? sin(3? + ?)

Scott92

Scott92

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Im got bit stuck/wanting to know if this is even correct think ive started correctly but not sure how to finish...
The third harmonic of a sound wave is given by
4 cos(3?) − 6 sin(3?b) ............................................................. what is that 'b'. what is Θ?
Express this sound wave in the form; ? sin(3? +?)
R (sin(30)cos(b)+cos(30)sin(b)) .............................................where did this come from
R sin(30)cos(b)+Rcos(30)sin(b)=4cos(30)-6sin(30) .............................................where did this come from
R cos(b)sin(30)+Rsin(b)cos(30)=4 cos(30)-6sin(30)
R sin(b)=4 ............................................................................................ Why?
R2sin2(b)=16 (as 4x4=16)
Rcos(b)=-6
R2cos2(b)=36
cos2+sin2=1...................................................................................Why?
R2(cos2(b)+sin2(b))=R2x1=R2
R2=36+16
R=36+16=213
R sin(b)Rcos(b)=4/-6 (fraction)=tan(b)=-2/3(fraction) then convert to a whole number is -0.667 to 3 significant numbers.
tan-1(-0.667)=-0.5882333373 or -0.588 to 3 significant numbers. Is this correct and not sure what trig function i need to find the vaule for b now thanks.
 
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Scott92 said:
Im got bit stuck/wanting to know if this is even correct think ive started correctly but not sure how to finish...
The third harmonic of a sound wave is given by
4 cos(3?) − 6 sin(3?b) ............................................................. what is that 'b'.
Express this sound wave in the form; ? sin(3? +?)
R (sin(30)cos(b)+cos(30)sin(b)) .............................................where did this come from
R sin(30)cos(b)+Rcos(30)sin(b)=4cos(30)-6sin(30) .............................................where did this come from
R cos(b)sin(30)+Rsin(b)cos(30)=4 cos(30)-6sin(30)
R sin(b)=4 ............................................................................................ Why?
R2sin2(b)=16 (as 4x4=16)
Rcos(b)=-6
R2cos2(b)=36
cos2+sin2=1...................................................................................Why?
R2(cos2(b)+sin2(b))=R2x1=R2
R2=36+16
R=36+16=213
R sin(b)Rcos(b)=4/-6 (fraction)=tan(b)=-2/3(fraction) then convert to a whole number is -0.667 to 3 significant numbers.
tan-1(-0.667)=-0.5882333373 or -0.588 to 3 significant numbers. Is this correct and not sure what trig function i need to find the vaule for b now thanks.
Click to expand...
Please look at the questions that has been asked modify your response.
 
The third harmonic of a sound wave is given by 4 cos(30) − 6 sin(30)
Express this sound wave in the form
? sin(30+ ?)

It seems some of the answer has been changed to b when it should be ? if that helps some and the weird 0 should just be a normal one.
cos2+sin2=1 was more of a note to myself as i was told that should always equal 1? maybe wrongly? Possibly got the sin4 from reading the question as sin 4 not cos so ive made a right old mess of it. Confused now what to do all together? Thanks for reply
 
Scott92 said:
Im got bit stuck/wanting to know if this is even correct think ive started correctly but not sure how to finish...
The third harmonic of a sound wave is given by
4 cos(3?) − 6 sin(3?b) ............................................................. what is that 'b'. what is Θ?
Express this sound wave in the form; ? sin(3? +?)
R (sin(30)cos(b)+cos(30)sin(b)) .............................................where did this come from
R sin(30)cos(b)+Rcos(30)sin(b)=4cos(30)-6sin(30) .............................................where did this come from
R cos(b)sin(30)+Rsin(b)cos(30)=4 cos(30)-6sin(30)
R sin(b)=4 ............................................................................................ Why?
R2sin2(b)=16 (as 4x4=16)
Rcos(b)=-6
R2cos2(b)=36
cos2+sin2=1...................................................................................Why?
R2(cos2(b)+sin2(b))=R2x1=R2
R2=36+16
R=36+16=213
R sin(b)Rcos(b)=4/-6 (fraction)=tan(b)=-2/3(fraction) then convert to a whole number is -0.667 to 3 significant numbers.
tan-1(-0.667)=-0.5882333373 or -0.588 to 3 significant numbers. Is this correct and not sure what trig function i need to find the vaule for b now thanks.
Click to expand...
A couple of notes for readability:
1) If you don't want to type [imath]\theta[/imath] all the time use a "t" not "0."

2) To square a number use ^2.

3) sin^2 + cos^2 is not 1. sin^2(t) + cos^2(t) = 1. You really do need to write the angle here.

So. [imath]R^2 = 36 + 16[/imath]. Why did you follow up and say R = 36 + 16?

You already have an equation for b: [imath]b = tan \left ( \dfrac{4}{-6} \right )[/imath]. When you take the inverse tangent you get a reference angle for b. This means that b is either in QII or QIV. In this case all you need is that b is negative so you are good.

So fix R and write down the answer. You are almost done.

-Dan
 
topsquark said:
A couple of notes for readability:
1) If you don't want to type [imath]\theta[/imath] all the time use a "t" not "0."

2) To square a number use ^2.

3) sin^2 + cos^2 is not 1. sin^2(t) + cos^2(t) = 1. You really do need to write the angle here.

So. [imath]R^2 = 36 + 16[/imath]. Why did you follow up and say R = 36 + 16?

You already have an equation for b: [imath]b = tan \left ( \dfrac{4}{-6} \right )[/imath]. When you take the inverse tangent you get a reference angle for b. This means that b is either in QII or QIV. In this case all you need is that b is negative so you are good.

So fix R and write down the answer. You are almost done.

-Dan
Click to expand...
Thanks for the tips
The r should be square root of the 36+16 its just lost the sq sign and the answer is square rooted as well 2sq13
Not sure i get what you mean by QII or QIV
Thanks again
 
Scott92 said:
Thanks for the tips
The r should be square root of the 36+16 its just lost the sq sign and the answer is square rooted as well 2sq13
Not sure i get what you mean by QII or QIV
Thanks again
Click to expand...
Q refers to - Quadrant. QI refers to the "first quadrant" and QII refers to the second quadrant.
 
Scott92 said:
So if i did this pi--0.588=3.729592654 is that what you mean? or am i missing the point thanks
Click to expand...
The negative value of arctangent will return a reference angle in QII or QIV. That means your angle b will be in either Quadrant II or Quadrant IV. ie. Given the reference angle [imath]\theta = tan^{-1} \left | \dfrac{4}{-6} \right |[/imath] then angle b will either be [imath]b = \pi - \theta[/imath] or [imath]b = 2 \pi - \theta[/imath].

(Do everything in terms of reference angles. Once you get used to it life gets much easier.)

-Dan
 
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