Prelim 2016 Maths Help Thread (1 Viewer)

InteGrand · May 1, 2016

Rathin said:
for the question;
let t=tan(θ/2), solve for the equation 12tan(θ)=5 for 180<θ<270
I got the answers as t=1/5 or t=-5 but the answer given by the textbook is only -5..why?

Since we are only interested in 180° < θ < 270°, clearly 90° < θ/2 < 135°, so tan(θ/2) (i.e. t) has to be negative (since θ/2 is in the second quadrant).

(And of course, that equation very easy to solve without t-method.)

Green Yoda · May 1, 2016

InteGrand said:
Since we are only interested in 180° < θ < 270°, clearly 90° < θ/2 < 135°, so tan(θ/2) (i.e. t) has to be negative (since θ/2 is in the second quadrant).

(And of course, that equation very easy to solve without t-method.)

ahh ok

eyeseeyou · May 9, 2016

How do I do the following:

1. If tan(theta)=t, express cos (squared) (theta) in terms of t
2.a. Given n x=2/3, find the possible value of cosx and cosecx
b. Given sinA=-0.6, find the psooible value of secA and tanA
c. If cot(theta)=-40/9 and 270<(theta)<360 find sin (theta) and cos (theta)
d. If sinb=s, express cosb and tanb in terms of s
e. If x is acute and sinx=1/(squareroot 5) show that tanx/1-secx = -2-(squareroot of 5). Find the value of this fraction when x is obtuse
f. If 5tanx=2, find the value of 5cosx-2sinx/cosx

leehuan · May 9, 2016

$1) \quad \cos^2{\theta}=\frac{1}{\sec^2{\theta}}=\frac{1}{1+\tan^2{\theta}}=\frac{1}{1+t^2}$

For some questions you should really use more brackets. (Also this post will be edited - I'll do a few more)
_______
$$2b) Firstly, since sine is negative, we are in either the third or the fourth quadrant. Now draw a right angled triangle.$\\ $The opposite side of the triangle has length 0.6 and the hypotenuse 1$\\ $By Pythagoras' theorem, the adjacent side has length $\sqrt{1^2-0.6^2}=0.8$

$$ Tangent can be positive in the third quadrant, OR negative in the fourth quadrant. So the two values are $\pm \frac{0.6}{0.8} = \frac{3}{4}$
Do the same for sec, noting that cosine is positive in the fourth quadrant or negative in the third. Note that secant is then hypotenuse/adjacent
_______
$$2c) The process is similar, however this time we are STRICTLY in the fourth quadrant.$\\ $Draw a right angled triangle with adjacent side length 40 and opposite side length 9 (because cot is 1/tan which is adj./opp.)$\\$By Pythagoras, the hypotenuse has length $\sqrt{40^2+9^2}=41$
$$Sine is negative in the fourth quadrant, so $\sin{\theta}=-\frac{9}{41}\\$Cosine is positive in the fourth quadrant, so $\cos{\theta}=\frac{40}{41}$
_______
$$2d) This really depends on what quadrant we're in. Because it's arbitrary I have to leave the plus or minus there$\\ $From the Pythagorean identity, we have $\\ \cos^2{\theta}=1-\sin^2{\theta} \Rightarrow \cos{\theta}=\pm \sqrt{1-\sin^2{\theta}} \\ \Rightarrow \cos{\theta}=\pm \sqrt{1-s^2}$
Then just use tan=sin/cos

eyeseeyou · May 9, 2016

leehuan said:
$1) \quad \cos^2{\theta}=\frac{1}{\sec^2{\theta}}=\frac{1}{1+\tan^2{\theta}}=\frac{1}{1+t^2}$

For some questions you should really use more brackets. (Also this post will be edited - I'll do a few more)

Not used to using latex lol

leehuan · May 9, 2016

eyeseeyou said:
Not used to using latex lol

Don't have to LaTeX it, take Q2f. Not sure if you meant 5cos2x - 2sinx/cosx
Or
(5cos2x-2sinx)/cosx

eyeseeyou · May 13, 2016

Bump did no one bother to help me?

leehuan · May 13, 2016

eyeseeyou said:
Bump did no one bother to help me?

