Ah no wait. I figured it out

But please help with this one:

Prove that (sin2x + cos 2x)^2 = 1 + Sin 4x

This is what i did:

Left side = (cos^2x - 2cos.sin.2x + sin^2x)

= (2x(sin^2 + cos^2) - 2cos.sin.2x)

= 2x - 2cos.sin.2x

Results 1 to 12 of 12

- May 5th 2007, 01:16 AM #1

- May 5th 2007, 01:55 AM #2
Ah no wait. I figured it out

But please help with this one:

Prove that (sin2x + cos 2x)^2 = 1 + Sin 4x

This is what i did:

Left side = (cos^2x - 2cos.sin.2x + sin^2x)

= (2x(sin^2 + cos^2) - 2cos.sin.2x)

= 2x - 2cos.sin.2x

- May 5th 2007, 03:31 AM #3

- May 5th 2007, 03:37 AM #4
This is not an identity. And how did you get that LHS?

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

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

So

sin(2x) + cos(2x) = 1 +sin(4x)

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

2sin(x)cos(x) + 1 - 2sin^2(x) = 1 + 4sin(x)cos(x)(1 - 2sin^2(x))

2sin(x)cos(x) + 1 - 2sin^2(x) = 1 + 4sin(x)cos(x)(1 - 2sin^2(x))

2sin(x)cos(x) - 4sin(x)cos(x)(1 - 2sin^2(x)) = 2sin^2(x)

2cos(x) - 4cos(x)(1 - 2sin(x)) = 2sin(x) <-- We must consider sin(x) = 0 as one of the solutions when we cancel it here.

cos(x)[2 - 4 + 8sin^2(x)] = 2sin(x) <-- Square both sides

cos^2(x)[-2 + 8sin^2(x)]^2 = 4sin^2(x)

(1 - sin^2(x))[4 - 32sin^2(x) + 64sin^4(x)] = 4sin^2(x)

Let's do ourselves a favor now and let y = sin^2(x)

(1 - y)[4 - 32y + 64y^2] = 4y <-- Divide both sides by 4

(1 - y)[1 - 8y + 32y^2] = y

1 - 8y + 32y^2 - y + 8y^2 - 32y^3 = y

32y^3 - 40y^2 + 10y - 1 = 0

This does NOT have a pleasant solution. Numerically I get only one real solution: y = 0.957794.

So

y = sin^2(x) = 0.957794

sin(x) = 0.97867

x = 1.36388, 1.77771 rad

or

x = 0.434137*(pi), 0.565836*(pi)

But we also have

sin(x) = 0

so

x = 0, (pi) rad

-Dan

- May 5th 2007, 04:31 AM #5

- May 5th 2007, 04:40 AM #6

- May 5th 2007, 09:48 AM #7

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Hello, janvdl!

This is a tricky one . . .

Prove that:

. . . . . . . . . . . . . . . .1 - sin(2x)

sin(x) - cos(x) . = . ------------------

. . . . . . . . . . . . . . sin(x) - cos(x)

I cross-multiplied ... just to verify the identity.

. . [I know it's a illegal move in the proof of an identity.]

But that gave me the Big Hint that we're all looking for.

Watch this . . .

[sin(x) - cos(x)]² .= .sin²(x) - 2·sin(x)·cos(x) + cos²(x) .= .1 - 2·sin(2x)

Get it? .That numerator is a perfect square . . .

. . . . . . . . . . . . . . .1 - sin(2x) . . . . . [sin(x) - cos(x)]²

The right side is: . ----------------- . = . -------------------- . = . sin(x) - cos(x)

. . . . . . . . . . . . . sin(x) - cos(x) . . . . . sin(x) - cos(x)

- May 5th 2007, 09:57 AM #8

- May 5th 2007, 10:44 AM #9
Goeien avond,

I can't see Soroban around, therefore I'll give you a hint:

Soroban wrote:

[sin(x) - cos(x)]² = sin²(x) - 2·sin(x)·cos(x) + cos²(x) = 1 - 2·sin(2x) Now change the order of the summands:

[sin(x) - cos(x)]² =**sin²(x) + cos²(x)**- 2·sin(x)·cos(x)= 1 - 2·sin(2x)

I assume that you know the value of**sin²(x) + cos²(x)**

- May 5th 2007, 10:48 AM #10

- May 5th 2007, 11:01 AM #11

- May 5th 2007, 11:13 AM #12

- Joined
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Hello, janvdl!

You are doing*strange*things to the equation . . .

Prove that: .(sin2x + cos 2x)² .= .1 + sin4x

Square out the left side:

. . . . sin²(2x) + 2·sin(2x)·cos(2x) + cos²(2x)

. .= . sin²(2x) + cos²(2x) + 2·sin(2x)·cos(2x)

. - - - \______. .______/ . . \_____ -______/

. . . . . . . . . . \/ . . . . . . . . . . . . .\/

. .= . . . . . . . 1 . . . . . . + . . . . sin(4x)