be 6?
Title: Re: Compound Angle & Vector Questions
Post by: TrueTears on April 09, 2010, 12:16:14 am: sure would :)