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anonymous

  • one year ago

what's the trigonometric subtraction formula for sine ?

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  1. anonymous
    • one year ago
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    It's not (sin(a)(cos(b)) - (cos(a)sin(b)) ?

  2. UnkleRhaukus
    • one year ago
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    Do you mean this one?\[\sin(\theta-\phi)=\sin\theta\cos\phi-\cos\theta\sin\phi\]

  3. anonymous
    • one year ago
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    yeah , my notes give it with a and b variables . I just wanted to make sure it was that one though

  4. anonymous
    • one year ago
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    thank you

  5. anonymous
    • one year ago
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    how would you use that formula to verify \[\sin(\frac{ \pi }{ 2 }-x)=cosx\]

  6. UnkleRhaukus
    • one year ago
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    \[\sin(\tfrac\pi2-x)\\ =\sin\tfrac\pi2\cos x-\cos\tfrac\pi2\sin x\\[3ex] \qquad\qquad(\sin\tfrac\pi2=\sin90°=1,\qquad \cos\tfrac\pi2=\cos 90°=0\ )\\[3ex] =\]

  7. anonymous
    • one year ago
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    oooooh okay , it's somewhat making sense

  8. UnkleRhaukus
    • one year ago
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    \[\sin(\tfrac\pi2-x)\\ =\sin\tfrac\pi2\cos x-\cos\tfrac\pi2\sin x\\[3ex] \qquad\qquad(\sin\tfrac\pi2=\sin90°=1,\qquad \cos\tfrac\pi2=\cos 90°=0\ )\\[3ex] =(1)\times\cos x-(0)\times\sin x\\ =\]

  9. anonymous
    • one year ago
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    so it'd be 1 , meaning that it is an identity ?

  10. UnkleRhaukus
    • one year ago
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    what would be 1?

  11. anonymous
    • one year ago
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    im so confused :/

  12. UnkleRhaukus
    • one year ago
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    the does the last line (of the above) simplify to become?

  13. anonymous
    • one year ago
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    idk

  14. UnkleRhaukus
    • one year ago
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    \[=\vdots\\=(1)\times\cos x-(0)\times\sin x\\ =\] What is \(1\times \cos x\)?

  15. anonymous
    • one year ago
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    cosx

  16. UnkleRhaukus
    • one year ago
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    good, and what is \(0\times \sin x\)

  17. anonymous
    • one year ago
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    0

  18. UnkleRhaukus
    • one year ago
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    yes, so what is \[=\quad\vdots\\[3ex]=(1)\times\cos x-(0)\times\sin x\\ =\cos x-0\\ =\]

  19. anonymous
    • one year ago
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    cosx ?

  20. UnkleRhaukus
    • one year ago
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    yeah. So. We have shown that: \[\sin(\tfrac\pi2-x) = \cos x\]

  21. UnkleRhaukus
    • one year ago
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    All clear?

  22. anonymous
    • one year ago
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    I think I'm just tired and that's why I'm not comprehending it completely lol

  23. UnkleRhaukus
    • one year ago
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    \[\sin(\tfrac\pi2-x) =\sin\tfrac\pi2\cos x-\cos\tfrac\pi2\sin x\\%[3ex] %\qquad\qquad(\sin\tfrac\pi2=\sin90°=1,\qquad \cos\tfrac\pi2=\cos 90°=0\ )\\[3ex] \qquad\qquad\,\, =(1)\times\cos x-(0)\times\sin x\\ \qquad\qquad\,\, =\cos x\]

  24. anonymous
    • one year ago
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    and the 1 and 0 are the points correct ?

  25. UnkleRhaukus
    • one year ago
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    the 1 and 0, are the values of the sine and cosine of the the first term in the sine of the difference: (pi/2-x), since the first term in this difference is a constant (pi/2), the sines and cosines of this term are necessarily constants. [and can be determined from points on a unit circle]

  26. anonymous
    • one year ago
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    yeah , that's what I meant by points . On the unit circle lol

  27. UnkleRhaukus
    • one year ago
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    i was hoping so .

  28. anonymous
    • one year ago
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    I just looked at the one that have right now to make sure that's what it's from

  29. UnkleRhaukus
    • one year ago
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    i suppose they could also be points on the plots of the function sine and cosine: |dw:1440307933477:dw|

  30. anonymous
    • one year ago
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    My tutor said they're points so that's what I've stuck to calling them lol

  31. UnkleRhaukus
    • one year ago
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    |dw:1440308112848:dw|

  32. UnkleRhaukus
    • one year ago
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    |dw:1440308149960:dw| actually they are more like the lengths on the unit circle, and points on the functions

  33. UnkleRhaukus
    • one year ago
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    do you see what i mean?, on the unit circle, at π/2 the length of the cosine side of the triangle goes to 0, as the length of the sine side , goes to equal 1. whereas on the function plots, the points at π/2 on cosine and sine 0, and 1 respectively

  34. UnkleRhaukus
    • one year ago
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    the point on the unit circle is at theta = π/2 ie (0,1)

  35. anonymous
    • one year ago
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  36. anonymous
    • one year ago
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    thats what I'm looking at

  37. UnkleRhaukus
    • one year ago
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    where each point is \((x,y)=(\cos\theta,\sin\theta)\)

  38. anonymous
    • one year ago
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    yeah

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