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jmartinez638

  • one year ago

How do you find all the solutions for 'sin(x/2 - pi/4) = sqrt2/2'?

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  1. phi
    • one year ago
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    solutions imply an equation equations have an = sign your expression does not. is that a typo?

  2. jmartinez638
    • one year ago
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    Yes, So sorry = sqrt2/2

  3. jmartinez638
    • one year ago
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    Edited original.

  4. phi
    • one year ago
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    do you know how to "undo" sin ?

  5. jmartinez638
    • one year ago
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    Like using identities?

  6. phi
    • one year ago
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    more like, do \(\sin^{-1} \) to both sides

  7. phi
    • one year ago
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    \[ \sin^{-1} \left(\sin\left(\frac{x}{2} - \frac{\pi}{4}\right) \right)=\sin^{-1} \frac{\sqrt2}{2} \]

  8. phi
    • one year ago
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    on the left side, the inverse sin of the sin undoes the sin we are left with \[ \frac{x}{2} - \frac{\pi}{4}=\sin^{-1} \frac{\sqrt2}{2} \]

  9. phi
    • one year ago
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    on the right side it is asking for an angle what angle is it where sin of that angle = sqr(2)/2 we want it in radians and it is an angle people memorize (so you should too)

  10. jmartinez638
    • one year ago
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    45 degrees or \[\pi\]/4

  11. phi
    • one year ago
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    yes. but because the question asks for *all* solutions we should eyeball the graph for sin |dw:1440367395779:dw|

  12. phi
    • one year ago
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    so one family of solutions is pi/4 + 2pi n and the other is 3pi/4 + 2pi n where n is any integer

  13. jmartinez638
    • one year ago
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    Oh that makes sense. In relation to the graph especially...

  14. phi
    • one year ago
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    so we should try to solve \[ \frac{x}{2} - \frac{\pi}{4}=\frac{\pi}{4} + 2\pi n\] and also \[ \frac{x}{2} - \frac{\pi}{4}=\frac{3\pi}{4} + 2\pi n\]

  15. jmartinez638
    • one year ago
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    \[(8n+5)\pi \div2\] for the first one

  16. jmartinez638
    • one year ago
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    Is that a viable answer?

  17. phi
    • one year ago
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    \[ \frac{x}{2} - \frac{\pi}{4}=\frac{\pi}{4} + 2\pi n \\ \frac{x}{2} = \frac{\pi}{2} + 2\pi n \\ x= \pi + 4 \pi n \]

  18. jmartinez638
    • one year ago
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    Oops yeah, I must've looked at the second one and not the first :3 Let me try the first...

  19. jmartinez638
    • one year ago
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    I got x=(4n+1)*pi

  20. phi
    • one year ago
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    yes , that looks good

  21. phi
    • one year ago
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    and for the 2nd equation?

  22. jmartinez638
    • one year ago
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    Does (8n+5)π÷2 work?

  23. phi
    • one year ago
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    I don't see how.

  24. jmartinez638
    • one year ago
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    Ah, I seem to have made a mistake. How about \[x=2(2n+1)\pi\]?

  25. phi
    • one year ago
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    yes, that looks good. though people would probably write it as 2pi(2n+1)

  26. phi
    • one year ago
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    and of course we have to say n is any integer

  27. jmartinez638
    • one year ago
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    Ok, of course. So those two equations, with 'n' designated as any integer, would be the solutions to that equation?

  28. phi
    • one year ago
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    yes

  29. jmartinez638
    • one year ago
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    That makes a lot of sense, thank you so much!

  30. phi
    • one year ago
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    yw

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