[20 Questions] Trigonometry, Locus&Parabola, Polynomials + More! :( (1 Viewer)

P!xel · Feb 11, 2009

All clear! ;D Thanks to everyone! Rep added :]
Question 1) - Solved, thanks to roadrage75.

Question 2) - Solved, thanks to youngminii.

Question 3) - Solved, thanks to kurt.physics + jetblack2007.

Question 4) - Solved, thanks to gurmies.

Question 5) - Solved, thanks to jetblack2007.

Question 6) - Solved, thanks to jetblack2007.

Question 7) - Solved, thanks to tommykins.

Question 8) - Solved, thanks to Prashant Sallan.

Question 9) - Solved, thanks to kurt.physics,

Question 10) - Solved, thanks to gurmies + jetblack2007.

Question 11) - Solved, thanks to gurmies + kurt.physics.

Question 12) - Solved, thanks to jetblack2007.

Question 13) - Solved, thanks to Trebla.

Question 14) - Solved, thanks to kurt.physics,

Question 15) - Solved, thanks to jetblack2007.

Question 16) - Solved, thanks to jetblack2007 + Prashant Sallan + tacogym27101990.

Question 17) - Solved, thanks to jetblack2007.

Question 18) - Solved, thanks to roadrage75 + greentsquare.

Question 19) - Solved, thanks to jetblack2007.

Question 20) - Solved, thanks to jetblack2007.

I need all working out T_T.

This came from a test that I pretty much failed >:|
So i'll need to learn from my mistakes ^_^.

Suggestion: The LaTeX Equation Editor down there looks very neat, I suggest you use that. ^^

jet · Feb 11, 2009

Didn't youre teacher give you worked solutions? We always got them. Otherwise, they can't work out which marks to give you.

I'll take question 6
$\text{if you can imagine the diagram/draw one, you'll get my solution much easier} \\ \text{Using the point due east of the tower, } tan45 = \frac{h}{PT} \, \text{, where T is the point at the bottom of the tower.} \\ \therefore PT = h \\ Similarly, \sqrt{3}h = QT \\ \text{Now, since Q is due south of T and P is due east, } \angle PTQ = 90^\circ \\ \text{Using Pythagoras' Theorem: } PT^2 + QT^2 = PQ^2 \\ \therefore h^2 + 3h^2 = 1600 \\ h^2 = 400 \\ h = 20 \, metres$

jet · Feb 11, 2009

Question 20:
$\text{From the polynomial, } \\ \Sigma \alpha = \alpha + \beta + \gamma = 0 \\ \Sigma \alpha = \alpha \beta + \alpha \gamma + \beta \gamma = \frac{-5}{2} \\ \Pi \alpha = \alpha \beta \gamma = \frac{1}{2} \\ \text{Now, } \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \\ = \frac{\alpha \beta + \alpha \gamma + \beta \gamma}{\alpha \beta \gamma} \\ = \frac{-5}{2} \times 2 = -5$

Thats all I can do for now

lolokay · Feb 11, 2009

..surely you can do some of these?

roadrage75 · Feb 11, 2009

P!xel said:
Question 1)

i'll do the first question.

sin3x = sin(x+2x) = sin(2x)cos(x) + cos(2x)sin(x)
= 2sin(x)cos^2(x) + sin(x)(cos^2(x) - sin^2(x)) --1

so using equation 1,

sin3x - sinx = sinx(2cos^2(x) + cos^2(x) - sin^2(x) - 1)

= sinx(3cos^2(x) - sin^2(x) - cos^2(x) - sin^2(x))
= sinx(2cos^2(x) - 2sin^2(x))
= 2sin(x)*cos(2x) ---2 (double angle formula)

now cos(3x) = cos(2x + x)
= cos(2x)cos(x) - sin(2x)sin(x)
= cos(2x)cos(x) - 2sin^2(x)cos(x) <---- (double angle form)

