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Atkinsoha

  • one year ago

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

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

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

  4. getusel
    • one year ago
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    sure

  5. getusel
    • one year ago
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    sorry

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

  7. getusel
    • one year ago
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    is this the question

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

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

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

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

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

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

  14. getusel
    • one year ago
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    done

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

  16. getusel
    • one year ago
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    for sure

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

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

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

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

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

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

  23. getusel
    • one year ago
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    does it make sense

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

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

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

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

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

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

  30. getusel
    • one year ago
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    it is verified

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

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

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

  34. getusel
    • one year ago
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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.

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

  36. getusel
    • one year ago
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    it is my pleasure

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