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Atkinsoha

  • 2 years ago

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

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

  3. Atkinsoha
    • 2 years ago
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    Can you show me how to figure that out?

  4. getusel
    • 2 years ago
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    sure

  5. getusel
    • 2 years ago
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    sorry

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

  7. getusel
    • 2 years ago
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    is this the question

  8. getusel
    • 2 years ago
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    because in the picture the question does not seem right

  9. getusel
    • 2 years ago
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    if this is the question

  10. getusel
    • 2 years 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
    • 2 years ago
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    having this in mind, let us insert it in the original expression

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

  13. getusel
    • 2 years ago
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    cosx/sinx=cotx=1/tanx

  14. getusel
    • 2 years ago
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    done

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

  16. getusel
    • 2 years ago
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    for sure

  17. Atkinsoha
    • 2 years ago
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    okay, hold on let me get it..

  18. Atkinsoha
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Here is the final answer

  21. Atkinsoha
    • 2 years ago
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    So what's the final answer?

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

  23. getusel
    • 2 years ago
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    does it make sense

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

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

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

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

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

  29. getusel
    • 2 years 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
    • 2 years ago
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    it is verified

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

  32. getusel
    • 2 years 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
    • 2 years ago
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    How do you know if it has significance to simplify it or not?

  34. getusel
    • 2 years 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
    • 2 years ago
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    ohh okay. thanks.

  36. getusel
    • 2 years ago
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    it is my pleasure

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is replying to Can someone tell me what button the professor is hitting...

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