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chels

  • 3 years ago

provide that this equation is an identity. sin(x)cos(x) / tan(x) = 1 - sin^2(x)

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  1. lokisan
    • 3 years ago
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    \[\frac{\sin x \cos x}{\tan x } =\frac{\sin x \cos x}{\frac{\sin x }{\cos x}}=\frac{\cos x}{\frac{1}{\cos x}}=\cos^2 x = 1-\sin^2 x\]

  2. chels
    • 3 years ago
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    i can't scroll..

  3. lokisan
    • 3 years ago
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    Try refreshing the page, or log out and log back in.

  4. beens
    • 3 years ago
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    sinx cosx/tanx=sinx cosx cosx/sinx =cosxcosx =cos^2x =1-sin^x

  5. beens
    • 3 years ago
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    sorry...1-sin^2x

  6. chels
    • 3 years ago
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    how does sinxcosx/tanx = sinx?

  7. lokisan
    • 3 years ago
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    It doesn't. It equals cos^2x

  8. chels
    • 3 years ago
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    just kidding, i understand how you wrote it now

  9. lokisan
    • 3 years ago
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    I like how you stalked me then...lol

  10. chels
    • 3 years ago
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    haha sorry. i got confused.

  11. chels
    • 3 years ago
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    1 / sec(x) - 1 minus 1 / sec(x) + 1 = 2cot^2(x)

  12. lokisan
    • 3 years ago
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    Can you type out the question using the equation editor?

  13. chels
    • 3 years ago
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    \[1\div \sec(x-1) - 1 \div \sec(x+1) = 2\cot^2x\]

  14. lokisan
    • 3 years ago
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    okay

  15. chels
    • 3 years ago
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    in the denominator it's supposed to be sec(x) - 1 and sec(x) + 1

  16. lokisan
    • 3 years ago
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    Yeah...I was thinking that.

  17. chels
    • 3 years ago
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    yeahh.

  18. lokisan
    • 3 years ago
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    Okay, it's easy now...

  19. lokisan
    • 3 years ago
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    \[\frac{1}{\sec x -1}-\frac{1}{\sec x +1}=\frac{\sec x +1 - (\sec x -1)}{\sec^2 x-1}=\frac{2}{\tan^2 x}\]

  20. lokisan
    • 3 years ago
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    \[=2\cot^2 x\]

  21. lokisan
    • 3 years ago
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    Afte\[1+\tan^2 x=\sec^2 x \rightarrow 1-\sec^2x = \tan^2 x\]r putting it over a common denominator, I use the fact that

  22. lokisan
    • 3 years ago
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    Then cot^2(x) is just from the definition, \[\cot x = \frac{1}{\tan x}\]

  23. lokisan
    • 3 years ago
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    Once you're done, let me know if it works for you or not.

  24. chels
    • 3 years ago
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    okay, i understand that one

  25. chels
    • 3 years ago
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    \[\cos(\pi/2 - x)\cos(-x) = \sin(2x) \div 2\]

  26. lokisan
    • 3 years ago
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    Okay, you have to identify a few things here. (1)\[\cos(\frac{\pi}{2}-x)=\sin x\](2)\[\cos(-x)=\cos x\]Then the left-hand side becomes,\[\cos(\frac{\pi}{2}-x)\cos(-x)=\sin x \cos x \]If you multiply both sides by 2, you get\[2\sin x \cos x = \sin 2x\]which is true (since this is the expansion of sin(2x)).

  27. chels
    • 3 years ago
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    how did you identify (1)?

  28. lokisan
    • 3 years ago
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    If you want to see why cos(pi/2 - x) = sin(x), draw coordinate axes, the unit circle, and swing the radial arm from 0 to pi (180 degrees) and then reverse a bit (that bit is equal to x). The angle in the triangle ...I'm doing that now...

  29. lokisan
    • 3 years ago
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    is x. I'll draw a pic.

  30. lokisan
    • 3 years ago
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    Actually, instead of a geometrical explanation (which will take forever online) do you know your double angle formulae?

