Sine/Cosine Addition Formulae

[MathsDude69]

Homework Statement

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.

Homework Equations

Sine and Cosine Addition Formulae

sin(A + B) = sinAcosB + cosAsinB

cos(A + B) = cosAcosB - sinAsinB


Can anyone help with this. Its driving me crazy!

 

[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).01

 

[MathsDude69]

Oh sorry. Heres where I am 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)

I have never tried additional sine/cosine formulae before. Am I going the right way about this?

P.S. Cheers for the assitance

 

[MathsDude69]

R = 3sin3t + 5cos3t/sin3tcosø + cos3tsinø

Cancelling out in the division that gives:

R = 2sin3t + 4cos3t/cosø + sinøIm not too sure where to go after here. Any suggestions?

 

[qntty]

Cancelling out in the division that gives:

R = (2sin3t + 4cos3t)/(cosø + sinø)

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
[tex]
\frac{4\cos{4t}}{2cos{4t}} = \frac{4}{2} \not= \frac{3\cos{4t}}{cos{4t}}[/tex]
and
[tex]
\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}[/tex]

 

[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.

 

[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.

 

[yeongil]

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.

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

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,
[tex](R cos(\theta))^{2} + (R sin(\theta))^{2} = 3^{2} + 5^{2}[/tex]
[tex]R^{2} cos^{2}(\theta) + R^{2} sin^{2}(\theta) = 9 + 25[/tex]
[tex]R^{2}(cos^{2}(\theta) + sin^{2}(\theta)) = 34[/tex]
[tex]R^{2} = 34[/tex]
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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