Verify the identity. cos (x + pi/2) = -sin x

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Verify the identity. cos (x + pi/2) = -sin x

Mathematics
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\[\cos \left( x + \frac{ \pi }{ 2 } \right) = - \sin x\]
\[\color{red}{\cos(a+b) = \cos(a)\cos(b)-\sin(a)\sin(b)}\]
cos ( x + pi/2 ) = cos (x) cos (pi/2) - sin (x) sin (pi/2) ?

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Good. now at \(\dfrac{\pi}{2}\) what is the value of \(\cos(\theta)\) and \(\sin(\theta)\)?
think of the unit circle.
cos (0) sin (1) ? or is it the other way around?
|dw:1440138751966:dw|
so cos (1) sin (0) ?
No.
\[\cos(90^\circ ) = \cos\left(\frac{\pi}{2}\right)=0\]\[\sin(90^\circ ) =\sin\left(\frac{\pi}{2}\right) = 1\]
sine functions represent y-values, and cosine functions represent x-values. Remember that.
i see, i see.
going back to our function , \(\color{red}{\cos(x+\frac{\pi}{2} ) = \cos(x)\cos(\frac{\pi}{2})-\sin(x)\sin(\frac{\pi}{2})}\) can you replace the newfound values and solve for it?
cos ( x + pi/2 ) = cos (x) cos (90) - sin x sin (90) ?
Yes, and we sound what cos(90) and sin(90) were, so substitute those in. \[\color{red}{\cos(90^\circ )} = \cos\left(\frac{\pi}{2}\right)=\color{red}{0}\] \[\color{red}{\sin(90^\circ )} =\sin\left(\frac{\pi}{2}\right) = \color{red}{1}\]
found*
cos ( x + pi/2) = cos (x) cos (0) - sin (x) sin(1)?
No, replace the values, 0 and 1, in the appropriate places.
No sin and cos needed. They EQUAL eachother, therefore sin(90) and cos(90) can be REPLACED by 0 and 1.
cos (x + pi/2) = 0 - 1
I don't think you understand what im saying...
\[\color{red}{\cos(90^\circ )} = \cos\left(\frac{\pi}{2}\right)=\color{red}{0}\]\[\color{red}{\sin(90^\circ )} =\sin\left(\frac{\pi}{2}\right) = \color{red}{1}\] \[\cos \left(x+\frac{\pi}{2}\right) = \cos (x)(\color{red}{0}) - \sin (x) (\color{red}{1})\]
Do you see what I mean now?
yes i do!
Can you simplify the rest from here?
what would i do to simplify?
Think about what anything multiplied by 0 is, and what happens when you multiply a number by 1.
cos ( x + pi/2 ) = - sin x i see.
Yay.
sorry if i gave you a hard time, but thank you so much for your help!
That's ok. As long as you understand the method used so you can ask new questions instead of ones where you're applying the same method over and over again.

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