I was waiting for you to fix up your brackets

eyeseeyou · May 13, 2016

leehuan said:
I was waiting for you to fix up your brackets

Oh yeah shit

For which questions?

leehuan · May 13, 2016

eyeseeyou said:
Oh yeah shit

For which questions?

Just fix up any brackets to questions I didn't attempt. I just didn't want to trial and error.

If no brackets are necessary at all to distinguish between fractions and etc then leave untouched.

eyeseeyou · May 13, 2016

eyeseeyou · May 15, 2016

Need help (leehuan or Integrand). I seem to be confused with this shit for some reason

1. \frac{2x+1}{x}<1
2. \frac{x}{2x+3}\leq5
3. \frac{-1}{t-1}\leqt
4. \frac{10t+1}{4t-1}\leq t+1
5. \frac{t-1}{t+1}<t
6. 3t-2\leq \frac{6t}{t+1}

Fucking hell it took me too long

eyeseeyou · May 15, 2016

eyeseeyou said:
Need help (leehuan or Integrand). I seem to be confused with this shit for some reason

1. \frac{2x+1}{x}<1
2. \frac{x}{2x+3}\leq5
3. \frac{-1}{t-1}\leqt
4. \frac{10t+1}{4t-1}\leq t+1
5. \frac{t-1}{t+1}<t
6. 3t-2\leq \frac{6t}{t+1}

Fucking hell it took me too long

wtf happened leehuan/integrand help me fix it up

InteGrand · May 15, 2016

eyeseeyou said:
wtf happened leehuan/integrand help me fix it up

$\noindent You need to type them in between tex tags.$

Type [tex.] [/tex.] (without the full stops). Then place your codes inside those. Also, it's good to leave spaces between things, like put a space between \leq and 5 for example, otherwise the code can mess up here sometimes.

Nailgun · May 15, 2016

eyeseeyou said:
Need help (leehuan or Integrand). I seem to be confused with this shit for some reason

$1. \frac{2x+1}{x}<1$

$2. \frac{x}{2x+3}\leq5$

$3. \frac{-1}{t-1}\leq t$

$4. \frac{10t+1}{4t-1}\leq t+1$

$5. \frac{t-1}{t+1}<t$

$6. 3t-2\leq \frac{6t}{t+1}$

Fucking hell it took me too long

leehuan · May 15, 2016

eyeseeyou said:
Need help (leehuan or Integrand). I seem to be confused with this shit for some reason

1. \frac{2x+1}{x}<1
2. \frac{x}{2x+3}\leq5
3. \frac{-1}{t-1}\leqt
4. \frac{10t+1}{4t-1}\leq t+1
5. \frac{t-1}{t+1}<t
6. 3t-2\leq \frac{6t}{t+1}

Fucking hell it took me too long

Lol yeah you have to wrap your latex around the $tags. For the questions though they're classic multiplying-by-the-denominator-[B]squared[/B] questions$

eyeseeyou · May 15, 2016

leehuan said:
Lol yeah you have to wrap your latex around the $tags. For the questions though they're classic multiplying-by-the-denominator-[B]squared[/B] questions[/QUOTE] Yeah I tried that but then I ended up fcking the answers Wait...how do I do that?$

Nailgun · May 15, 2016

eyeseeyou said:
Yeah I tried that but then I ended up fcking the answers

Wait...how do I do that?

multiply both sides by the square of the denominator to get rid of it
it should be trivial to solve then

leehuan · May 15, 2016

eyeseeyou · May 15, 2016

leehuan said:
$\begin{align*}\frac{2x+1}{x}&<1\\2x^2+x&<x^2\\x^2+x&<0\\x(x+1)&<0\end{align*}$
$-1<x<0$

What about the others

Also leehuan could you please show me the working to the rates of change question

Prelim 2016 Maths Help Thread (1 Viewer)

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