Precalculus challenge: can we just cancel out the sine?

We can solve sin(2x)=sin(x) by using the double angle identity for sine but how do we solve the trig equation sin(πx)=sin(x)? Here I will show you a way to "cancel out the sine" when we have sine on both sides of an equation. Lastly, make sure you try cos(πθ)=cos(θ)

0:00 A challenge for all precalc and trig students
0:35 Q1 sin(2θ)=sin(θ)
4:12 Q2 sin(πθ)=sin(θ)
11:07 Check out Brilliant to learn more!
12:04 Question for you! cos(πθ)=cos(θ)

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Thank you,
blackpenredpen

[avi]

I always use these formulas:
Cos(a)=cos(b)
=> a=2npi+-b.
Sin(a)=sin(b)
=> a=npi+b(-1)^n
Tan(a)=tan(b)
=> a=npi + b

Really useful and a life changer for me.

[kingarthr]

Pre-calc feels so long ago yet it's only been 2 years but I'm doing abstract algebra and analysis now
Time flies when you're having fun

[jamesn.]

I haven't watched bprp for almost 2 years back when the sub count wasnt this high. Really happy this channel is doing well. And also really surprised about the beard, but you do look wizard though

[TheRealMaster]

The usage of +2πn and n E Z allows the cancellation of the sin function by maintaining a representation of all valid solutions which would otherwise be lost when trying to take the inverse of a function that is not one to one.

[jatem_rabei]

What i have learnt in school, using the unit circle, was what you showed on the second equation.
I always use that strategy solving trig equations
sin(a)=sin(b)
a=b+2pik or a=pi-b+2pik,
k being an integer

Similar thing for cosine and tangent

[VerSalieri]

sina=sinb implies that:
a = b + 2kpi or a = pi - b + 2kpi. (sine function is symmetric with respect to y-axis, which has polar equation of theta=pi/2, so a=b or a=2theta -b.... with a period of 2pi)

cosa = cosb implies that:
a = b + 2kpi. or a = -b + 2kpi. (Cosine function is symmetric with respect to x-axis which has the equation theta =0...that is why we get + or - theta)


tana= tanb implies that a = b + kpi. (tangent function is symmetric wrt to origin O)


no need to use double angle formula.... no need for sketches also.. but the sketches help see it, rather than memorizing techniques if solving.

Example:
sin2x = sinx,
2x=x+2kpi. or 2x=pi-x+2kpi
x=2kpi. or x=pi/3 +2kpi/3

[hemantbhosale]

You would have shifted sin(theta) on the left you will get sin(πtheta)-sin(theta) solve according to sin(a)-sin(b) formula and generalize the solution your way is also amazing but this is also a way of solving

[minhdoantuan]

cos(pi*O)=cos(O) (O is theta)
pi*O= O+2k*pi or pi*O= -O+2k*pi (k is integer)
(pi-1)*O=2k*pi or (pi+1)*O=2k*pi
O=(2k*pi)/(pi-1) or O=(2k*pi)/(pi+1)

[christianpalumbo]

BPRP! I have an easy solution for this.
When two sines are equal, it means either that their arguments are equal or that they add up to π.

1) sin(2x) = sin(x)
So 2x = x + 2πn, which means x = 2πn
or 2x = π - x + 2πn, which means x = π/3 + 2πn/3 = π(2n + 1)/3.

2) sin(πx) = sin(x)
So πx = x + 2πn, which leads to x = 2πn/(π - 1)
or πx = π - x + 2πn, which leads to x = π/(π + 1) + 2πn/(π + 1) = π(2n + 1)/(π + 1).

[MushookieMan]

I solved it by thinking of the intersection of a horizontal line with the sine curve. That allowed me to catch both cases.

[tsa_gamer]

1. Theta=60
2.Theta=180/(π+1)
You may check in calculator for
1.sin(theta)=sin(2theta)
2.sin(πtheta) =sin(theta)

[ilickcatnip]

Hey there! Yes we have an identity that can be used to solve the second problem. Whenever we reach to sinø=sin@, we can simply say
ø = nπ +(-1)ⁿ@ . Use this for your second problem and you'll get the answer real quick. It will automatically cover both the cases depending upon whether n is even or odd.

[GRBtutorials]

I’m by no means a precalc student, but I didn’t use calculus, so here’s the set of solutions I found: {k ∈ ℤ | 2kπ/(π-1) ∨ π - 2kπ/(π-1)}.

[Adomas_B]

To quote a song
Pre calculus didn't help me prepare for calculus, for calculus, help meee

[nabayansaha]

For second one,
sin(πa)-sina= 0
2 cos( (πa+a)/2) sin((πa-a)/2)= 0
so,
either, sin term or cos term is zero . Hence I get 2 solution sets .

[prabhatdubey]

Hey, I have one

π(theta)=nπ + ((-1)^n)theta
Where n is an integer

[MichaelRothwell]

Here are the 3 solutions for solving sin A = sin B I have seen so far:
1. (from video) B=2nπ+A or B=(2n+1)π-A
2. (from a comment) B=nπ+((-1)^n)A
3. (from several comments) 0=sin A - sin B =2cos((A+B)/2)sin((A-B)/2) ["sum to product" formula]; now solve cos((A+B)/2)=0 and sin((A-B)/2)=0 [straight forward]

[sayandeogharia]

we can also use SinA-SinB = 2Cos((A+B)/2)Sin((A-B)/2) then compute..Lil bit of overkill i guess ...

[yogpanjarale]

0:26
In trigonometric equations chapter class 11
We are taught identity for sin (x) = sin (y)
Then x = n(pi) + (-1)to power n

Using x and y it is theta and alpha

[bensenliu]

could we use sum to product for question 2 as well?