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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    \[\cos \left( x + \frac{ \pi }{ 2 } \right) = - \sin x\]

  2. Jhannybean
    • one year ago
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    \[\color{red}{\cos(a+b) = \cos(a)\cos(b)-\sin(a)\sin(b)}\]

  3. anonymous
    • one year ago
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    cos ( x + pi/2 ) = cos (x) cos (pi/2) - sin (x) sin (pi/2) ?

  4. Jhannybean
    • one year ago
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    Good. now at \(\dfrac{\pi}{2}\) what is the value of \(\cos(\theta)\) and \(\sin(\theta)\)?

  5. Jhannybean
    • one year ago
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    think of the unit circle.

  6. anonymous
    • one year ago
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    cos (0) sin (1) ? or is it the other way around?

  7. Jhannybean
    • one year ago
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    |dw:1440138751966:dw|

  8. anonymous
    • one year ago
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    so cos (1) sin (0) ?

  9. Jhannybean
    • one year ago
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    No.

  10. Jhannybean
    • one year ago
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    \[\cos(90^\circ ) = \cos\left(\frac{\pi}{2}\right)=0\]\[\sin(90^\circ ) =\sin\left(\frac{\pi}{2}\right) = 1\]

  11. Jhannybean
    • one year ago
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    sine functions represent y-values, and cosine functions represent x-values. Remember that.

  12. anonymous
    • one year ago
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    i see, i see.

  13. Jhannybean
    • one year ago
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    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?

  14. anonymous
    • one year ago
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    cos ( x + pi/2 ) = cos (x) cos (90) - sin x sin (90) ?

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

  16. Jhannybean
    • one year ago
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    found*

  17. anonymous
    • one year ago
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    cos ( x + pi/2) = cos (x) cos (0) - sin (x) sin(1)?

  18. Jhannybean
    • one year ago
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    No, replace the values, 0 and 1, in the appropriate places.

  19. Jhannybean
    • one year ago
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    No sin and cos needed. They EQUAL eachother, therefore sin(90) and cos(90) can be REPLACED by 0 and 1.

  20. anonymous
    • one year ago
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    cos (x + pi/2) = 0 - 1

  21. Jhannybean
    • one year ago
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    I don't think you understand what im saying...

  22. Jhannybean
    • one year ago
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    \[\color{red}{\cos(90^\circ )} = \cos\left(\frac{\pi}{2}\right)=\color{red}{0}\]\[\color{red}{\sin(90^\circ )} =\sin\left(\frac{\pi}{2}\right) = \color{red}{1}\] \[\cos \left(x+\frac{\pi}{2}\right) = \cos (x)(\color{red}{0}) - \sin (x) (\color{red}{1})\]

  23. Jhannybean
    • one year ago
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    Do you see what I mean now?

  24. anonymous
    • one year ago
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    yes i do!

  25. Jhannybean
    • one year ago
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    Can you simplify the rest from here?

  26. anonymous
    • one year ago
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    what would i do to simplify?

  27. Jhannybean
    • one year ago
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    Think about what anything multiplied by 0 is, and what happens when you multiply a number by 1.

  28. anonymous
    • one year ago
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    cos ( x + pi/2 ) = - sin x i see.

  29. Jhannybean
    • one year ago
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    Yay.

  30. anonymous
    • one year ago
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    sorry if i gave you a hard time, but thank you so much for your help!

  31. Jhannybean
    • one year ago
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    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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