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Atkinsoha Group Title

Verify, then find which is NOT equivalent : (Picture below)

  • one year ago
  • one year ago

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  1. Atkinsoha Group Title
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    • one year ago
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  2. getusel Group Title
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    \[\frac{ cosx }{ sinx }\]

    • one year ago
  3. Atkinsoha Group Title
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    Can you show me how to figure that out?

    • one year ago
  4. getusel Group Title
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    sure

    • one year ago
  5. getusel Group Title
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    sorry

    • one year ago
  6. getusel Group Title
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    \[\sin(\pi/2-x)\csc\]

    • one year ago
  7. getusel Group Title
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    is this the question

    • one year ago
  8. getusel Group Title
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    because in the picture the question does not seem right

    • one year ago
  9. getusel Group Title
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    if this is the question

    • one year ago
  10. getusel Group Title
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    \[\sin(\frac{ \pi }{2 }-x)=\sin(\frac{ \pi }{ 2 })cosx-\cos(\frac{ \pi }{ 2 })sinx=1.cosx-0.sinx=cosx\]

    • one year ago
  11. getusel Group Title
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    having this in mind, let us insert it in the original expression

    • one year ago
  12. getusel Group Title
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    \[\sin(\frac{ \pi }{2 }-x)cscx=cosxcscx=cosx.\frac{ 1 }{ sinx }\]

    • one year ago
  13. getusel Group Title
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    cosx/sinx=cotx=1/tanx

    • one year ago
  14. getusel Group Title
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    done

    • one year ago
  15. Atkinsoha Group Title
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    Thank you! I have another one that maybe you could help me with?

    • one year ago
  16. getusel Group Title
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    for sure

    • one year ago
  17. Atkinsoha Group Title
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    okay, hold on let me get it..

    • one year ago
  18. Atkinsoha Group Title
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    If you could do it all in one message that would be great so I can follow along easier.

    • one year ago
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  19. getusel Group Title
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    \[\frac{ cscA-sinA }{ \cot ^{2}A }=\frac{\frac{ 1 }{sinA }-sinA }{ \frac{ \cos ^{2}A }{ \sin ^{2}A } }=\frac{ \frac{ 1-\sin ^{2}A }{sinA } }{ \frac{ \cos ^{2}A }{\sin ^{2} A} }\] \[\frac{ 1-\sin ^{2}A }{sinA }.\frac{ \sin ^{2} A}{ \cos ^{2}A }=\frac{ \cos ^{2}A sinA}{ cosA }=sinAcosA=\frac{ 1 }{2 }\sin(2A)\]

    • one year ago
  20. getusel Group Title
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    Here is the final answer

    • one year ago
  21. Atkinsoha Group Title
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    So what's the final answer?

    • one year ago
  22. getusel Group Title
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    \[\frac{ 1 }{ 2 }\sin(2A)\]

    • one year ago
  23. getusel Group Title
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    does it make sense

    • one year ago
  24. Atkinsoha Group Title
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    uhmm.. yeah sort of. im terrible at this whole simplifying thing. up for a few more?

    • one year ago
  25. getusel Group Title
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    If you practice, it is not that hard my friend

    • one year ago
  26. Atkinsoha Group Title
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    i got all of the verifying ones but this one..

    • one year ago
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  27. getusel Group Title
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    I am going for launch. If you stay for 30 minutes i will get back to you.

    • one year ago
  28. Atkinsoha Group Title
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    I'll be here all evening. Thank you! Have a good time :)

    • one year ago
  29. getusel Group Title
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    \[\frac{ \cos \alpha-\cos \beta }{ \sin \alpha+\sin \beta }+\frac{ \sin \alpha-\sin \beta }{ \cos \alpha+\cos \beta}=\] \[\frac{ \cos ^{2}\alpha-\cos ^{2}\beta+\sin ^{2}\alpha-\sin^2\beta }{(\sin \alpha+\sin \beta)(\cos \alpha +\cos \beta)}\] =\[\frac{ \cos^2 \alpha+\sin^2\alpha-(\sin^2\beta +\cos^2\beta) }{(\sin \alpha+\sin \beta)(\cos \alpha+\cos \beta) }=\frac{ 1-1 }{(\sin \alpha+\sin \beta) (\cos \alpha+\cos \beta) }=0\]

    • one year ago
  30. getusel Group Title
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    it is verified

    • one year ago
  31. Atkinsoha Group Title
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    how is the denominator verified? nothing was done to it?

    • one year ago
  32. getusel Group Title
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    since it does not have any significance to simplify the denominator, we can leave it as it is.

    • one year ago
  33. Atkinsoha Group Title
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    How do you know if it has significance to simplify it or not?

    • one year ago
  34. getusel Group Title
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    In this particular problem, the purpose is to simplify the right hand side to zero. That tells us the numerator of the RHS has to be siplified to zero. our focus should be mostly on working out the numerator.

    • one year ago
  35. Atkinsoha Group Title
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    ohh okay. thanks.

    • one year ago
  36. getusel Group Title
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    it is my pleasure

    • one year ago
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is replying to Can someone tell me what button the professor is hitting...

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