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Bellabambina223

  • 4 years ago

Find the exact value of sin(T -pi/6) when sin(t)=1/3 and cos(t)=Rad8/3

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  1. amistre64
    • 4 years ago
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    \[sin(a+b)=sin(a)cos(b)+sin(b)cos(a)\]

  2. amistre64
    • 4 years ago
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    in this case: a = T and b = -pi/6 but not sure if that fits too well

  3. amistre64
    • 4 years ago
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    that might work, even with the funny looking decimal

  4. JoãoVitorMC
    • 4 years ago
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    ops there is a error there i will correct that

  5. myininaya
    • 4 years ago
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    what funny looking decimal?

  6. myininaya
    • 4 years ago
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    T=t?

  7. JoãoVitorMC
    • 4 years ago
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    \[\sin(T - \pi/6) = \sin(T)\cos(\pi/6)-\sin(\pi/6)\cos(T)\]\[\sqrt{3}/6 - 0,0232 = 0,2654\]

  8. Bellabambina223
    • 4 years ago
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    so basically use the double angle formula?

  9. JoãoVitorMC
    • 4 years ago
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    yes

  10. myininaya
    • 4 years ago
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    \[\frac{1}{3} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2} \cdot \frac{\sqrt{8}}{3}\]

  11. Bellabambina223
    • 4 years ago
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    oh did you get that from a chart myininaya?

  12. JoãoVitorMC
    • 4 years ago
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    Remember that \[\sin(a \pm b)= \sin(a)\cos(b) \pm \sin(b)\cos(a)\]

  13. Bellabambina223
    • 4 years ago
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    ok.. yes

  14. myininaya
    • 4 years ago
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    yep bella \[\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}; \sin(\frac{\pi}{6})=\frac{1}{2}\]

  15. Bellabambina223
    • 4 years ago
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    ok thank you!

  16. amistre64
    • 4 years ago
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    across the pond they use the "," as a "."

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