# Tutor Hunt Questions

“Express 3sinθ+5sinθ in the form Rsin(θ+α)

2 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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## 11 Answers

3 of something plus 5 of something is just 8 of something

So 3sinθ + 5sinθ = 8sinθ

So 3sinθ + 5sinθ = 8sinθ

Answered by Michael | 21 months ago

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.

Answered by Imogen | 20 months ago

Hi Matt. I think you made an error when posting it, one factor should be in cos(theta).

Answered by Octavian | 2 years ago

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

Answered by Tahseen | 18 months ago

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

________________________________________________________________

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)

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

________________________________________________________________

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)

Answered by David | 2 years ago

3sinΘ + 5sinΘ express to Rsin(θ+α)

Rsin(θ + α) can be expand to Rsinθ + Rsinα

let x = sinθ

so 3x + 5x = 8x

replace x for sinθ

3sinθ + 5sinθ = 8sinθ

In the form Rsin(θ + α) = Rsinθ + Rsinα where R is 8 so what is sinα?

if sinα is equals zero what is α?

Rsinθ + Rsinα = 8sinθ + 8sinα

if sinα = 0

α = 〖sin〗^(-1)0 = 0 or 180 because sin(0) = 0 or sin(180) = 0

sin(α) = sin(0) or sin(180) where α is neither 0 or 180

Rsinθ + Rsinα = Rsin(θ + α) where R = 8

so

8sinθ + 8sinα = 8sin(θ + 0) or 8sin(θ + 180)

Hope this helps?

Rsin(θ + α) can be expand to Rsinθ + Rsinα

let x = sinθ

so 3x + 5x = 8x

replace x for sinθ

3sinθ + 5sinθ = 8sinθ

In the form Rsin(θ + α) = Rsinθ + Rsinα where R is 8 so what is sinα?

if sinα is equals zero what is α?

Rsinθ + Rsinα = 8sinθ + 8sinα

if sinα = 0

α = 〖sin〗^(-1)0 = 0 or 180 because sin(0) = 0 or sin(180) = 0

sin(α) = sin(0) or sin(180) where α is neither 0 or 180

Rsinθ + Rsinα = Rsin(θ + α) where R = 8

so

8sinθ + 8sinα = 8sin(θ + 0) or 8sin(θ + 180)

Hope this helps?

Answered by Augusta | 21 months ago

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.

Answered by Farid | 2 years ago

3sin# + 5 sin# = 8 sin#

So problem solved.

I think the question was incorrect.

So problem solved.

I think the question was incorrect.

Answered by Graham | 2 years ago

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!

Answered by George | 2 years ago