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

P!xel

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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. ^^
 
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jet

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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
 

jet

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Question 20:


Thats all I can do for now
 
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roadrage75

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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

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

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

Sorry, I'm stupid >.<"
 
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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...
 

P!xel

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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

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Question 4













Question 10

















Question 11

a)







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

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

sin(A + B) = sinAcosB + cosAsinB

sin2x = 2sinxcosx









 
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P!xel

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

sin(A + B) = sinAcosB + cosAsinB

sin2x = 2sinxcosx









LOL. I like your style of humor :).

Thanks for answering ! ^^

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

gurmies

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

sin(A + B) = sinAcosB + cosAsinB

sin2x = 2sinxcosx









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...

 

P!xel

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

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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

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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
 

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