Prove 2cosx=sin2x(sinx)?

Answer 1

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

Start with the more complex side:

RHS = sin(2x)sin(x)

sin(2x) = 2sin(x)cos(x), so

RHS = [2sin(x)cos(x)] [sin(x)]
RHS = 2sin^2(x)cos(x)
RHS = 2[1 - cos^2(x)]cos(x)
RHS = 2cos(x) - cos^3(x)

This isn't an identity, so it's not a proof.

If, however, you meant "Solve for x", look below.

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

Use the property that sin(2x) = 2sin(x)cos(x)

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

Simplify the right hand side.

2cos(x) = 2sin^2(x)cos(x)

Move everything to the left hand side.

2cos(x) - 2sin^2(x)cos(x) = 0

Factor out 2cos(x)

2cos(x) [1 - sin^2(x)] = 0

Now that we have a product equated to 0, equate each factor to 0.

2cos(x) = 0
1 - sin^2(x) = 0

Let's solve these individually. I'm going to assume an interval of [0, 2pi)

2cos(x) = 0 means
cos(x) = 0, which means x = {pi/2, 3pi/2}

1 - sin^2(x) = 0 means
[1 - sin(x)] [1 + sin(x)] = 0, which in turn means

1 - sin(x) = 0
1 + sin(x) = 0

sin(x) = 1, which occurs when x = pi/2
sin(x) = -1, which occurs when x = 3pi/2

Therefore, our solutions are
x = {pi/2, 3pi/2}

Answer 2

Um 2cosx does not equal sin2x(sinx)

no matter how confusing your parentheses are...

2cos0 = 2
sin2*0 = 0
sin0 = 0

2 does not equal 0

edit: oooh, you wanted to solve for x, sorry

Answer 3

This statement isn't true. Note that if x=0, we get 2*1 = 0*0, which isn't true.

If we wanted to find the values of x where this WAS true (in which case we'd just be solving this for x), use the identity sin(2x) = 2sin(x)cos(x). This gives us:

2cos(x) = (2sin(x)cos(x))sin(x))
2cos(x) = 2sin²(x)cos(x)
cos(x) = sin²(x)cos(x)
2 = 2sin²(x)
1 = sin²(x)

So x = n*pi/2 for odd values of x. Note that this teachnically meant having to divide by zero along the way, but the answer fits the original equation because 2cos(n*pi/2) for odd n is going to be 0, and sin(2n*pi/2) = sin(n*pi), which is going to be zero too.