= cos(x)(cos2x-2sin^2(x))

so cos(x) - cos(3x) = cos(x) ( 1- (cos(2x) - 2sin^2(x)))
= cos(x)(1-cos(2x) + 2sin^2(x))
= cos(x)( 1-(cos^2(x) - sin^2(x)) + 2sin^2(x))
= cos(x)( sin^2(x) + cos^2(x) - cos^2(x) + sin^2(x) + 2sin^2(x))

(** note sin^2(x) + cos^2(x) = 1)

and so = cos(x)(4sin^2(x)) = 4cos(x)*sin(x)*sin(x) = 2sin(2x)*sin(x)

using two and three it can be seen that (2sin(x)*cos(2x))/(2sin(2x)sin(x)))

can be simplfied to cos(2x)/sin(2x) = cot(2x)

M@ster P · Feb 11, 2009

im not very good at 3unit but alot of these questions look doable, you sure you can't do any of them?

Prashant Sallan · Feb 11, 2009

Question 8)

A)
30 45 60

Sin 1/2 1/root 2 root3/2

Cos root3/2 1/root 2 1/2

Tan 1/root 3 1 root 3

B) Sin (45 +30) = sin 75

But
Sin (45 + 30) = Sin45Cos30 + Cos45Sin30 (identity expansion)

(1/root2 X root3/2) + (1/root2 x 1/2)

root3 + 1
2root2

P!xel · Feb 11, 2009

I absolutely suck with trig :|
All I can remember is perhaps the exact values.

Sorry, I'm stupid >.<"

Prashant Sallan · Feb 11, 2009

P!xel said:
I absolutely suck with trig :|
All I can remember is perhaps the exact values.

Sorry, I'm stupid >.<"

dont worri- i suc k at its application aswell-- just try to remember the identities.. that would be a start...

tommykins · Feb 11, 2009

P!xel said:
Question 7)

$\frac{1-cos 2x}{1+cos2x}\\ =\frac{1-(cos^{2}x - sin^{2}x)}{1+(cos^{2}x-sin^{2}x)}\\ =\frac{1- cos^{2}x + sin^{2}x}{1+cos^{2}x - sin^{2}x}\\ =\frac{sin^{2}x + sin^{2}x}{cos^{2}x+cos^{2}x} =\frac{2sin^{2}x}{2cos^{2}x} = \tan ^{2}x \text{ as required.}\\ \text{Now, tan x is the square root of the given equation, hence } \tan 22\frac{1}{2}^{\circ} \text{ is equal to}\\ \frac{1-cos 45}{1+cos45} \Rightarrow \frac{\sqrt2 - 1}{\sqrt2+1}$

P!xel · Feb 11, 2009

Prashant Sallan said:
dont worri- i suc k at its application aswell-- just try to remember the identities.. that would be a start...

Thanks for the encouragement :|
Those identities are one of the things I hate the most >: ( .

But i'll try! I'll start memorizing them now!

gurmies · Feb 11, 2009

Question 4

$\cos 2A = \cos A$

$2\cos^{2}A-1=\cos A$

$2\cos^{2}A - \cos A - 1 = 0$

$(2\cos A + 1)(\cos A - 1) = 0$

$\cos A = -\frac{1}{2}\,\,\,\,\,\,\,or\,\,\,\,\,\,\, \cos A = 1$

$\therefore A = 120^{\circ}, 240^{\circ}, 0^{\circ}, 360^{\circ}$

Question 10

$2\sin^{2}\theta = \sin2\theta$

$2\sin^{2}\theta - 2\sin\theta \cos\theta = 0$

$2\sin\theta (\sin\theta-\cos\theta) = 0$

$\sin\theta = 0 \,\,\,\,\,\,\, or \,\,\,\,\,\,\, \sin\theta = \cos\theta$

$\theta = 0, \pi, 2\pi$

$Back \,\, to \,\, \sin\theta = \cos\theta$

$\tan\theta = 1$

$\theta = \frac{\pi }{4}, \frac{5\pi }{4}$

Question 11

a) $x^{2} + y^{2}-4x + 2y =0$

$(x-2)^{2} - 4 + (y+1)^{2} - 1 = 0$

$(x-2)^{2} + (y+1)^{2} = 5$

$\therefore Centre \,\, is \,\, (2, -1)\,\, and\,\,radius\,\,is\,\,\sqrt{5}\,\,units.$

