Question

How do you solve `Sin 3 theta = sin theta`?

Answer 1

`theta=tan^(-1)1=pi/4 +k*2pi or (5pi)/4+m*2pi` ; `k,m in ZZ`.

Explanation:

`sin3theta=sintheta`

`therefore sin(2theta+theta)=sintheta`

`thereforesin2thetacostheta+cos2thetasintheta=sintheta`

`therefore2sinthetacosthetacostheta+(cos^2theta-sin^2theta)sintheta=sintheta`

`therefore2cos^2theta+cos^2theta-sin^2theta=1`

`therefore2cos^2theta-sin^2theta=1-cos^2theta=sin^2theta`

`therefore2cos^2theta=2sin^2theta`

`thereforecostheta=sintheta`

`therefore 1=(sintheta)/(costheta)=tantheta>0`

`therefore theta=tan^(-1)1=pi/4 +k*2pi or pi+pi/4+m*2pi` ; `k,m in ZZ`.

Answer 2

There are two sets of solutions:
`theta = pi n`
`theta = pi/4+pi/2 n`
(where `n` - any integer number)

Explanation:

Recall the derivation of a formula for `sin 3theta`:
`sin 3theta = sin(theta+2theta)`
`= sin theta * cos 2theta + sin 2theta * cos theta`
`= sin theta * (cos^2 theta-sin^2 theta) + 2sin theta*cos theta*cos theta`
`= sin theta * (1-2sin^2 theta) + 2sin theta * (1-sin^2 theta)`
`= 3sin theta - 4 sin^3 theta`

Applying this to our equation, we get
`3sin theta - 4 sin^3 theta = sin theta`
or
`2sin theta - 4sin^3 theta = 0`
or
`sin theta*(1-2sin^2 theta) = 0`

This equation has one set of solutions when `sin theta = 0`, that is
(Solution 1.1) `theta = 0+2pi n` and
Solution 1.2) `theta=pi+2pi n`,
which can be combined into
(Solution 1') `theta = pi n`, where `n` - any integer number.

Another set of solution is from
`1-2sin^2 theta = 0`
or
`sin theta = +-sqrt(2)/2`
With "`+`" sign the solutions are
(Solution 2.1) `theta = pi/4 +2pi n` and
(Solution 2.2) `theta = (3pi)/4+2pi n`
With "`-`" sign the solutions are
(Solution 3.1) `theta = -pi/4+2pi n` and
(Solution 3.2) `theta = -(3pi)/4 + 2pi n`
In both cases `n` is any integer number.

NOTE: Solutions 2.1, 2.2, 3.1 and 3.2 can be combined into one expression: `theta = pi/4+pi/2n`, where `n` - any integer number.

CHECK (we can ignore `2pi n` since `2pi` is a period for all participating functions)
Solution 1.1:
Left side equals
`sin (3*0) = 0`
Right side equals
`sin 0 = 0`
Solution 1.2:
Left side equals
`sin (3pi) =` [since `2pi` is a period] `= sin (3pi-2pi) = sin(pi) = 0`
Right side equals
`sin(pi) = 0`
Solution 2.1:
Left side equals
`sin (3*pi/4) = sin (pi-pi/4) =` [since `sin phi=sin(pi-phi)`] `= sin(pi/4) = sqrt(2)/2`
Right side equals
`sin(pi/4) = sqrt(2)/2`
Solution 2.2:
Left side equals
`sin (3*(3pi)/4) = sin ((9pi)/4)=sin((9pi)/4-2pi) = sin(pi/4) = sqrt(2)/2`
Right side equals
`sin ((3pi)/4) = sqrt(2)/2` (see 2.1 above)
Solution 3.1:
Left side equals
`sin (3*(-pi)/4) =`[since `sin(-phi)=-sin(phi)`] `= -sin ((3pi)/4) = - sqrt(2)/2`
Right side equals
`sin((-pi)/4) =`[since `sin(-phi)=-sin(phi)`] `=-sqrt(2)/2`
Solution 3.2:
Left side equals
`sin (3*(-3pi)/4) = sin ((-9pi)/4)=sin((-9pi)/4+2pi) = sin(-pi/4) = -sqrt(2)/2`
Right side equals
`sin ((-3pi)/4) = -sin((3pi)/4) = -sqrt(2)/2`

Answer 3

Either

`theta=(npi)/2+pi/4,"where "n in ZZ`

Or
`theta=kpi,"where "k in ZZ`

Explanation:

The given equation

`sin3theta=sintheta`

`=>sin3theta-sintheta=0`

`=>2cos((3theta+theta)/2)sin((3theta-theta)/2)=0`

`=>2cos(2theta)sintheta=0`

Either
`:.cos(2theta)=0`

`2theta=(2n+1)pi/2=npi+pi/2`

`=>theta=(npi)/2+pi/4,"where "n in ZZ`

Or,

`sintheta=0`

`theta=kpi,"where "k in ZZ`