Trig Proofs....ugh

HopeWelch

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Oct 31, 2011
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sin(2x) = -2sinx sin(x - Pi/2)

Ok so I know that sin(2x) is like saying

sin(x+x) which is sinx cosx + cosx sinx

and I know that Pi/2 is 90 degrees

and I know that cosx = sin(90-x) but the right side of the equation is sin(x - Pi/2) not sin(Pi/2 - x)

and I've put all this back into the proof and it's not going anywhere...

What am I missing here? Help would be greatly appreciated.

Edited this post because I forgot to mention I'm only allowed to manipulate the left side of the equation.
 
It is easier than that. Note that \(\displaystyle sin(x-\frac{\pi}{2})=-cos(x)\)

And \(\displaystyle sin(2x)=2sin(x)cos(x)\)

Now see it?.
 
Wow that was easy. Why is sin(x - Pi/2) = -cos(x) ?

Is it because sin(x - Pi/2) = sin[-(Pi/2 - x)] which is then -sin(Pi/2 - x) or -cosx ?

Just making sure that the way I see this is correct.

And thank you for your help!
 
So here is the proof, does this make sense algebraically?

sin(2x) = -2sinx sin(x - Pi/2)

2sinx cosx = " "

2sinx (-cosx) = " "

2sinx sin[-(90 - x)] = " "

-2sinx sin(x -Pi/2) = -2sinx sin(x - Pi/2)
 
Hello, HopeWelch!

You are expected to know these identities:

. . \(\displaystyle \sin(A - B) \:=\:\sin A\cos B - \cos A\sin B\)

. . \(\displaystyle \sin2A \:=\:2\sin A\cos A\)

\(\displaystyle \text{Prove: }\:\sin 2x \:=\: -2\sin x \sin(x - \frac{\pi}{2})\)

On the right:.\(\displaystyle -2\sin x\sin\left(x - \frac{\pi}{2}\right)\)

. . . . . . . \(\displaystyle = \;-2\sin x\left(\sin x\cos\frac{\pi}{2} - \cos x\sin\frac{\pi}{2}\right)\)

. . . . . . . \(\displaystyle =\;-2\sin x\big[\sin x\cdot 0 - \cos x\cdot 1\big]\)

. . . . . . . \(\displaystyle =\;2\sin x\cos x\)

. . . . . . . \(\displaystyle =\;\sin2x\)
 
I know those identities but I can only manipulate the left side. I'm not allowed to touch the right. I get it now though looking at your steps. I was trying to use the sinx = cos(90 - x) instead of just doing the cosine of 90 degrees is 0...I see it now. Thank you!!
 
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