Question

Solve: sin(x/2) = 1 - cos x

Answer 1

try using half angle formulas

Answer 2

sin(x/2) can be derived

Answer 3

A couple of hints:

\(\sin(2x) = 2\sin(x)\cos(x) \implies \sin(x) = 2\sin(x/2)\cos(x/2)\)

\(\cos(2x) = 1 - 2\sin^{2}(x) \implies \cos(x) = 1 - 2\sin^{2}(x/2)\)

\(\cos(2x) = 2\cos^{2}(x) - 1 \implies \cos(x) = 2\cos^{2}(x/2) - 1\)

Answer 4

I attempted to use sqare root of 1-cos /2 , then sqare both sides, then make the left side equal to zero so i can factor but i don't know what to do after that so
i suspect i'm doing it wrong..

Answer 5

I rather like the middle one. That seems to solve it directly.

Answer 6

I don't understand y cos(2x) came into play for this problem

Answer 7

for the middle response..

Answer 8

There are quite a few identities that we tried to drum into your head in your trig class. These are just a few of them. In this case, we need to relate, somehow, sin(x/2) with cos(x), both given in the problem statement. I just happen to have these identities floating around in my head and that one seemed helpful for this problem.

The Problem Statement: sin(x/2) = 1 - cos x
The Identity cos(x) = 1 - 2sin^2(x/2)
The identity, slightly modified: 2sin^2(x/2) = 1 - cos(x)

Well then, if they are both equal to 1 - cos(x), the transitive property of equals suggests:

sin(x/2) = 2sin^{2}(x/2)

or

sin(x/2) - 2sin^{2}(x/2) = 0

or

sin(x/2)(1 - 2sin(x/2)) = 0

Are we getting anywhere?

Answer 9

Note: When I wrote the three identities, I first wrote them in the way they usually appear and the way I tend to have them memorized. Then I rewrote them in a form that might be more useful to this specific problem.

Answer 10

The middle identity can also be the source of the half-angle identity that you attempted to use. That should work fine. Something just went wrong in your algebra.

Answer 11

My identities say that sin(x/2)=sqroot(1-cosx/2). So I substituted that for sin(x/2).
Now it reads that sqroot(1-cosx/2)=1-cosx.
Then I squared both sides to eliminate the sqroot.
Now it reads: 1-cosx/2=(1-cosx)(1-cosx).
Then I did some algebra and ended with 2cos^2x+5cosx+1=0, but I was unable to factor this out, unless the quadratic formula is needed.
Is this a valid process to solve this problem?

Answer 12

Asked and answered. No reason that shouldn't work. Let's see...

\(\sqrt{\dfrac{1-\cos(x)}{2}} = 1-\cos(x)\)

Squaring, since everything is positive, this is not dangerous.

\(\dfrac{1-\cos(x)}{2} = (1-\cos(x))^{2} = 1 - 2\cos(x) + \cos^{2}(x)\)

\(1-\cos(x) = 2 - 4\cos(x) + 2\cos^{2}(x)\)

\(2\cos^{2}(x) - 3\cos(x) + 1 = 0\) -- Looks like you missed a little, there, on the middle term.

Should be able to factor that.

Compare it with my previous process: \(\sin(x/2)(1 - 2\sin(x/2)) = 0\)

Doesn't much look the same, does it? :-)

Answer 13

Thank you! Silly me can now soundly sleep tonight