  31. chels
    • 3 years ago
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    yeah, i get it now. i just had to go through it all

  32. lokisan
    • 3 years ago
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    Okay

  33. chels
    • 3 years ago
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    \[\tan(x \div 2) \cos^2(x) - \tan(x \div 2) = \sin(x) \div \sec(x) minus \sin (x)\]

  34. chels
    • 3 years ago
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    sin(x) is subtracted from sin(x) / sec(x) in the second equation.

  35. lokisan
    • 3 years ago
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    Okay, sorry, had to do something. Here we go...

  36. lokisan
    • 3 years ago
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    Recognize, on the left-hand side, you can factor as\[\tan^2\frac{x}{2}(\cos^2 x-1)=\tan^2 x (-\sin^2 x)\]

  37. chels
    • 3 years ago
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    ok, yep.

  38. lokisan
    • 3 years ago
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    Also, on the right-hand side, \[\frac{\sin x }{\sec x}-\sin x = \sin x (\cos x - 1)\]We have then\[\tan^2 \frac{x}{2}(-\sin^2 x)=\sin x (\cos x -1)\]

  39. chels
    • 3 years ago
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    got it so far..

  40. lokisan
    • 3 years ago
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    Divide both sides by (-sin^2(x)) to get\[\tan^2\frac{x}{2}=\frac{1-\cos x}{\sin x}=\frac{1-(\cos^2 \frac{x}{2}- \sin ^2 \frac{x}{2})}{2\sin \frac{x}{2}\cos \frac{x}{2}}\]

  41. lokisan
    • 3 years ago
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    But, in the numerator, 1-cos^2 (x/2) = sin^2(x/2). So after expanding the numerator and doing this, we have\[\tan \frac{x}{2}=\frac{2\sin ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\]which is, tan(x/2) by definition.

  42. lokisan
    • 3 years ago
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    Forget the tan^2(x/2) thing I started with - transcribing notes to this site leads to screw-ups.

  43. chels
    • 3 years ago
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    ok

  44. chels
    • 3 years ago
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    find exact value of (-pi/12)

  45. lokisan
    • 3 years ago
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    function?

  46. chels
    • 3 years ago
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    sin(-pi/12)

  47. chels
    • 3 years ago
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    sorry

  48. lokisan
    • 3 years ago
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    Use the fact that 2 x pi/12 = pi/6 which is 30 degrees. We know sin(-30 deg) = -sin(30 deg) is. You can use the double angle formula to link the two.

  49. chels
    • 3 years ago
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    square root of 3 over 4?

  50. lokisan
    • 3 years ago
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    Do you want to try that and see how you go. I need a shower.

  51. lokisan
    • 3 years ago
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    The sine of 30 degrees is 1/2. Were you ever shown how to construct the basic angles from triangles? You can always work out 30, 45 and 60, and various combinations, using two triangles. One is a right-angled one with sides 1, 1 and hypotenuse sqrt(2). The angles are 90, 45 and 45. You can construct the 30 and 60 from taking an equilateral triangle of side length 2. Then cut it in half. You'll two right-angled triangles. The angles in each will be 30, 60 and 90. The sides will be 1, sqrt(3) and 2 (2 is the hypotenuse).

  52. lokisan
    • 3 years ago
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    You can read off your sine, cosine and tangents from them.

  53. lokisan
    • 3 years ago
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    chels, you can do this with the double angle formula for sine:\[\sin(-\frac{\pi}{12})=\sin(\frac{\pi}{6}-\frac{\pi}{4})\]\[=\sin \pi/6 \cos \pi/4 - \cos \pi/6 \sin \pi/4\]\[=\frac{1}{2}\frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}}=\frac{1-\sqrt{3}}{2\sqrt{2}}=\frac{\sqrt{2}-\sqrt{6}}{4}\]

  54. lokisan
    • 3 years ago
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    You could also use the combination,\[\frac{\pi}{4}-\frac{\pi}{3}\]

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