1. Apr 15, 2009

### MathsDude69

1. The problem statement, all variables and given/known data

1, Express x = 3sin3t + 5cos3t in the form of x=Rsin(3t + ø)

2, Express x = 2sin(0.5t + 3) + 3cos(0.5t + 1) in the form x = Rsin(0.5t + ø)

All working must be done in radians.

2. Relevant equations

sin(A + B) = sinAcosB + cosAsinB

cos(A + B) = cosAcosB - sinAsinB

Can anyone help with this. Its driving me crazy!!

2. Apr 15, 2009

### yeongil

You need to show some work first before we can help you!

One hint is to use the sin(A + B) formula to expand x=Rsin(3t + ø).

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Last edited: Apr 15, 2009
3. Apr 15, 2009

### MathsDude69

Oh sorry. Heres where im at: Using the sine addition formula for x = Rsin(3t + ø):

sin(A + B) = sinAcosB + cosAsinB

sin(3t + ø) = sin3tcosø + cos3t+sinø

and if x = 3sin3t + 5cos3t then:

3sin3t + 5cos3t = R (sin3tcosø + cos3tsinø)

P.S. Cheers for the assitance

4. Apr 15, 2009

### MathsDude69

R = 3sin3t + 5cos3t/sin3tcosø + cos3tsinø

Cancelling out in the division that gives:

R = 2sin3t + 4cos3t/cosø + sinø

Im not too sure where to go after here. Any suggestions?

Last edited: Apr 15, 2009
5. Apr 15, 2009

### qntty

Careful with this step. You cannot cancel anything. You can only cancel things when they appear in both terms of the numerator and denominator. And when you do cancel, you get ride fo the sin3t and cos3t completely, don't subtract cos3t. Eg
$$\frac{4\cos{4t}}{2cos{4t}} = \frac{4}{2} \not= \frac{3\cos{4t}}{cos{4t}}$$
and
$$\frac{4\cos{4t} + 4\sin{4t}}{2cos{4t}} = \frac{4\cos{4t}}{2cos{4t}} + \frac{4\sin{4t}}{2cos{4t}} = \frac{4}{2} + \frac{4\sin{4t}}{2cos{4t}} \not= \frac{4 + 4\sin{4t}}{2}$$

Last edited: Apr 15, 2009
6. Apr 15, 2009

### MathsDude69

Fair point. I can see that now and thanks for your help. I can still see no further than the this point:

R = 3sin3t + 5cos3t/sin3tcosø + cos3tsinø

I am presuming that the question is designed such that via using the sine/cosine addition formulae that the constants R and ø can be attained by simplifing the rest of the equation. I think there is some additional theorem/property of this type of equation I am missing.

7. Apr 15, 2009

### MathsDude69

I have managed to find the additional info required for question 1 and solved it:-

asinx + bcosx can be assumed as

Rsin(x + ø)

where:

Tanø = b/a

and

R = sqrt(a2 + b2)

Works out x = 5.83sin(3t + 1.03)

Still stuck on q2 though if anybody has any ideas.

8. Apr 15, 2009

### yeongil

I did things differently for #1. I used the Pythagorean identity, $$sin^{2}(x) + cos^{2}(x) = 1$$.

x = R sin(3t + ø)
= R[sin(3t)cos(ø) + cos(3t)sin(ø)]
= Rcos(ø)sin(3t) + Rsin(ø)cos(3t)
But also
x = 3sin3t + 5cos3t
So R cos(ø) = 3 and R sin(ø) = 5.

Then,
$$(R cos(\theta))^{2} + (R sin(\theta))^{2} = 3^{2} + 5^{2}$$
$$R^{2} cos^{2}(\theta) + R^{2} sin^{2}(\theta) = 9 + 25$$
$$R^{2}(cos^{2}(\theta) + sin^{2}(\theta)) = 34$$
$$R^{2} = 34$$
So R is sqrt(34), which you got.
Finally, pick either R cos(ø) = 3 and R sin(ø) = 5 to solve for ø, which you got.

For #2, you'll have to use the formulas again for sin(0.5t + 3) and cos(0.5t + 1).
x = 2sin(0.5t + 3) + 3cos(0.5t + 1)
= 2[sin(0.5t)cos(3) + cos(0.5t)sin(3)] + 3[cos(0.5t)cos(1)-sin(0.5t)sin(1)]
distribute, then group.

Also, remember that
x = Rsin(0.5t + ø)
= R[sin(0.5t)cos(ø) + cos(0.5t)sin(ø)]

Can you take it from there?

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