Trig. (1 Viewer)

[joesmith1975]

Show sin3x in tems of sin x

 

[icycloud]

Normal 3U method:

Sin[3x] = Sin[2x+x]
= Sin[2x]Cos[x] + Sin[x]Cos[2x]
= 2Sin[x]Cos[x]Cos[x] + Sin[x](1-2Sin²[x])
= 2Sin[x]Cos²[x] + Sin[x] - 2Sin³[x]
= 2Sin[x](1-Sin²[x]) + Sin[x] - 2Sin³[x]
= 2Sin[x] - 2Sin³[x] + Sin[x] - 2Sin³[x]
= 3Sin[x] - 4Sin³[x]

4U Method:
Or if you want to be fancy, use De Moivre's theorem (in the 4U syllabus):

Cis[3x] = (Cis[x])³

Let c = Cos[x], s = Sin[x]
Cos[3x] + iSin[3x] = (c+is)³
= c³ + 3c²is + 3c(is)² + (is)³
= (c³-3cs) + i(3c²s-s³)

Comparing coefficients, Sin[3x] = 3Cos²[x]Sin[x] - Sin³[x]
= (1-Sin²[x])3Sin[x] - Sin³[x]
= 3Sin[x] - 4Sin³[x] (same answer we got before)

 

[joesmith1975]

Thanks for the help as i was stuck on a couple of lines from the bottom, also i didnt reliease could do it using de mov theorm. Thanks anyways

 

[joesmith1975]

sorry another question i cant get around
show:

sin 7x cos 3x = 1/2 [sin(7x +3x) + sin (7x -3x)]

 

[SoulSearcher]

Ok, now I'm using a method here that is no longer in the syllabus so I wonder why you have been given this question, but I'll do it anyway.

general rule for sums of sin is
sin + sin = 2 sin(1/2 sum) * cos (1/2 difference)
so comparing that to the question,
sin 7x cos 3x
= 1/2 * 2 sin 7x cos 3x
=1/2 [sin (7x+3x) + sin (7x-3x)]

but as I said, this method is no longer in the syllabus so I shouldn't really be using this method.

 

[joesmith1975]

i didnt get wat u did lol and its in a textbook maths in focus Ex 5.11 Q 8

 

[Sober]

show:

sin 7x cos 3x = 1/2 [sin(7x +3x) + sin (7x -3x)]

RHS = ½ [sin (7x+3x) + sin (7x-3x)]

= ½ [ sin(7x)·cos(3x) + sin(3x)·cos(7x) + sin(7x)·cos(3x) - sin(3x)·cos(7x) ]

= ½ [ 2·sin(7x)·cos(3x) ]

= sin(7x)·cos(3x)

= LHS

 

[pLuvia]

Ok, now I'm using a method here that is no longer in the syllabus so I wonder why you have been given this question, but I'll do it anyway.

general rule for sums of sin is
sin + sin = 2 sin(1/2 sum) * cos (1/2 difference)
so comparing that to the question,
sin 7x cos 3x
= 1/2 * 2 sin 7x cos 3x
=1/2 [sin (7x+3x) + sin (7x-3x)]

but as I said, this method is no longer in the syllabus so I shouldn't really be using this method.

This method is a 4unit method I believe

 

[Templar]

It's really easy to just expand RHS and show it's the same.

1/2(Sin[7x+3x]+Sin[7x-3x])
=1/2(Sin7xCos3x+Sin3xCos7x+Sin7xCos3x-Sin3xCos7x)
=1/2(2Sin7xCos3x)
=Sin7xCos3x

 

[joesmith1975]

Hence find intergal between pi/4 and 0 of sin7x cos 3x dx

 

[Riviet]

Hence find intergal between pi/4 and 0 of sin7x cos 3x dx

Using your identity found in the previous bit, the integral will become:
π/4
∫ 1/2 [sin(10x) + sin (4x)] dx
0

You can take out the 1/2 out of the integral, since it's a constant; it will also make your working a little easier. Also remember that ∫sin[ax] dx = [-cos(ax)]/a.

 

[joesmith1975]

thanks riviet

 

[Mountain.Dew]

This method is a 4unit method I believe

mmmmmmm not really...its quite simple to show those formulae at the 3U level...

just the application of: sin(A-B) = sinAcosB - cosAsinB ...(1)

sin(A+B) = sinAcosB + cosAsinB....(2)

then, (1) + (2) ==> sin(A-B) + sin(A+B) = sinAcosB - cosAsinB + sinAcosB + cosAsinB

therefore, sinAcosB = 1/2[sin(A-B) + sin(A+B)]

the same method applies for cosAcosB as well, not to mention (1) - (2) ==> similar results.

 

[Riviet]

The sums of products and products of sums formulae are not in the syllabus, but you can sometimes be asked to prove one; of course they give you questions to lead you through the process. The last part of the question usuallys gets you to [hence] use the formula proved in the question to simplify a trig expression.

 

[NiMm]

i was just wondering if these were the correct expansions

R sin (θ – α) = a sin θ – b cos θ

R cos (θ - α) = a cos θ + b sin θ

R cos (θ + α) = a cos θ – b sin θ

R sin (θ + α) = a sin θ + b cos θ


- correct me if im wrong

 

[Mountain.Dew]

i was just wondering if these were the correct expansions

R sin (θ – α) = a sin θ – b cos θ

R cos (θ - α) = a cos θ + b sin θ

R cos (θ + α) = a cos θ – b sin θ

R sin (θ + α) = a sin θ + b cos θ

- correct me if im wrong

ummmm yes, they do look very wrong...

where is the R's and α's in the expansion? or have u substituted them for a and b?

 

[pLuvia]

R sin (θ – α) = a sin θ – b cos θ

R cos (θ - α) = a cos θ + b sin θ

R cos (θ + α) = a cos θ – b sin θ

R sin (θ + α) = a sin θ + b cos θ

Correct ones are

R sin(θ + α) = Rsinθcosα + Rcosθsinα

R sin(θ - α) = Rsinθcosα - Rcosθsinα

R cos(θ + α) = Rcosθcosα - Rsinθsinα

R cos(θ - α) = Rcosθcosα + Rsinθsinα

 

[Riviet]

I like to remember it as sine is straight and cos being gay. This means that in the two expansions of sine, sin and cos go together because sine is straight, but in cosine expansions, the cos's like to go together since they're gay, hence you get the cos.cos. Also remember that the sign stays the same in sine expansions and changes for cosine expansions.