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anonymous
 one year ago
Verify the identity.
cos (x + pi/2) = sin x
anonymous
 one year ago
Verify the identity. cos (x + pi/2) = sin x

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\cos \left( x + \frac{ \pi }{ 2 } \right) =  \sin x\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4\[\color{red}{\cos(a+b) = \cos(a)\cos(b)\sin(a)\sin(b)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos ( x + pi/2 ) = cos (x) cos (pi/2)  sin (x) sin (pi/2) ?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4Good. now at \(\dfrac{\pi}{2}\) what is the value of \(\cos(\theta)\) and \(\sin(\theta)\)?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4think of the unit circle.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos (0) sin (1) ? or is it the other way around?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4dw:1440138751966:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so cos (1) sin (0) ?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4\[\cos(90^\circ ) = \cos\left(\frac{\pi}{2}\right)=0\]\[\sin(90^\circ ) =\sin\left(\frac{\pi}{2}\right) = 1\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4sine functions represent yvalues, and cosine functions represent xvalues. Remember that.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4going 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos ( x + pi/2 ) = cos (x) cos (90)  sin x sin (90) ?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4Yes, and we sound what cos(90) and sin(90) were, so substitute those in. \[\color{red}{\cos(90^\circ )} = \cos\left(\frac{\pi}{2}\right)=\color{red}{0}\] \[\color{red}{\sin(90^\circ )} =\sin\left(\frac{\pi}{2}\right) = \color{red}{1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos ( x + pi/2) = cos (x) cos (0)  sin (x) sin(1)?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4No, replace the values, 0 and 1, in the appropriate places.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4No sin and cos needed. They EQUAL eachother, therefore sin(90) and cos(90) can be REPLACED by 0 and 1.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos (x + pi/2) = 0  1

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4I don't think you understand what im saying...

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4\[\color{red}{\cos(90^\circ )} = \cos\left(\frac{\pi}{2}\right)=\color{red}{0}\]\[\color{red}{\sin(90^\circ )} =\sin\left(\frac{\pi}{2}\right) = \color{red}{1}\] \[\cos \left(x+\frac{\pi}{2}\right) = \cos (x)(\color{red}{0})  \sin (x) (\color{red}{1})\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4Do you see what I mean now?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4Can you simplify the rest from here?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what would i do to simplify?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4Think about what anything multiplied by 0 is, and what happens when you multiply a number by 1.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos ( x + pi/2 ) =  sin x i see.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry if i gave you a hard time, but thank you so much for your help!

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.4That'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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