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starkeyfish

  • 2 years ago

Verify the identity: (1-sinx)/cosx = cosx/(1+sinx)

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  1. RolyPoly
    • 2 years ago
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    LS = (1-sinx)/cosx = 1/cosx - sinx/cosx RS = \(\large\frac{cosx}{1+sinx}\) = \(\large\frac{cosx(1-sinx)}{(1+sinx)(1-sinx)}\) = \( \large\frac{cosx}{1-sin^2x} - \frac{cosxsinx}{1-sin^2x}\) =.... Note: \[1 = cos^2x + sin^2x \]

  2. Jhannybean
    • 2 years ago
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    why is it so tiny!

  3. .Sam.
    • 2 years ago
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    Because its a spoiler

  4. starkeyfish
    • 2 years ago
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    I'm very bad at basic math , so I'm finding all of this to be very foreign.

  5. RolyPoly
    • 2 years ago
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    Actually, simply working from the left is ok :D \[RS = \frac{cosx}{(1+sinx)}\]\[=\frac{cos}{1+sinx}\times \frac{1-sinx}{1-sinx}\]By the identity a^2 - b^2 = (a+b)(a-b), we get\[=\frac{cos(1-sinx)}{1-sin^2x}\]Then, \(sin^2x+cos^2x =1\) to express \(1-sin^2x\) in terms of \(cos^2x\). Finally, cancel the common factor, and you'll get the left side!

  6. starkeyfish
    • 2 years ago
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    so 1-sin^2 = cos^2x?

  7. RolyPoly
    • 2 years ago
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    Oh, yes.

  8. starkeyfish
    • 2 years ago
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    okay, thank you!

  9. RolyPoly
    • 2 years ago
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    You're welcome~~

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