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snowflakepixie Group Title

Verify (1/sinx+1) + (1/cscx+1) = 1

  • one year ago
  • one year ago

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  1. zepdrix Group Title
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    \[\large \frac{1}{1+\sin x}+\frac{1}{\csc x+1}\] Let's do some manipulation. We're going to take advantage of two of the three basic square identities that you first learn in trig.\[\large \color{royalblue}{\sin^2x+\cos^2x=1 \qquad \qquad \cot^2x+1=\csc^2x}\] In our problem, let's multiply the first term by the `conjugate` of the bottom.\[\large \color{orangered}{\frac{1-\sin x}{1-\sin x}}\frac{1}{1+\sin x}+\frac{1}{\csc x+1}\color{orangered}{\frac{\csc x-1}{\csc x-1}}\] When we multiply conjugates, we end up with the `Difference of squares` on the bottoms.\[\large \frac{1-\sin x}{1-\sin^2x}+\frac{\csc x-1}{\csc^2x-1}\]

    • one year ago
  2. zepdrix Group Title
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    Now we can use our blue identities,\[\large \color{royalblue}{\cos^2x=1-\sin^2x \qquad \qquad \cot^2x=\csc^2x-1}\] Using these identities gives us,\[\large \frac{1-\sin x}{\color{royalblue}{1-\sin^2x}}+\frac{\csc x-1}{\color{royalblue}{\csc^2x-1}} \qquad \rightarrow \qquad \frac{1-\sin x}{\color{royalblue}{\cos^2x}}+\frac{\csc x-1}{\color{royalblue}{\cot^2x}}\]

    • one year ago
  3. zepdrix Group Title
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    Recalling that, \[\large \color{brown}{\cot^2x=\frac{\cos^2x}{\sin^2x}}\]Gives us,\[\large \frac{1-\sin x}{\cos^2x}+\frac{\csc x-1}{\color{brown}{\dfrac{\cos^2x}{\sin^2x}}}\]Now notice that we're dividing by a fraction, which we can rewrite as multiplication if we `flip` the bottom fraction.\[\large \frac{1-\sin x}{\cos^2x}+\frac{\color{brown}{\sin^2x}(\csc x-1)}{\color{brown}{\cos^2x}}\]

    • one year ago
  4. zepdrix Group Title
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    Then distribute the sin^2x to each term in the brackets. There's only a little bit of simplification to do after that. Lemme know if you're still stuck on something.

    • one year ago
  5. snowflakepixie Group Title
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    It took a while to read that since most of the formatting didn't work, but I think I've got it now. Thanks c:

    • one year ago
  6. zepdrix Group Title
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    Formatting didn't work? :c If you refresh the page it usually fixes it. That stinks :d

    • one year ago
  7. Azteck Group Title
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    I feel that the faster way of doing it without any identities is this. \[\csc x=\frac{1}{\sin x}\] \[\frac{ 1 }{ \sin x+1 }+\frac{ 1 }{ \csc x +1 }=\frac{ 1 }{ \sin x+1 }+\frac{ 1 }{ \dfrac{ 1 }{ \sin x } +1 }\] \[\frac{ 1 }{ \sin x+1 }+\frac{ 1 }{ \csc x +1 }=\frac{ 1 }{ \sin x+1 }+\frac{ 1 }{ \dfrac{ 1+\sin x }{ \sin x } }\] \[=\frac{ 1 }{ \sin x+1 }+\frac{ \sin x }{1+\sin x}\] \[=\frac{1(1+\sin x+\sin x(\sin x +1)}{(\sin x+1)^2}\] \[=\frac{1+\sin x+\sin^2 x +\sin x}{(\sin x +1)^2}\] \[=\frac{\sin^2 x+2\sin x +1}{(\sin x +1)^2}\] Factorise the numerator \[=\frac{(\sin x +1)^2}{(\sin x +1)^2}\] \[=1\]

    • one year ago
  8. Azteck Group Title
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    \[1(1+\sin x+\sin x(\sin x+1)\] The numerator on one the lines was supposed to have a bracket somewhere. \[1(1+\sin x)+\sin x(\sin x+1)\]

    • one year ago
  9. Azteck Group Title
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    on one of*

    • one year ago
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