Totally can't be bothered with the rest at this moment in time

youngminii · Feb 11, 2009

That is a crapload of questions.
Here's 2.

sin(A + B) = sinAcosB + cosAsinB

sin2x = 2sinxcosx

$sin2x = 2cos^2x$

$2sinxcosx = 2cos^2x$

$sinxcosx = cos^2x$

$sinx = cosx$

$\therefore x = \frac{\pi}{4} + \pi k, k\epsilon Z$

P!xel · Feb 11, 2009

youngminii said:
That is a crapload of questions.
Here's 2.

sin(A + B) = sinAcosB + cosAsinB

sin2x = 2sinxcosx

$sin2x = 2cos^2x$

$2sinxcosx = 2cos^2x$

$sinxcosx = cos^2x$

$sinx = cosx$

$\therefore x = 45 HOW THE FUCK DO YOU DO DEGREES ON THIS THING + 180k, k\epsilon Z$

LOL. I like your style of humor

.

Thanks for answering ! ^^

But as you said, still a crap load of questions :|

gurmies · Feb 11, 2009

youngminii said:
That is a crapload of questions.
Here's 2.

sin(A + B) = sinAcosB + cosAsinB

sin2x = 2sinxcosx

$sin2x = 2cos^2x$

$2sinxcosx = 2cos^2x$

$sinxcosx = cos^2x$

$sinx = cosx$

$\therefore x = 45 HOW THE FUCK DO YOU DO DEGREES ON THIS THING + 180k, k\epsilon Z$

When you divided by cos x in that last line, you assumed that cos x ≠ 0.

sinxcosx - cos²x = 0

cosx(sinx - cosx) = 0

As you can see, cosx = 0 is a solution...

tommykins · Feb 11, 2009

P!xel said:
LOL. I like your style of humor .

Thanks for answering ! ^^

But as you said, still a crap load of questions :|

^{\circ}

P!xel · Feb 11, 2009

Never mind, I just made myself look even more stupid.

gurmies · Feb 11, 2009

I divided both sides by 2

youngminii · Feb 11, 2009

gurmies said:
When you divided by cos x in that last line, you assumed that cos x ≠ 0.

sinxcosx - cos²x = 0

cosx(sinx - cosx) = 0

As you can see, cosx = 0 is a solution...

Oh crap lol, totally forgot about that.
In that case, x = pi/4 + kpi, where k is an integer, or pi/2 + kpi, where k is an integer.

Trebla · Feb 11, 2009

Try not to post so many questions next time...some of these you can ask your teacher or a tutor for help where you will probably benefit more than from here...

Q13
$1^2+3^2+....+(2n-1)^2 = \dfrac{n(2n-1)(2n+1)}{3}\\$For $ n=1\\LHS = 1^2 = 1\\RHS = \dfrac{3}{3} = 1\\$Statement true for $n = 1\\$Assume the truth of $n = k\\1^2+3^2+....+(2k-1)^2 = \dfrac{k(2k-1)(2k+1)}{3}\\$Consider $n = k + 1\\1^2+3^2+....+(2k+1)^2 = \dfrac{(k+1)(2k+1)(2k+3)}{3}\\LHS = 1^2+3^2+....+(2k+1)^2\\= \dfrac{k(2k-1)(2k+1)}{3} + (2k+1)^2\\= \dfrac{2k+1}{3}\left (k(2k-1)+3(2k+1)\right)\\= \dfrac{2k+1}{3}\left (2k^2+5k+3\right)\\= \dfrac{(k+1)(2k+1)(2k+3)}{3}\\= RHS\\$Statement is true for $n = k + 1$ if it is true $n = k$. Since it is true for $n = 1$ it inductively holds true for all integers $ n \geq 1$

[20 Questions] Trigonometry, Locus&Parabola, Polynomials + More! :( (1 Viewer)

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