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rox13kh

  • 2 years ago

sin0 + cos0 = 7/6 what is sin0 - cos0 = ? 0 represents theta

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  1. Hyun11
    • 2 years ago
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    \[2(\pi n+\arctan(\frac{1}{13}(6-\sqrt{23})))\]

  2. Hyun11
    • 2 years ago
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    \[n\in \mathbb{Z}\]

  3. Callisto
    • 2 years ago
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    \[sin\theta + cos\theta = \frac{7}{6}\]\[(sin\theta + cos\theta)^2 = (\frac{7}{6})^2\]\[sin^2\theta + cos^2\theta+2sin\theta cos\theta= \frac{49}{36}\]\[1+2sin\theta cos\theta= \frac{49}{36}\]\[2sin\theta cos\theta= \frac{13}{36}\] \[sin\theta - cos\theta\]\[=\sqrt{(sin\theta - cos\theta)^2}\]\[=\sqrt{(sin^2\theta-2sin\theta\cos\theta+\cos^2\theta)}\]You know what \(sin^2\theta + cos^2\theta\) and \(2sin\theta\cos\theta\) are, so you can find the answer.

  4. Callisto
    • 2 years ago
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    @Hyun11 Can you explain how you get your answer?

  5. xlegendx
    • 2 years ago
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    ^ http://wolframalpha.com

  6. shubhamsrg
    • 2 years ago
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    sin0 + cos0 = 7/6 let sin0 - cos0 = x square both sides of both eqns and add them, you'll get your answer in the very next step.

  7. rox13kh
    • 2 years ago
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    how if you are asked \[(\sin \theta)^2 - (\cos \theta)^2\] ? @Callisto

  8. myininaya
    • 2 years ago
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    Do you know how to solve for sin(theta) and cos(theta) given the two equations above?

  9. myininaya
    • 2 years ago
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    a+b=7/6 a-b=(callisto basically gave you this answer) Solve the system of linear equations

  10. myininaya
    • 2 years ago
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    Plug into a^2-b^2 and you got it

  11. rox13kh
    • 2 years ago
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    thx, that helped a lot @myininaya

  12. myininaya
    • 2 years ago
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    np :)

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