# need a hand please (1 Viewer)

#### hscuserr

##### New Member
A particle is moving in a straight line according to the equation
x = 5 + 6 cos 2t + 8 sin 2t,
where x is the displacement in metres and t is the time in seconds.
(i) Prove that the particle is moving in simple harmonic motion by showing 2
2 that x satisfies an equation of the form x = −n ( x − c).
(ii) When is the displacement of the particle zero for the first time? 3

in questions (ii),

how do you Rewrite 6cos2t + 8sin2t = −5 as 10cos(2t − α) = −5
is there any reason why it becomes cos(2t - a)

#### calamebe

##### Active Member
Well, you can expand 10cos(2t-α) to 10cos(2t)cos(α)+10sin(2t)sin(α), and then equate one of the terms. So for example, the coefficient of sin(2t) on both sides is 10sin(α)=8, so α=arcsin(4/5). You could also write this is α=arccos(3/5) or α=arctan(4/3), and if you want to see why you can draw a right angled triangle with sides 3, 4 and 5. The reason we put it in that cos(2t-α) or sin(2t+α) form is that it is so much easier to work with.

• astroman

#### InteGrand

##### Well-Known Member
A particle is moving in a straight line according to the equation
x = 5 + 6 cos 2t + 8 sin 2t,
where x is the displacement in metres and t is the time in seconds.
(i) Prove that the particle is moving in simple harmonic motion by showing 2
2 that x satisfies an equation of the form x = −n ( x − c).
(ii) When is the displacement of the particle zero for the first time? 3

in questions (ii),

how do you Rewrite 6cos2t + 8sin2t = −5 as 10cos(2t − α) = −5
is there any reason why it becomes cos(2t - a)

,

Last edited:

#### davidgoes4wce

##### Well-Known Member
A particle is moving in a straight line according to the equation
x = 5 + 6 cos 2t + 8 sin 2t,
where x is the displacement in metres and t is the time in seconds.
(i) Prove that the particle is moving in simple harmonic motion by showing 2
2 that x satisfies an equation of the form x = −n ( x − c).
(ii) When is the displacement of the particle zero for the first time? 3

in questions (ii),

how do you Rewrite 6cos2t + 8sin2t = −5 as 10cos(2t − α) = −5
is there any reason why it becomes cos(2t - a)
Do you have an answer for it? Just want to compare my answers

#### calamebe

##### Active Member
Do you have an answer for it? Just want to compare my answers
You'd get the general solution being : t = (3arcsin(4/5) + 2pi(3n±1))/6, and from that the first time is t = (3arcsin(4/5) + 2pi)/6, or roughly t = 1.51.

#### Nexa579

##### New Member
Are we (3U) allowed to just write and without showing the equating coefficients with the trig identities?

Unsure because pretty much all the worked solutions don't use the above way but I personally use it, so not sure if HSC will accept?

#### calamebe

##### Active Member
Are we (3U) allowed to just write and without showing the equating coefficients with the trig identities?

Unsure because pretty much all the worked solutions don't use the above way but I personally use it, so not sure if HSC will accept?
That's what my teacher taught us, so I'm guessing it's allowed.

#### InteGrand

##### Well-Known Member
You also generally need to be careful with the arctangent.