anonymous
  • anonymous
Verify the identity. Show your work. cos(α - β) - cos(α + β) = 2 sin α sin β
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@ganeshie8 @pooja195 @hartnn @triciaal
Nnesha
  • Nnesha
do you know the identity cos(x-y) = ??
anonymous
  • anonymous
No

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Nnesha
  • Nnesha
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 \[\huge\rm cos(\color{Red}{x}+\color{blue}{y})=cos\color{Red}{x} \cos \color{blue}{y} - \sin \color{Red}{x} \sin\color{blue}{ y}\]
anonymous
  • anonymous
okay but how would I solve it?
Nnesha
  • Nnesha
replace cos(a-b) and cos (a+b) with those identities
Nnesha
  • Nnesha
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)
anonymous
  • anonymous
okay but still how would I solve it
Nnesha
  • Nnesha
replace cos (a+b) with this identity and then distribute it by -1 \[\huge\rm cos(\color{Red}{x}+\color{blue}{y})=cos\color{Red}{x} \cos \color{blue}{y} - \sin \color{Red}{x} \sin\color{blue}{ y}\]
Nnesha
  • Nnesha
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}
Nnesha
  • Nnesha
\[\large\rm \color{ReD}{cos\color{Red}{\alpha } \cos \color{blue}{\beta} + \sin \color{Red}{\alpha } \sin\color{blue}{ \beta } }- (cos \alpha cos \beta - sin \alpha sin \beta)\] distribute parentheses by negative one you will get the answer
anonymous
  • anonymous
This makes no sense to me
Nnesha
  • Nnesha
\[\large\rm \color{ReD}{cos\color{Red}{(\alpha) } \cos \color{blue}{(\beta)} + \sin \color{Red}{(\alpha) } \sin\color{blue}{ (\beta) } }- [cos (\alpha) cos (\beta) - sin( \alpha) sin (\beta)]\]
anonymous
  • anonymous
It's an identity, there's a proof for it, but you just have to remember that \[\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)\]\[\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)\]
anonymous
  • anonymous
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
anonymous
  • anonymous
go around not ground
Nnesha
  • Nnesha
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 ?
anonymous
  • anonymous
@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?
Nnesha
  • Nnesha
http://www.purplemath.com/modules/idents.htm here ar all of the identities you should write theses down n ur notes
Nnesha
  • Nnesha
these*
anonymous
  • anonymous
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
anonymous
  • anonymous
@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..
Nnesha
  • Nnesha
do you mean show how cos(a+b) = cos (a)cos(b) - sin (a) sin(b) ???? this identity proof ?
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
No this one.... cos(a-b) - cos(a+b) = 2 sin a sin b
anonymous
  • anonymous
\[\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)\]\[\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\] Using these 2 identities we get \[\cos(a-b)-\cos(a+b)=(\cos(a)\cos(b)+\sin(a)\sin(b))-(\cos(a)\cos(b)-\sin(a)\sin(b))\]
Nnesha
  • Nnesha
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 ?? :=)
anonymous
  • anonymous
after that its just cancelling some terms adding some terms and u can get the result
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @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 ?
anonymous
  • anonymous
No just don't worry about it, thanks for trying!
Nnesha
  • Nnesha
i'll give it a last try :=) let's say we have two equations |dw:1444074368037:dw|
Nnesha
  • Nnesha
same concept for those identities |dw:1444074468250:dw|
Nnesha
  • Nnesha
ooops the pink box supposed to be cos(a-b)
Nnesha
  • Nnesha
distribute just like in algebra\[\large\rm -a(b+c)=-a*b + (-a) * c = -ab -ac\]
Nnesha
  • Nnesha
\[- [\cos(a)\cos(b) -\sin(a)\sin(b)] = ??\] just multiply by negative one
Nnesha
  • Nnesha
remember `terms ` are separated by negative or plus sign so cos(a)cos(b) is only one term
anonymous
  • anonymous
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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