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nbedwards934 Group Title

tanθ+cotθ=cscθxsecθ (I will draw what I have so far)

  • one year ago
  • one year ago

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  1. nbedwards934 Group Title
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    |dw:1354253316580:dw|

    • one year ago
  2. jidery Group Title
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    |dw:1354253520763:dw|

    • one year ago
  3. zordoloom Group Title
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    Are you trying to prove this identity??

    • one year ago
  4. nbedwards934 Group Title
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    yes I am. I've started working on the side cscθxsecθ but I'm stuck. I drew as far as I got in the first drawing.

    • one year ago
  5. zordoloom Group Title
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    Would you like further help?

    • one year ago
  6. nbedwards934 Group Title
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    yes please

    • one year ago
  7. zordoloom Group Title
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    Very well. Let me start from the beginning them and go step by step.

    • one year ago
  8. nbedwards934 Group Title
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    alright. Thank you

    • one year ago
  9. zordoloom Group Title
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    so. i'll use x instead of theta. start by replacing tanx with (sin/cos) and cotx with (cos/sin) so... (cosx/sinx)+(sinx/cosx)=cscxsecx now replace cscx with (1/sinx) and secx with (1/cos) (cosx/sinx)+(sinx/cosx)=(1/cosx)(1/sinx)

    • one year ago
  10. zordoloom Group Title
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    Are you following so far?

    • one year ago
  11. nbedwards934 Group Title
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    you're working with both sides? My professor told us to only work with one side and leave the other as is.

    • one year ago
  12. zordoloom Group Title
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    Alright, If you want, I can work only on one side.

    • one year ago
  13. nbedwards934 Group Title
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    that would be better, thank you. I started work on the cscθxsecθ side

    • one year ago
  14. zordoloom Group Title
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    okay, would you mind if I work from the left side?

    • one year ago
  15. nbedwards934 Group Title
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    I actually might have figured this one out. Give me just a minute to finish it

    • one year ago
  16. zordoloom Group Title
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    ok

    • one year ago
  17. zordoloom Group Title
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    tanx+cotx=cscxsecx Replace tanx with an equivalent expression (sinx)/(cosx) using the fundamental identities. (sinx)/(cosx)+cotx=cscxsecx Replace (cosx)/(sinx) using the fundamental identities. (sinx)/(cosx)+(cosx)/(sinx)=cscxsecx To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is cosxsinx. Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions. (sinx)/(cosx)*(sinx)/(sinx)+(cosx)/(sinx)*(cosx)/(cosx)=cscxsecx Multiply sinx by sinx to get sinx^(2). (sin^(2)x)/(sinx*cosx)+(cosx)/(sinx)*(cosx)/(cosx)=cscxsecx Multiply cosx by sinx to get sinxcosx. (sin^(2)x)/(sinxcosx)+(cosx)/(sinx)*(cosx)/(cosx)=cscxsecx Multiply cosx by cosx to get cosx^(2). (sin^(2)x)/(sinxcosx)+(cos^(2)x)/(cosx*sinx)=cscxsecx Multiply sinx by cosx to get sinxcosx. (sin^(2)x)/(sinxcosx)+(cos^(2)x)/(sinxcosx)=cscxsecx Combine the numerators of all expressions that have common denominators. (sin^(2)x+cos^(2)x)/(sinxcosx)=cscxsecx Replace sin^(2)x+cos^(2)x with 1 using the identity sin^(2)(x)+cos^(2)(x)=1. (1)/(sinxcosx)=cscxsecx Replace the expressions with an equivalent expression using the fundamental identities. (1)/(sinxcosx)....Replace with cscxsecx Therefore cscxsecx=cscxsecx

    • one year ago
  18. zordoloom Group Title
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    Does that help?

    • one year ago
  19. nbedwards934 Group Title
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    I'll be honest, it's very difficult to interpret the equations you have typed up there, but I'm trying haha

    • one year ago
  20. zordoloom Group Title
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    Here. This might help a bit. It's solved by a computer.

    • one year ago
  21. zordoloom Group Title
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    • one year ago
  22. nbedwards934 Group Title
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    Thank you so much! I didn't even think about changing the sin^2θ+cos^2θ to 1. I have the formula right in front of me, but it didn't register. But now it makes sense. Thanks!

    • one year ago
  23. zordoloom Group Title
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    Yep

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

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