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Mathematics 7 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

\[\cos \left( x + \frac{ \pi }{ 2 } \right) = - \sin x\]

OpenStudy (jhannybean):

\[\color{red}{\cos(a+b) = \cos(a)\cos(b)-\sin(a)\sin(b)}\]

OpenStudy (anonymous):

cos ( x + pi/2 ) = cos (x) cos (pi/2) - sin (x) sin (pi/2) ?

OpenStudy (jhannybean):

Good. now at \(\dfrac{\pi}{2}\) what is the value of \(\cos(\theta)\) and \(\sin(\theta)\)?

OpenStudy (jhannybean):

think of the unit circle.

OpenStudy (anonymous):

cos (0) sin (1) ? or is it the other way around?

OpenStudy (jhannybean):

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OpenStudy (anonymous):

so cos (1) sin (0) ?

OpenStudy (jhannybean):

No.

OpenStudy (jhannybean):

\[\cos(90^\circ ) = \cos\left(\frac{\pi}{2}\right)=0\]\[\sin(90^\circ ) =\sin\left(\frac{\pi}{2}\right) = 1\]

OpenStudy (jhannybean):

sine functions represent y-values, and cosine functions represent x-values. Remember that.

OpenStudy (anonymous):

i see, i see.

OpenStudy (jhannybean):

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?

OpenStudy (anonymous):

cos ( x + pi/2 ) = cos (x) cos (90) - sin x sin (90) ?

OpenStudy (jhannybean):

Yes, and we sound what cos(90) and sin(90) were, so substitute those in. \[\color{red}{\cos(90^\circ )} = \cos\left(\frac{\pi}{2}\right)=\color{red}{0}\] \[\color{red}{\sin(90^\circ )} =\sin\left(\frac{\pi}{2}\right) = \color{red}{1}\]

OpenStudy (jhannybean):

found*

OpenStudy (anonymous):

cos ( x + pi/2) = cos (x) cos (0) - sin (x) sin(1)?

OpenStudy (jhannybean):

No, replace the values, 0 and 1, in the appropriate places.

OpenStudy (jhannybean):

No sin and cos needed. They EQUAL eachother, therefore sin(90) and cos(90) can be REPLACED by 0 and 1.

OpenStudy (anonymous):

cos (x + pi/2) = 0 - 1

OpenStudy (jhannybean):

I don't think you understand what im saying...

OpenStudy (jhannybean):

\[\color{red}{\cos(90^\circ )} = \cos\left(\frac{\pi}{2}\right)=\color{red}{0}\]\[\color{red}{\sin(90^\circ )} =\sin\left(\frac{\pi}{2}\right) = \color{red}{1}\] \[\cos \left(x+\frac{\pi}{2}\right) = \cos (x)(\color{red}{0}) - \sin (x) (\color{red}{1})\]

OpenStudy (jhannybean):

Do you see what I mean now?

OpenStudy (anonymous):

yes i do!

OpenStudy (jhannybean):

Can you simplify the rest from here?

OpenStudy (anonymous):

what would i do to simplify?

OpenStudy (jhannybean):

Think about what anything multiplied by 0 is, and what happens when you multiply a number by 1.

OpenStudy (anonymous):

cos ( x + pi/2 ) = - sin x i see.

OpenStudy (jhannybean):

Yay.

OpenStudy (anonymous):

sorry if i gave you a hard time, but thank you so much for your help!

OpenStudy (jhannybean):

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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