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

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- anonymous

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

- schrodinger

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- anonymous

- Nnesha

do you know the identity cos(x-y) = ??

- anonymous

No

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## More answers

- 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

okay but how would I solve it?

- Nnesha

replace cos(a-b) and cos (a+b) with those identities

- 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

okay but still how would I solve it

- 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

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

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

- anonymous

This makes no sense to me

- 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

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

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

go around not ground

- 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

@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

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

- Nnesha

these*

- 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

@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

do you mean show how cos(a+b) = cos (a)cos(b) - sin (a) sin(b) ???? this identity proof ?

- 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

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

No this one.... cos(a-b) - cos(a+b) = 2 sin a sin b

- 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

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

after that its just cancelling some terms adding some terms and u can get the result

- 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

No just don't worry about it, thanks for trying!

- Nnesha

i'll give it a last try :=)
let's say we have two equations
|dw:1444074368037:dw|

- Nnesha

same concept for those identities |dw:1444074468250:dw|

- Nnesha

ooops the pink box supposed to be cos(a-b)

- Nnesha

distribute just like in algebra\[\large\rm -a(b+c)=-a*b + (-a) * c = -ab -ac\]

- Nnesha

\[- [\cos(a)\cos(b) -\sin(a)\sin(b)] = ??\] just multiply by negative one

- Nnesha

remember `terms ` are separated by negative or plus sign
so cos(a)cos(b) is only one term

- 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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