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“Express 3sinθ+5sinθ in the form Rsin(θ+α)
7 years ago
Maths Question asked by Matt

Hi there,
Did you mean either 3sinθ + 5cosθ? Or 3cosθ + 5sinθ?
Chris
Are you sure that the question gives 3sinθ+5sinθ, and not 3sinθ+5cosθ?
I changed the question to what I think was meant.
3 of something plus 5 of something is just 8 of something
So 3sinθ + 5sinθ = 8sinθ
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10 Answers
3 of something plus 5 of something is just 8 of something
So 3sinθ + 5sinθ = 8sinθ
So 3sinθ + 5sinθ = 8sinθ
3sinW +5sinW = Rsin(W + a): I take it that the question is :
3sinW + 5cosW = Rsin(W + a)
So Rsin(W + a) = R(sinW*cosa + cosW*sina)
= (Rcosa)sinW + (Rsina)cosW
Compare 3sinW + 5cosW with (Rcosa)sinW + (Rsina)cosW
Clearly Rsina = 5
Rcosa = 3
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so (Rsina)*(Rsina) + (Rcosa)*(Rcosa) = 3*3 + 5*5
then R*R( sina*sina + cosa *cosa) = 34 NB. ( sina*sina + cosa *cosa) = 1
thus R*R = 34
R = square root of 34 =1.73
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Rsina/Rcosa =5/3
sina/cosa = 5/3
tana = 5/3
a = inverse tan (5/3) = 59.0 degrees
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so 3sinW + 5CosW = 1.73sin( W + 59.0)
3sinW + 5cosW = Rsin(W + a)
So Rsin(W + a) = R(sinW*cosa + cosW*sina)
= (Rcosa)sinW + (Rsina)cosW
Compare 3sinW + 5cosW with (Rcosa)sinW + (Rsina)cosW
Clearly Rsina = 5
Rcosa = 3
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so (Rsina)*(Rsina) + (Rcosa)*(Rcosa) = 3*3 + 5*5
then R*R( sina*sina + cosa *cosa) = 34 NB. ( sina*sina + cosa *cosa) = 1
thus R*R = 34
R = square root of 34 =1.73
______________________________________________________________-
Rsina/Rcosa =5/3
sina/cosa = 5/3
tana = 5/3
a = inverse tan (5/3) = 59.0 degrees
________________________________________________________________
so 3sinW + 5CosW = 1.73sin( W + 59.0)
As I believe your original question was
"Express 3sinθ + 5cosθ in the form Rsin(θ+α)"
this is how you would work it out:
1. Expand Rsin(θ+α) differently using the double angle formula:
Rsinθcosα + Rsinαcosθ
2. (Rcosα)= 3 and (Rsinα)= 5.
"Express 3sinθ + 5cosθ in the form Rsin(θ+α)"
this is how you would work it out:
1. Expand Rsin(θ+α) differently using the double angle formula:
Rsinθcosα + Rsinαcosθ
2. (Rcosα)= 3 and (Rsinα)= 5.
If we decide to call Sinθ, X instead, just to make things look simpler, we can write the equation as this. Sinθ=X so 3X + 5x.
Now we know 3X just equals X+X+X and 5x just equals X+X+X+X+X, so if we add them together we get 8X!
If we think about what we did in the beginning, we can just change X back into Sinθ, like this, 8Sinθ.
The questions asks for the form Rsin(θ+a), so all we have to do is look at our answer, which is 8sinθ, and see what R and a equal. R is clearly 8, and theta has nothing added to it, so a must be zero.
Hope this helps, if you have any questions feel free to ask!
Now we know 3X just equals X+X+X and 5x just equals X+X+X+X+X, so if we add them together we get 8X!
If we think about what we did in the beginning, we can just change X back into Sinθ, like this, 8Sinθ.
The questions asks for the form Rsin(θ+a), so all we have to do is look at our answer, which is 8sinθ, and see what R and a equal. R is clearly 8, and theta has nothing added to it, so a must be zero.
Hope this helps, if you have any questions feel free to ask!
Hi Matt. I think you made an error when posting it, one factor should be in cos(theta).
I think you have made a mistake with your question, I think you meant to write 3sinθ+5cosθ. If that`s what you meant then if you expand Rsin(θ+α)=Rsinθcosα +Rcosθsinα
Then equate coefficients:
3sinθ=Rsinθcosα therefore 3=Rcosα
and 5cosθ = Rcosθsinα therefore 5= Rsinα
dive the last two equations Rsinα / Rcosα =tanα = 5/3
square and add you get 9+25 = R squared (since square of sine + square of cos is 1), so R=root of 34
If your original question was correct then 3sinθ+5sinθ = 8sinθ
so R=8 and α = 0
Good luck with your studies.
Then equate coefficients:
3sinθ=Rsinθcosα therefore 3=Rcosα
and 5cosθ = Rcosθsinα therefore 5= Rsinα
dive the last two equations Rsinα / Rcosα =tanα = 5/3
square and add you get 9+25 = R squared (since square of sine + square of cos is 1), so R=root of 34
If your original question was correct then 3sinθ+5sinθ = 8sinθ
so R=8 and α = 0
Good luck with your studies.
3sin# + 5 sin# = 8 sin#
So problem solved.
I think the question was incorrect.
So problem solved.
I think the question was incorrect.
I think `α` is meant for the periodicity, so its 180 degree. So the answer is
3*sinθ + 5*sinθ = 8*sin(θ+180)
BW
T
3*sinθ + 5*sinθ = 8*sin(θ+180)
BW
T