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CEBS95

  • 3 years ago

tan^2(60) + 4cos^2(45) - 9tan^2(30) - 8sin^2(30) The answer is zero but why!? i need step by step solution please

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  1. Shadowys
    • 3 years ago
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    why can't you just plug the whole thing into your calculator?

  2. CEBS95
    • 3 years ago
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    because in the exam they already give you the answer, you have to do the procedure

  3. Callisto
    • 3 years ago
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    tan60 = \(\sqrt{3}\) cos45 = \(\frac{1}{\sqrt2}\) tan30 = \(\frac{1}{\sqrt3}\) sin30 = 1/2 These are all special angles and you should remember them.

  4. CEBS95
    • 3 years ago
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    yes i know them but i just continue to get 6 instead of 0

  5. Shadowys
    • 3 years ago
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    I'm getting zero just fine though. note the squares.

  6. CEBS95
    • 3 years ago
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    tan^2(60) = 3 ? 4cos^2(45) = 4 ?

  7. Callisto
    • 3 years ago
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    4cos^2(45) = 4 (1/2) = ...?

  8. CEBS95
    • 3 years ago
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    my math base is not good, that's the reason i wrote step by step solution

  9. Callisto
    • 3 years ago
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    \[cos45 =\frac{1}{\sqrt2}\]\[cos^245 =(\frac{1}{\sqrt2})^2 = \frac{1}{2}\]\[4cos45 =4(\frac{1}{2}) = ...?\]

  10. CEBS95
    • 3 years ago
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    4 (1/2) = 2

  11. Callisto
    • 3 years ago
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    Yes.

  12. winterfez
    • 3 years ago
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    \[\frac{ \sin ^{2} (60)}{ \cos ^{2}(60)}+4\cos ^{2}(45)-9\frac{ \sin ^{2}(30) }{ \cos ^{2}(30) }-8\sin ^{2}(30)\]

  13. CEBS95
    • 3 years ago
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    so, tan^2(60) = 3

  14. Callisto
    • 3 years ago
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    tan^2(60) = 3 <- Yes

  15. winterfez
    • 3 years ago
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    |dw:1355048071288:dw|

  16. CEBS95
    • 3 years ago
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    i found in my book that cos 45 = \[\frac{ \sqrt{2} }{ 2}\]

  17. Callisto
    • 3 years ago
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    \[\frac{1}{\sqrt2}=\frac{1}{\sqrt2}\times \frac{\sqrt2}{\sqrt2}=\frac{\sqrt2}{\sqrt2^2} = \frac{\sqrt2}{2}\] Just rationalization.

  18. CEBS95
    • 3 years ago
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    9tan^2(30) = \[\tan( 30) = \frac{ \sqrt{3} }{ 3 } ; \tan ^{2}(30) = 3/9 = 1/3 \]

  19. CEBS95
    • 3 years ago
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    times 9 = 3 right?

  20. Callisto
    • 3 years ago
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    Yes.

  21. CEBS95
    • 3 years ago
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    8 sin ^2 (30) = 4 ?

  22. CEBS95
    • 3 years ago
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    no, i think it 8

  23. CEBS95
    • 3 years ago
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    9tan^2(30) - 8sin^2(30) = 3 - 8 = -5, at the end that will give me 5- (-5) that equals 10

  24. Callisto
    • 3 years ago
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    8 sin ^2 (30) = 8 (1/2)^2 = 8 (1/4) = ...

  25. CEBS95
    • 3 years ago
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    2

  26. CEBS95
    • 3 years ago
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    still, 3-2 = 1, 5-1 = 4

  27. winterfez
    • 3 years ago
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    |dw:1355049445375:dw| look at this one to see how i get 0

  28. Callisto
    • 3 years ago
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    tan^2(60) + 4cos^2(45) - 9tan^2(30) - 8sin^2(30) tan^2(60) = \(\sqrt3^2\) = ... 4cos^2(45) = 4\((\frac{1}{\sqrt2})^2\) = 4(1/2) = ... - 9tan^2(30) = -9 \((\frac{1}{\sqrt3})^2\) = -9 (1/3) = ... - 8sin^2(30) = -8 (1/2)^2 = -8 (1/4) = ... Sum up the four answers.

  29. winterfez
    • 3 years ago
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    |dw:1355049643294:dw|

  30. CEBS95
    • 3 years ago
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    zero! nice thanks calisto

  31. Callisto
    • 3 years ago
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    You're welcome :)

  32. Callisto
    • 3 years ago
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    | 30 | 45 | 60 ------------------------------------ sin | \(\frac{1}{2}\) | \(\frac{1}{\sqrt2} /\frac{\sqrt2}{2} \) | \(\frac{\sqrt3}{2}\) ------------------------------------ cos | \(\frac{\sqrt3}{2}\) | \(\frac{1}{\sqrt2} /\frac{\sqrt2}{2} \) | \(\frac{1}{2}\) ------------------------------------ tan | \(\frac{1}{\sqrt3} /\frac{\sqrt3}{3} \) | 1 | \(\sqrt{3}\)

  33. Callisto
    • 3 years ago
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    Memorize the table..

  34. CEBS95
    • 3 years ago
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    yes, with practice i will memorise the terms

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