Double angle formula

[igwun]

Using the double angle formula sin2?=sin?cos? find all ? such that sin2?=sin?.

I cannot figure this out. This is what I have done so far:

sin?cos?-sin?=0
sin?(cos?-1)=0

What do I do next? Thanks!

 

[Deleted member 4993]

Using the double angle formula sin2?=sin?cos? find all ? such that sin2?=sin?.

I cannot figure this out. This is what I have done so far:

2sin?cos?-sin?=0 << corrected
sin?(2cos?-1)=0<< corrected

sin? = 0 ? ? = n * ? (where n = 0, ± 1, ±2,...)

or

2cos? - 1 = 0 ? cos? = 1/2 ? ???
What do I do next? Thanks!

 

[soroban]

Hello, igwun!

That formula is incorrect.

I'll assume that the answers are: .\(\displaystyle 0 \leq \theta < 2\pi\)

Using the double-angle formula \(\displaystyle \sin2\theta \:=\) 2\(\displaystyle \sin\theta \cos\theta\)
find all \(\displaystyle \theta\) such that: .\(\displaystyle \sin2\theta \:=\:\sin\theta\)

We have: .\(\displaystyle 2\sin\theta\cos\theta \:=\:\sin\theta \quad\Rightarrow\quad 2\sin\theta\cos\theta - \sin\theta \:=\:0\)

Factor: .\(\displaystyle \sin\theta(2\cos\theta - 1) \:=\:0\)


Set each factor equal to zero and solve . . .

. . \(\displaystyle \sin\theta \:=\:0 \quad\Rightarrow\quad\boxed{ \theta \:=\:0.\:\pi}\)

. . \(\displaystyle 2\cos\theta -1 \:=\:0 \quad\Rightarrow\quad \cos\theta \:=\:\tfrac{1}{2} \quad\Rightarrow\quad \boxed{\theta \:=\:\tfrac{\pi}{3},\:\tfrac{5\pi}{3}}\)