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IloveCharlie

  • 2 years ago

cos67degrees 30'= ?

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  1. IloveCharlie
    • 2 years ago
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  2. ash2326
    • 2 years ago
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    @IloveCharlie what's 30' in degrees?

  3. GOODMAN
    • 2 years ago
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    Divide 30 seconds by 60. After that, add it to 67. Then take the cosine of it all.

  4. IloveCharlie
    • 2 years ago
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    Hmmm... I got .382. Can you please confirm?

  5. ash2326
    • 2 years ago
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    yeah, it's right but could you use calculator for this? @IloveCharlie

  6. IloveCharlie
    • 2 years ago
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    I just don't know which one it matches up with :/ From the options given

  7. IloveCharlie
    • 2 years ago
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    Whoa @jim_thompson5910 you're typing a lot

  8. jim_thompson5910
    • 2 years ago
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    Yes it's a lot, but it at least gives you the exact answer 67 degrees 30' = 67 degrees + 30/60 = 67+0.5 = 67.5 degress Notice how 2*67.5 = 135 and \[\Large \cos(135) = -\frac{\sqrt{2}}{2}\] (using the unit circle) Now turn to the identity \[\Large \cos(x) = \sqrt{ \frac{ \cos(2x) + 1 }{2} }\] If we let x = 67.5, then \[\Large \cos(x) = \sqrt{ \frac{ \cos(2x) + 1 }{2} }\] \[\Large \cos(67.5) = \sqrt{ \frac{ \cos(2*67.5) + 1 }{2} }\] \[\Large \cos(67.5) = \sqrt{ \frac{ \cos(135) + 1 }{2} }\] \[\Large \cos(67.5) = \sqrt{ \frac{ -\frac{\sqrt{2}}{2} + 1 }{2} }\] \[\Large \cos(67.5) = \sqrt{ \frac{ -\frac{\sqrt{2}}{2} + \frac{2}{2} }{2} }\] \[\Large \cos(67.5) = \sqrt{ \frac{ \frac{-\sqrt{2}+2}{2} }{2} }\] \[\Large \cos(67.5) = \sqrt{ \frac{ \frac{2-\sqrt{2}}{2} }{2} }\] \[\Large \cos(67.5) = \sqrt{ \left( \frac{2-\sqrt{2}}{2} \right)\left(\frac{1}{2}\right) }\] \[\Large \cos(67.5) = \sqrt{ \frac{(2-\sqrt{2})*1}{2*2} }\] \[\Large \cos(67.5) = \sqrt{ \frac{2-\sqrt{2}}{4} }\] \[\Large \cos(67.5) = \frac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}\] \[\Large \cos(67.5) = \frac{\sqrt{2-\sqrt{2}}}{2}\]

  9. jim_thompson5910
    • 2 years ago
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    sry i had a typo, but i fixed it

  10. IloveCharlie
    • 2 years ago
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    Wow, thanks so much for the awesome step by step explanation! Really helps!

  11. jim_thompson5910
    • 2 years ago
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    you're welcome

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