C4 Solve: sin4theta = cos 2theta

Newyork Nagaram...

solve the following equation for -Pi<0<Pi
sin4theta = cos 2theta :frown:

I tried sin(2theta+2theta) but still :frown:

Please provide full working.
Thx :smile:

Reply 1

alan.accordion

I reckon this'll work; I'll be using x instead of theta as it's more concise.

sin(4x)=cos(2x)
expand LHS using sin double angle formula
2sin(2x)cos(2x)=cos(2x)
divide both sides by cos(2x)
2sin(2x)=1
divide by two
sin(2x)=1/2

can you take it from here? (hint: I think there are four solutions in the range...)

Note: I think this is right, but haven't done this for a while now...
If it is wrong the error will be from dividing by cos(2x) - do reply and let me know if this is or isn't valid.

Hope this helps!

(edited 13 years ago)

Reply 2

TwilightKnight

Original post by alan.accordion

You can't (or shouldn't) divide by cos2x here. It's part of the answer you're looking for, so by dividing through by it, you're elminating (potentially) 2 solutions.

I would keep extrapolating by using the double angle formulae. It will reduce, eventually, to an equation containing only one circle function.

Reply 3

Zuzuzu

Original post by alan.accordion

I reckon this'll work; I'll be using x instead of theta as it's more concise.

sin(4x)=cos(2x)
expand LHS using sin double angle formula
2sin(2x)cos(2x)=cos(2x)

Up to here is right, but remember cos(2x) could be equal to 0 so you can't divide through by it.

Reply 4

blue_shift86

Original post by mufas

solve the following equation for -Pi<0<Pi
sin4theta = cos 2theta :frown:

I tried sin(2theta+2theta) but still :frown:

Please provide full working.
Thx :smile:

did this in 30 secs :P. Not the solution but the method. Sin4A= 2sin(2A). Cos (2A)

now sin 4A = cos 2A, thus: 2sin 2A.cos2A-Cos2A = 0

Factorising: Cos 2A*(2Sin2A -1) = 0.

You solve this between the limits. :smile:

Reply 5

Newyork Nagaram...

Original post by blue_shift86

huh

? Didn't get it :frown:

Reply 6

TwilightKnight

Original post by mufas

huh

? Didn't get it :frown:

Essentially, he's saying you can take the Cos2x over to the other side, expand Sin4x (Sin(2x+2x)using the double angle identity, and take out Cos2x as a factor.

It's a legitimate way to do it. You just have to remember that each trig function you're doing is in the interval 0<x<4pi, so you're going to have a fair few solutions, (i.e, go through 2 periods of the respective trig graphs, or around the unit circle twice).

Reply 7

Piecewise

Hint: note that

\cos{2\theta} = \sin\left(2\theta+\frac{\pi}{2} \right)

and recall that if

\sin{\lambda} = \sin{\alpha}

then

\lambda = 2n\pi+\alpha

(2n+1)\pi-\alpha

Reply 8

blue_shift86

Original post by mufas

huh

? Didn't get it :frown:

Apologies, let me try again:

Q) sin (4x) = cos (2x)

Sin (A+B) = SinA. CosB + SinB. Cos A. [Sin two angle identity)
If you let A and B both be 2x, then you get:

LHS: Sin (2x+2x) = Sin(2x).Cos(2x) + Sin(2x).Cos(2x) = 2Sin(2x).Cos(2x)
RHS = cos (2x)

Now LHS=RHS

2Sin(2x). Cos(2x) = Cos (2x) [take cos 2x to other side)
2Sin(2x).Cos(2x) - Cos (2x) = 0
Now factorise out Cos (2x)

So: Cos(2x).[2Sin(2x) - 1] = 0

Now you just solve it normally between -pi and +pi.

I hope this helps. Sorry for being a bit late. I've been gardening for the last few hours :smile:

Reply 9

Zuzuzu