# 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.

#### 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)
$\bg_white \noindent It's the auxiliary angle transform. Just in general, we may write$

$\bg_white a \cos \theta + b\sin \theta = R\cos \left(\theta - \phi\right)$,

$\bg_white \noindent where R = \sqrt{a^{2} + b^{2}} and \phi is the angle made by the point (a,b) to the origin. In the setting of complex numbers (for those doing 4U), R = |a+bi| and \phi = \mathrm{Arg} \left(a+bi\right) (one way to obtain the above trigonometric identity is indeed via complex numbers. The 3U-friendly method is to expand the R.H.S. and equate coefficients as calamebe explained above.).$

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#### 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
$\bg_white \noindent It's the auxiliary angle transform. Just in general, we may write$

$\bg_white a \cos \theta + b\sin \theta = R\cos \left(\theta - \phi\right)$,

$\bg_white \noindent where R = \sqrt{a^{2} + b^{2}} and \phi is the angle made by the point (a,b) to the origin. In the setting of complex numbers (for those doing 4U), R = |a+bi| and \phi = \mathrm{Arg} \left(a+bi\right) (one way to obtain the above trigonometric identity is indeed via complex numbers. The 3U-friendly method is to expand the R.H.S. and equate coefficients as calamebe explained above.).$
Are we (3U) allowed to just write $\bg_white \noindent R = \sqrt{a^{2} + b^{2}}$ and $\bg_white \alpha = \tan^{-1}(\frac{b}{a})$ 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 $\bg_white \noindent R = \sqrt{a^{2} + b^{2}}$ and $\bg_white \alpha = \tan^{-1}(\frac{b}{a})$ 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.

#### InteGrand

##### Well-Known Member
$\bg_white \noindent (It suffices to take the \phi in my earlier post to be the \textsl{principal argument} \mathrm{Arg}\left(a+bi\right). If a is positive, then this is indeed just \tan^{-1}\left(\frac{b}{a}\right). So if you wanted to always use \tan^{-1}, you could just factor out a negative at the start if a were negative, but then your R would be negative. Otherwise, just use the standard method of finding the principal argument (using what is essentially the \textbf{atan2 function} in many computer languages, linked below for further reading).)$

• atan2 function: https://en.wikipedia.org/wiki/Atan2#Definition_and_computation .

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