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anonymous

  • one year ago

sin theta= 3/5 cos beta = 2sqrt6/5 Find sin(theta+beta). Is the answer 6sqrt6+4/25?

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  1. shamim
    • one year ago
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    R u able to find out Cos theta=?

  2. anonymous
    • one year ago
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    Yes 4/5

  3. UsukiDoll
    • one year ago
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    \[\sin (\theta) = \frac{3}{5}, \cos(\beta) = 2\sqrt{\frac{6}{5}}\] ? I'm confused on what cosine beta should be

  4. shamim
    • one year ago
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    Sin beta=?

  5. anonymous
    • one year ago
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    1/5

  6. UsukiDoll
    • one year ago
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    oh yeah we need sin beta and cosine beta otherwise the angle sum identity formula isn't going to work

  7. UsukiDoll
    • one year ago
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    \[\sin(\theta + \beta) = \sin(\theta)\cos(\beta)+\cos(\theta)\sin(\beta)\]

  8. UsukiDoll
    • one year ago
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    so far we have sin (theta) = 3/5 |dw:1433043667731:dw| so cos (theta) = 4/5

  9. anonymous
    • one year ago
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    Yep got that

  10. UsukiDoll
    • one year ago
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    ummm what was the cosine beta?

  11. UsukiDoll
    • one year ago
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    is it \[\cos(\beta) = 2 \frac{\sqrt{6}}{5} ?\] is the square root inside the 6/5? that part is throwing me off

  12. UsukiDoll
    • one year ago
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    \[\sin(\theta + \beta) = \frac{3}{5}\cos(\beta)+\frac{4}{5}\sin(\beta)\] I don't understand what cosine beta is... but I do know that I need another triangle so I can grab sin beta and we add this all together

  13. UsukiDoll
    • one year ago
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    is the 6/5 fraction inside the whole square root or is it just square root 6 / 5 ?

  14. UsukiDoll
    • one year ago
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    \[\cos(\beta) = 2 \frac{\sqrt{6}}{5} ? \] or \[\cos(\beta) = 2 \sqrt{\frac{6}{5}} \]

  15. UsukiDoll
    • one year ago
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    I need clarification on that part... otherwise I might end up with the wrong answer.

  16. UsukiDoll
    • one year ago
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    |dw:1433044545197:dw|

  17. UsukiDoll
    • one year ago
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    \[\sin(\theta + \beta) = \frac{3}{5}\frac{2\sqrt{6}}{5}+\frac{4}{5} (\frac{1}{5})\]

  18. UsukiDoll
    • one year ago
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    \[\frac{6\sqrt{6}}{25}+\frac{4}{25}\]

  19. UsukiDoll
    • one year ago
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    I relooked at cos beta and realized that 5 was the longest side of the triangle and that 2 sqrt{6} was the adjcent part... needed to find the opposite part of the triangle so I could find sin beta.

  20. UsukiDoll
    • one year ago
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    you got the right answer.

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