a1234
  • a1234
Use the value given and the trig identities to find the indicated function. cosθ = (sqrt13)/7 sinθ = (?) I don't get this. Is it 6/7?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Try: \[\sin^2 \theta + \cos^2 \theta = 1 \\ \sin^2 \theta + (\frac{\sqrt{13}}{7})^2 = 1\]
a1234
  • a1234
That's what I did. At the end there would be sin^2θ = 36/49, right?
anonymous
  • anonymous
Yea: \[\sin^2 \theta =1- (\frac{\sqrt{13}}{7})^2 =\frac{49}{49} - \frac{13}{49}=\frac{36}{49}\]

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a1234
  • a1234
then sinθ = 6/7 when i take the square root of both sides?
anonymous
  • anonymous
Sorry I got hung up for a second yes that should be the result
a1234
  • a1234
Thanks!
anonymous
  • anonymous
Your Welcome! :D

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