Prelim 2016 Maths Help Thread (3 Viewers)

Green Yoda · May 28, 2016

Find general solution to sin(3x)+sin(x)=0

leehuan · May 28, 2016

Rathin said:
Find general solution to sin(3x)+sin(x)=0

Use the fact that sin(3x)=3sin(x)-4sin³(x) to complete this question.

Proof is left as an exercise to the reader.....

InteGrand · May 28, 2016

Rathin said:
Find general solution to sin(3x)+sin(x)=0

An alternate method to leehuan's is to derive and use a sums-to-product identity.

Green Yoda · May 28, 2016

leehuan said:
Use the fact that sin(3x)=3sin(x)-4sin³(x) to complete this question.

Proof is left as an exercise to the reader.....

wot

InteGrand · May 28, 2016

Rathin said:
wot

To derive that triple angle formula, you can apply the compound angle formula to sin(2x + x).

Green Yoda · May 28, 2016

InteGrand said:
To derive that triple angle formula, you can apply the compound angle formula to sin(2x + x).

I tried that method..it didnt work out for me

InteGrand · May 28, 2016

Rathin said:
I tried that method..it didnt work out for me

sin(3x) = sin(2x + x)

= sin(2x)cos(x) + cos(2x)sin(x).

Now, use double angle formulas (for the cos(2x), use 1 – 2sin² x), and use Pythagorean identity when necessary.

Green Yoda · May 28, 2016

Answer only says nπ
but I also got nπ+(-1)^n x (π/2) or nπ+(-1)^n x (-π/2) as the gen sol

leehuan · May 28, 2016

Rathin said:
Answer only says nπ
but I also got nπ+(-1)^n x (π/2) or nπ+(-1)^n x (-π/2) as the gen sol

They combine to give n.pi

Think about it.

(If you don't see it, start writing out the values for n=0,1,2,3,4,5,6,...)

Green Yoda · May 28, 2016

leehuan said:
They combine to give n.pi

Think about it.

(If you don't see it, start writing out the values for n=0,1,2,3,4,5,6,...)

ah I see, just curious..in exam conditions if you dont pick up this connection..do you loose marks for writing 3 separate?

InteGrand · May 28, 2016

Your answer would be n*pi/2 rather than n*pi, as you can see by splitting up into cases of n even (n = 2k), or n odd (n = 2k-1).

Trebla · May 28, 2016

Rathin said:
Find general solution to sin(3x)+sin(x)=0

Another approach would be to note that this equation is equivalent to

sin 3x = sin(-x)

If you solve it within one 'revolution' (like you would for a standard trig equation) you get

3x = -x
or
3x = π - (-x)

Now generalise this to multiple 'revolutions'

3x = -x + 2πn
or
3x = π - (-x) + 2πn

This resolves to

x = πn/2
or
x = 3πn/2

However, x = 3πn/2 represents a subset of the values within x = πn/2 so the most general solution is x = πn/2

leehuan · May 29, 2016

Rathin said:
ah I see, just curious..in exam conditions if you dont pick up this connection..do you loose marks for writing 3 separate?

Doubt it.

And refer to what they^ said. I didn't actually check your answer nor the given ones.

fluffchuck · May 29, 2016

Yep, pretty much what Trebla said,

sin(3x) + sin(x) = 0

sin(3x) = -sin(x)

And since the sine function is odd, sin(-x) = -sin(x)

Hence, sin(3x) = sin(-x)

You should be able to solve from there, using your general solution formula.

eyeseeyou · May 29, 2016

More motion question for you all (this will be good for the current 3U kids):

This is motion defined by a=a(x)

