Jonathan T.

asked • 12/01/16

simplify the expression (sinAsinB)(cosAcosB)

simply simplify the expression. i have to find out if my answer i have is right or wrong

1 Expert Answer

By:

Mark M. answered • 12/01/16

Tutor
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Mathematics Teacher - NCLB Highly Qualified

Jonathan T.

thats what i have. what i got is SinA-SinB. 
 
i did it like this:
(SinA-SinB)(cosAcosB)
=sinAcosA+sinAcosB-sinBcosA-sinBcosB
=sinAcosA-sinBcosB, then i used product to sum identity and got
=1/2(sin(A+A) + sin(A-A)) - 1/2(sin(B+B) + sin(B-B))
=1/2sin(2a) - 1/2sin(2B)
=sinAsinB.
i feel like i may have gone wrong with either foiling it or doing the product to sum identity. 
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12/01/16

Mark M.

The first line of your solution is different that the product in the posting.
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12/01/16

Jonathan T.

you are absolute correct. sorry about that.
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12/01/16

Mark M.

The second line should only be a binomial unless the second parenthesis contains a minus sign.
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12/01/16

Jonathan T.

i wrote it wrong again. the problem is (sinA-sinB)(cosA+cosB)
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12/01/16

Mark M.

How did line 2 collapse to line 3?
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12/01/16

Jonathan T.

i guess thats where i went wrong. any idea what i did wrong ther?
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12/01/16

Mark M.

sinAcosA + sinAcosB - sinBcosA - sinBcosB
sinAcosB - sinBcosA + sinAcosB - sinBcosA   commute
sin(A - B) + sin(A - B)                                 difference formula
2sin(A - B)
 
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12/01/16

Jonathan T.

where does the SinAcosA and -SinBcosB go?
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12/04/16

Jonathan T.

or is (SinAcosB)-(SinBcosA) the same as (SinACosA-sinBcosB)
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12/04/16

Mark M.

My error
sinAcosA + sinAcosB - sinBcosA - sinBcosB
sinAcosB - sinBcosA + sinAcosA - sinB cosB   communite
sin(A - B) + sinAcosA - sinBcosB                   difference formula
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12/05/16

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