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anonymous

  • one year ago

Verify the identity. Show your work. cos(α - β) - cos(α + β) = 2 sin α sin β

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  1. anonymous
    • one year ago
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    @ganeshie8 @pooja195 @hartnn @triciaal

  2. Nnesha
    • one year ago
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    do you know the identity cos(x-y) = ??

  3. anonymous
    • one year ago
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    No

  4. Nnesha
    • one year ago
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    hmm alright use this \[\huge\rm cos(\color{Red}{x}-\color{blue}{y})=cos\color{Red}{x} \cos \color{blue}{y} + \sin \color{Red}{x} \sin\color{blue}{ y}\] and \[\huge\rm cos(\color{Red}{x}+\color{blue}{y})=cos\color{Red}{x} \cos \color{blue}{y} - \sin \color{Red}{x} \sin\color{blue}{ y}\]

  5. anonymous
    • one year ago
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    okay but how would I solve it?

  6. Nnesha
    • one year ago
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    replace cos(a-b) and cos (a+b) with those identities

  7. Nnesha
    • one year ago
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    like for cos(a-b)\[\huge\rm \color{ReD}{cos\color{Red}{\alpha } \cos \color{blue}{\beta} + \sin \color{Red}{\alpha } \sin\color{blue}{ \beta } }- cos (\alpha + \beta)\] use first identity for cos (a +b)

  8. anonymous
    • one year ago
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    okay but still how would I solve it

  9. Nnesha
    • one year ago
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    replace cos (a+b) with this identity and then distribute it by -1 \[\huge\rm cos(\color{Red}{x}+\color{blue}{y})=cos\color{Red}{x} \cos \color{blue}{y} - \sin \color{Red}{x} \sin\color{blue}{ y}\]

  10. Nnesha
    • one year ago
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    cos (a+b) = cos a cos b - sin a sin b so can substitute `cos(a+b)` for { cos a cos - sin a sin b}

  11. Nnesha
    • one year ago
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    \[\large\rm \color{ReD}{cos\color{Red}{\alpha } \cos \color{blue}{\beta} + \sin \color{Red}{\alpha } \sin\color{blue}{ \beta } }- (cos \alpha cos \beta - sin \alpha sin \beta)\] distribute parentheses by negative one you will get the answer

  12. anonymous
    • one year ago
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    This makes no sense to me

  13. Nnesha
    • one year ago
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    \[\large\rm \color{ReD}{cos\color{Red}{(\alpha) } \cos \color{blue}{(\beta)} + \sin \color{Red}{(\alpha) } \sin\color{blue}{ (\beta) } }- [cos (\alpha) cos (\beta) - sin( \alpha) sin (\beta)]\]

  14. anonymous
    • one year ago
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    It's an identity, there's a proof for it, but you just have to remember that \[\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)\]\[\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)\]

  15. anonymous
    • one year ago
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    just like how we have algebriac identities \[(a+b)^2=a^2+b^2+2ab\] we have numerous trigonometric identities, you don't need to ground proving them every time, they r something you have to learn

  16. anonymous
    • one year ago
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    go around not ground

  17. Nnesha
    • one year ago
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    it's just like algebra plugin the identities let's say cos a = x and cos b = y sin a = w sin b =z so \[\huge\rm x y+ wz - (xy - wz ) = 2 wz\] can you prove this ?

  18. anonymous
    • one year ago
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    @Nishant_Garg Ok I get it, so can you explain the step by step process on proving the identity of cos(a-b) - cos(a+b) = 2 sin a sin b?

  19. Nnesha
    • one year ago
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    http://www.purplemath.com/modules/idents.htm here ar all of the identities you should write theses down n ur notes

  20. Nnesha
    • one year ago
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    these*

  21. anonymous
    • one year ago
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    Sorry, I haven't gone through the proofs myself, in our time we were just told to learn these identities even though the proof was in the book, it was just there for satisfaction of the mind, they never ask u the proofs

  22. anonymous
    • one year ago
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    @Nnesha I get what you saying but can you just please explain the proof process? You telling me isn't helping cause I don't know what I'm doing.. I need someone to SHOW me..

  23. Nnesha
    • one year ago
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    do you mean show how cos(a+b) = cos (a)cos(b) - sin (a) sin(b) ???? this identity proof ?

  24. anonymous
    • one year ago
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    Weren't you taught about the proofs, if you are being given questions like these, then they must've also taught u the proofs(talking about your school), if not they must've at least taught u about these identities

  25. anonymous
    • one year ago
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    Like work it out step by step & just tell me how you got the answer so I can work out my other problems like this on my own.

  26. anonymous
    • one year ago
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    No this one.... cos(a-b) - cos(a+b) = 2 sin a sin b

  27. anonymous
    • one year ago
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    \[\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)\]\[\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\] Using these 2 identities we get \[\cos(a-b)-\cos(a+b)=(\cos(a)\cos(b)+\sin(a)\sin(b))-(\cos(a)\cos(b)-\sin(a)\sin(b))\]

  28. Nnesha
    • one year ago
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    i understand but i would like u to work with me i just .... don't feel comfortable doing all work... even if the question is simple like 2+2 = ? anywaz do you know how to distribute ?? :=)

  29. anonymous
    • one year ago
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    after that its just cancelling some terms adding some terms and u can get the result

  30. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Nnesha \[\large\rm \color{ReD}{cos\color{Red}{(\alpha) } \cos \color{blue}{(\beta)} + \sin \color{Red}{(\alpha) } \sin\color{blue}{ (\beta) } }- [cos (\alpha) cos (\beta) - sin( \alpha) sin (\beta)]\] \(\color{blue}{\text{End of Quote}}\) did you understand this step so we can do next stpe ?

  31. anonymous
    • one year ago
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    No just don't worry about it, thanks for trying!

  32. Nnesha
    • one year ago
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    i'll give it a last try :=) let's say we have two equations |dw:1444074368037:dw|

  33. Nnesha
    • one year ago
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    same concept for those identities |dw:1444074468250:dw|

  34. Nnesha
    • one year ago
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    ooops the pink box supposed to be cos(a-b)

  35. Nnesha
    • one year ago
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    distribute just like in algebra\[\large\rm -a(b+c)=-a*b + (-a) * c = -ab -ac\]

  36. Nnesha
    • one year ago
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    \[- [\cos(a)\cos(b) -\sin(a)\sin(b)] = ??\] just multiply by negative one

  37. Nnesha
    • one year ago
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    remember `terms ` are separated by negative or plus sign so cos(a)cos(b) is only one term

  38. anonymous
    • one year ago
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    so cos(α-β) - cos(α+β) = (cos(α)∙cos(β) + sin(α)∙sin(β)) - (cos(α)∙cos(β) - sin(α)∙sin(β)) cos(α-β) - cos(α+β) = 2∙sin(α)∙sin(β)

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