Solving trig equation with identies

**benny92000** · November 21st 2011, 02:17 PM

If sin(2x) =0 tanx - cos 2x, then sinx - cosx =?

I know of all the double angle identies for sin and cosine, and I know tan=sin/cos.
I still can't manage to find sinx-cosx
Perhaps you can get me in the right direction.. I tried subbing a few identites, but apparently not the right ones. And note that this problem isn't for class.... This is from Princeton Review math lvl 2 review.

**skeeter** · November 21st 2011, 02:32 PM

Originally Posted by benny92000

If sin(2x) =0 tanx - cos 2x, then sinx - cosx =?

I know of all the double angle identies for sin and cosine, and I know tan=sin/cos.
I still can't manage to find sinx-cosx
Perhaps you can get me in the right direction.. I tried subbing a few identites, but apparently not the right ones. And note that this problem isn't for class.... This is from Princeton Review math lvl 2 review.

makes no sense ... is that 0 times tanx ?

**benny92000** · November 21st 2011, 03:27 PM

I apologize. There is no 0.

**skeeter** · November 21st 2011, 03:41 PM

If sin(2x) = tanx - cos 2x, then sinx - cosx =?

$\sin(2x) = \tan{x}-\cos(2x)$

$\cos(2x) = \tan{x} - \sin(2x)$

$\cos^2{x} - \sin^2{x} = \tan{x} - \sin(2x)$

$(\cos{x}-\sin{x})(\cos{x}+\sin{x}) = \tan{x} - \sin(2x)$

$\cos{x} - \sin{x} = \frac{\tan{x} - \sin(2x)}{\cos{x}+\sin{x}}$

$\sin{x} - \cos{x} = \frac{\sin(2x) - \tan{x}}{\cos{x}+\sin{x}}$

**benny92000** · November 21st 2011, 04:49 PM

Makes sense. They are looking for a number value, however. Choices are negative square root of 2, 0, 1, 2 times the square root of 2, or not enough information.

**Soroban** · November 21st 2011, 07:21 PM

Hello, benny92000!

$\text{If }\sin 2x \:=\: \tan x - \cos 2x,\,\text{ then }\,\sin x - \cos x \:=$

. . $(a)\;\text{-}\sqrt{2} \qquad (b)\;0 \qquad (c)\;1 \qquad (d)\;2\sqrt{2} \qquad (e)\text{ not enough information}$

Note that: . $x \,\ne\,\tfrac{\pi}{2},\:\tfrac{3\pi}{2}$

We have: . $\sin2x + \cos2x \:=\:\tan x$

. $2\sin x\cos x + 1 - 2\sin^2x \:=\:\tan x$

. . . . $2\sin^2\!x - 2\sin x\cos x \:=\:1 - \tan x$

. . . . $2\sin x(\sin x - \cos x) \:=\: 1 - \frac{\sin x}{\cos x}$

. . . . $2\sin x(\sin x - \cos x) \;=\;\frac{\cos x - \sin x}{\cos x}$

Multiply by $\cos x\!:$
n . . . . . . . . . . . . $2\sin x\cos x(\sin x - \cos x) \;=\;\cos x - \sin x$

. . $2\sin x\cos x(\sin x - \cos x) - (\sin x - \cos x) \;=\;0$

n . . . . . . . . $(\sin x - \cos x)(2\sin x\cos x - 1) \;=\;0$

And we have two equations: . $\begin{Bmatrix}\sin x - \cos x &=& 0 \\ 2\sin x\cos x -1 &=& 0 \end{Bmatrix}$

Both equations have the solution: . $x \:=\:\tfrac{\pi}{4} + \pi n$

Therefore: . $\sin x - \cos x \;=\;0\;\;\;\text{ answer (b)}$

**benny92000** · November 22nd 2011, 03:03 PM

That's pretty detailed. Thank you. How did you immediately eliminate pi/2 and 3pi/2 for x?

**Prove It** · November 22nd 2011, 03:42 PM

Originally Posted by benny92000

That's pretty detailed. Thank you. How did you immediately eliminate pi/2 and 3pi/2 for x?

Because tan(x) = sin(x)/cos(x) and we can't have cos(x) = 0.

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