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mathsux4real

  • 3 years ago

Last problem I need help with! Please! Using the given data, find the values of sine, cosine, and tangent of 2 A and the quadrant in which 2 A terminates: angle A in Quadrant II, cos A = -7/25 cos 2 A = a. 527/625 b. -336/625 c. 336/625 d. -527/625

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  1. ajprincess
    • 3 years ago
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    \(\cos2A=2\cos A\sin A\) Find the value of sinA using \(\sin A=\sqrt{1-\cos^2A}\). Can u do nw?

  2. mathsux4real
    • 3 years ago
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    So 2(-0.28)(0.276)

  3. mathsux4real
    • 3 years ago
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    :/ It's not matching up

  4. ajprincess
    • 3 years ago
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    Find them in the fraction form. Vat do u get for sinA?

  5. mathsux4real
    • 3 years ago
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    69/250

  6. muhammad9t5
    • 3 years ago
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    cos=Adjacent/Hypotenuse and sin=Opposite/Hypotenuse and tan=Opposite/ajacent so here you have given just adjacent and hypotenuse and you can fine opposite by Pythagourus Theorum.

  7. ajprincess
    • 3 years ago
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    \(\sin A=\sqrt{1-\cos^2A}\) \(\sin A=\sqrt{1-(\large\frac{-7}{25})^2}\) \(\sin A=\sqrt{1-\large\frac{49}{625}}\) \(\sin A=\sqrt{\large\frac{625-49}{625}}\) \(\sin A=\sqrt{\large\frac{576}{625}}\) \(\sin A=\large\frac{24}{25}\) Nw can u find cos2A?

  8. muhammad9t5
    • 3 years ago
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    its hard method friend. find sin by sin=perp/hyp we need perp so from cos=base/hyp here base=-7 and hyp=25 so now use Pythagoras theorem to find perp.

  9. mathsux4real
    • 3 years ago
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    Yes! -336/625. Thank you so much for step by step explanation! @ajprincess Thank you to @muhammad9t5 as well. I appreciate the time and effort my friends

  10. ajprincess
    • 3 years ago
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    U r welcome.:)

  11. muhammad9t5
    • 3 years ago
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    hyp^2=base^+perp^2 25^2=-7^2+perp^2 625=49+perp^2 625-49=perp^2 576=perp^2 24=perp

  12. muhammad9t5
    • 3 years ago
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    thanks! you are always welcome friend.

  13. ajprincess
    • 3 years ago
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    I agree vth u @muhammad9t5. Ur method is easy.

  14. muhammad9t5
    • 3 years ago
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    yup. @ajprincess please fan me.

  15. ajprincess
    • 3 years ago
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    I fanned u. i fan all those who fan me.:)

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