Double-Angle Identities!

[PowerXtremeBaby]

Hi! We briefly covered double angle identities in class, and my professor says they are going to be on the final exam tomorrow and gave us some problems to look at in order to study them.

Problem is, the lesson was so brief that I don't understand them at all!

1) sin(2x)+sin(x)=0
all I know is that the identity is sin(2x)=2sin(x)cos(x)

2) sin(2x)cos(x)=sin(x)
with the same identity sin(2x)=2sin(x)cos(x)

The book gives like one example on double angle for sine, which is laid out confusingly and doesn't make any sense! Please help!

 

[Aladdin]

#1 ) sin(2x)+sin(x)=0 ---- sin(2x)=2sin(x)cos(x)

2sin(x)cos(x)+sin(x)=0

(sinx)(2cos(x)+1)=0

sinx=0 Solve x

2cos(x)=-1

cos(x)=-1/2 Solve x .

#2)sin(2x)cos(x)=sin(x)

Think about it .

 

[PowerXtremeBaby]

So would this be right?
1)
sin(2x)+sin(x)=0
2 sinxcosx+sinx=0
sinx(2cosx+1)=0
sinxx=0 or cosx=-1/2=cos(2pi/3)
x=kpi or 2kpi(+/-)2pi/3
where K is an element of Z(integers)

2)
sin(2x) = 2sin(x)cos(x)
so:
2sin(x)cos(x)^2 = sin(x)
divide both sides by sin(x),
2cos(x)^2 = 1
sin(x) = 0
so when sin(x) = 0, x = 0, pi,2pi... so k*pi, where K is an element of Z(integers)
When cos(x)^2 = 1/2
cos(x) = 1/root(2)
x = pi/4, 7pi/4, plus 2pi revolutions.

 

[Aladdin]

I think So , Well done ... :wink:

 

[PowerXtremeBaby]

Okay good, I think I understand it! Thank you!