Trigonometry Question (1 Viewer)

[kev-kun]

Hi all, just a bit stuck on this question, help would be great!
Part b (ii) that is.

 

[obliviousninja]

Hi all, just a bit stuck on this question, help would be great!
Part b (ii) that is.
View attachment 28602

Think about the amplitude of the sin curve for the max and min.

 

[kev-kun]

So it would be max = 1 and min = -1?

 

[obliviousninja]

So it would be max = 1 and min = -1?

For the given equation its max min values occur at 2, -2.
Therefore RHS must be the same, ie 2, -2.

Then solve for x.

 

[braintic]

So it would be max = 1 and min = -1?

No, the amplitude is 2.

 

[braintic]

For the given equation its max min values occur at 2, -2.
Therefore RHS must be the same, ie 2, -2.

Then solve for x.

What RHS?

 

[obliviousninja]

What RHS?

Right hand side.

 

[HSC2014]

As in, what is your equation to say "RHS"? There is no "given" equation - just an expression.
But yeah 2sin(x+120) = +-2 is what you meant

 

[kev-kun]

For the given equation its max min values occur at 2, -2.
Therefore RHS must be the same, ie 2, -2.

Then solve for x.

No, the amplitude is 2.

Uhh I never actually learnt in class how to find the amplitude and how to find the max/min in the function.... This just happened to appear in one of the past papers. HELP!!

 

[braintic]

Right hand side.

Yes thanks, I know what it stands for. I mean the right hand side of what?

 

[HSC2014]

Uhh I never actually learnt in class how to find the amplitude and how to find the max/min in the function.... This just happened to appear in one of the past papers. HELP!!

Graph y = sin x
You will see that it is periodic like a transverse wave in physics with "amplitude" 1 (i.e. range: -1 <= y <= 1 )
So when you have y = 2sin x (which is the form of the question), you will have twice the amplitude (i.e. range: -2 <= y <= 2)
So by equating 2sin(x+120) = +- 2, you will be able to find the x values for which that function reaches its "maximum/minimum"

 

[kev-kun]

Graph y = sin x
You will see that it is periodic like a transverse wave in physics with "amplitude" 1 (i.e. range: -1 <= y <= 1 )
So when you have y = 2sin x (which is the form of the question), you will have twice the amplitude (i.e. range: -2 <= y <= 2)
So by equating 2sin(x+120) = +- 2, you will be able to find the x values for which that function reaches its "maximum/minimum"

Ohhh okay now I get it; thanks!

 

[obliviousninja]

Yes thanks, I know what it stands for. I mean the right hand side of what?

2sin(x+120) = ? (RHS, u get the jist).

 

[braintic]

2sin(x+120) = ? (RHS, u get the jist).

The reason I asked was that you also had 2sin(x+120) = asinx + bcosx.
Also, I personally would write 2=2sin(x+120).
Without an equation present, it really is quite ambiguous.

 

[SharkeyBoy]

Sorry if it sounds stupid, but how do you do part i) ?
I get ii, just not i - I'm probably missing something quite obvious...

 

[braintic]

Sorry if it sounds stupid, but how do you do part i) ?
I get ii, just not i - I'm probably missing something quite obvious...

Use the sin(A+B) expansion, then the exact values for sin120 and cos120.

 

[SharkeyBoy]

Use the sin(A+B) expansion, then the exact values for sin120 and cos120.

Sweet, thanks