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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
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 @pooja195 @hartnn @triciaal

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2do you know the identity cos(xy) = ??

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2hmm 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
 one year ago
Best ResponseYou've already chosen the best response.0okay but how would I solve it?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2replace cos(ab) and cos (a+b) with those identities

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2like for cos(ab)\[\huge\rm \color{ReD}{cos\color{Red}{\alpha } \cos \color{blue}{\beta} + \sin \color{Red}{\alpha } \sin\color{blue}{ \beta } } cos (\alpha + \beta)\] use first identity for cos (a +b)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay but still how would I solve it

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2replace 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
 one year ago
Best ResponseYou've already chosen the best response.2cos (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
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \color{ReD}{cos\color{Red}{\alpha } \cos \color{blue}{\beta} + \sin \color{Red}{\alpha } \sin\color{blue}{ \beta } } (cos \alpha cos \beta  sin \alpha sin \beta)\] distribute parentheses by negative one you will get the answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This makes no sense to me

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \color{ReD}{cos\color{Red}{(\alpha) } \cos \color{blue}{(\beta)} + \sin \color{Red}{(\alpha) } \sin\color{blue}{ (\beta) } } [cos (\alpha) cos (\beta)  sin( \alpha) sin (\beta)]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's an identity, there's a proof for it, but you just have to remember that \[\cos(x+y)=\cos(x)\cos(y)\sin(x)\sin(y)\]\[\cos(xy)=\cos(x)\cos(y)+\sin(x)\sin(y)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just 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
 one year ago
Best ResponseYou've already chosen the best response.0go around not ground

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2it'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
 one year ago
Best ResponseYou've already chosen the best response.0@Nishant_Garg Ok I get it, so can you explain the step by step process on proving the identity of cos(ab)  cos(a+b) = 2 sin a sin b?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2http://www.purplemath.com/modules/idents.htm here ar all of the identities you should write theses down n ur notes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, 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
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.2do you mean show how cos(a+b) = cos (a)cos(b)  sin (a) sin(b) ???? this identity proof ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Weren'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
 one year ago
Best ResponseYou've already chosen the best response.0Like 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
 one year ago
Best ResponseYou've already chosen the best response.0No this one.... cos(ab)  cos(a+b) = 2 sin a sin b

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\cos(ab)=\cos(a)\cos(b)+\sin(a)\sin(b)\]\[\cos(a+b)=\cos(a)\cos(b)\sin(a)\sin(b)\] Using these 2 identities we get \[\cos(ab)\cos(a+b)=(\cos(a)\cos(b)+\sin(a)\sin(b))(\cos(a)\cos(b)\sin(a)\sin(b))\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2i 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
 one year ago
Best ResponseYou've already chosen the best response.0after that its just cancelling some terms adding some terms and u can get the result

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2\(\color{blue}{\text{Originally Posted by}}\) @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
 one year ago
Best ResponseYou've already chosen the best response.0No just don't worry about it, thanks for trying!

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2i'll give it a last try :=) let's say we have two equations dw:1444074368037:dw

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2same concept for those identities dw:1444074468250:dw

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2ooops the pink box supposed to be cos(ab)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2distribute just like in algebra\[\large\rm a(b+c)=a*b + (a) * c = ab ac\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2\[ [\cos(a)\cos(b) \sin(a)\sin(b)] = ??\] just multiply by negative one

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2remember `terms ` are separated by negative or plus sign so cos(a)cos(b) is only one term

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so cos(αβ)  cos(α+β) = (cos(α)∙cos(β) + sin(α)∙sin(β))  (cos(α)∙cos(β)  sin(α)∙sin(β)) cos(αβ)  cos(α+β) = 2∙sin(α)∙sin(β)
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