1. The acceleration of particle P, moving in a straight line, is given by x (with 2 dots on top of x)=2x-3 where x metres is the displacement from the origin O. Initially the particle is at O and its velocity v is 2 metres per second
a. Show that the velocity of the particle v^2=2x^2-6x+4
b. Calculate the velocity and acceleration of P at x=1 and briefly describe the motion of P after it moves from x=1
2. The acceleration of a particle is (2x-5) m/s^2, where x in the distance in metres from the origin
a. Find an expression for the velocity of this particle in terms of x, given that the particle is at rest one metre to the left of the origin initially
b. Describe the motion
3. A particle is moving along a straight line, with velocity v m/s and acceleration given by the expression k(4-x^2) m/s^2 where k is a constant
a. Show that v^2=4+Ae^(-2kx) where A is a constant satisifies the acceleration condition
b. If it starts from x=0 with a velocity of 7m/s, find the value of A
c. Does the particle ever change direction? Justify your answer
d. At x=1, v=4m/s, find the speed correct to two decimal places when x=2
e. As the motion continues, what happens, to the velocity and acceleration

InteGrand · May 29, 2016

eyeseeyou said:
More motion question for you all (this will be good for the current 3U kids):

This is motion defined by a=a(x)

1. The acceleration of particle P, moving in a straight line, is given by x (with 2 dots on top of x)=2x-3 where x metres is the displacement from the origin O. Initially the particle is at O and its velocity v is 2 metres per second
a. Show that the velocity of the particle v^2=2x^2-6x+4
b. Calculate the velocity and acceleration of P at x=1 and briefly describe the motion of P after it moves from x=1
2. The acceleration of a particle is (2x-5) m/s^2, where x in the distance in metres from the origin
a. Find an expression for the velocity of this particle in terms of x, given that the particle is at rest one metre to the left of the origin initially
b. Describe the motion
3. A particle is moving along a straight line, with velocity v m/s and acceleration given by the expression k(4-x^2) m/s^2 where k is a constant
a. Show that v^2=4+Ae^(-2kx) where A is a constant satisifies the acceleration condition
b. If it starts from x=0 with a velocity of 7m/s, find the value of A
c. Does the particle ever change direction? Justify your answer
d. At x=1, v=4m/s, find the speed correct to two decimal places when x=2
e. As the motion continues, what happens, to the velocity and acceleration

$\noindent 1) Here's some hints.$

$\noindent a) Use $\ddot{x} = \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{1}{2} v^2 \right)$.$

$\noindent b) Note from a) $v^2 = 2x^2-6x+4$. At $x=1$, we have acceleration $\ddot{x} = 2-3=-1$, and also $v^2 = 2-6+4 = 0 \Rightarrow$ the velocity is $0$. For the description part, at $x=1$, the acceleration is negative and velocity is $0$, so the particle will start to get a negative velocity after this, so it'll move backwards (i.e. to the left on the $x$-axis). As this happens, the $x$ value gets less, so the acceleration becomes even more negative, so the velocity gets more and more large negative (so faster speed in the left direction), and so the particle will move faster and faster to the left.$

Orwell · May 29, 2016

I'm so bad at visualising worded questions, especially bearings in trigonometry. Can anyone solve this and maybe send me a picture of the working out and the diagram they constructed?

davidgoes4wce · May 29, 2016

Orwell said:
I'm so bad at visualising worded questions, especially bearings in trigonometry. Can anyone solve this and maybe send me a picture of the working out and the diagram they constructed?

Seems like a familiar question. From Grove right?

Orwell · May 29, 2016

I found it in my Maths in Focus text book. Also, I'll probably be posting innumerable seemingly easy questions considering I have an exam tomorrow haha. I've got another question:

jathu123 · May 29, 2016

Orwell said:
I found it in my Maths in Focus text book. Also, I'll probably be posting innumerable seemingly easy questions considering I have an exam tomorrow haha. I've got another question:

cos(θ+10) = -1/2 [10<=θ+10<=370]
acute angle is 60, therefore
θ+ 10 = (180-60), (180+60) [All Stations To Central]
θ + 10 = 120, 240
θ = 110, 230

Prelim 2016 Maths Help Thread (3 Viewers)

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