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Mathematics 20 Online
OpenStudy (anonymous):

Simplify completely into an expression with sin(A) or cos(A) only: 1−(sin(A)+cos(A))^2=

OpenStudy (sidsiddhartha):

expand the square first

OpenStudy (anonymous):

is it 1-(sin^2A + sinAcosA + sinAcosA + cos^2A) ?

OpenStudy (sidsiddhartha):

yup now use the indentity \[\sin^2A+\cos^2A=1\]

OpenStudy (anonymous):

do you get 1-1+2sinAcosA ? so just 2sinAcosA

OpenStudy (sidsiddhartha):

yes and \[2sinA.cosA=\sin2A\]

OpenStudy (anonymous):

hmmm it says its wrong

OpenStudy (anonymous):

this one too

OpenStudy (sidsiddhartha):

there will be minus ,look we missed it

OpenStudy (sidsiddhartha):

now what it says?

OpenStudy (anonymous):

okay yeah that was the problem, thank you!

OpenStudy (sidsiddhartha):

wlcm :)

OpenStudy (anonymous):

wait how did you know there was supposed to be negative? @sidsiddhartha

OpenStudy (sidsiddhartha):

\[1-[sinA+cosA]^2=1-[\sin^2A+\cos^2a+2sinA.cosA]\\=1-[1+2sinA.cosA]=1-1-2sinA.cosA=-2sinA.cosA=-\sin2A\]

OpenStudy (sidsiddhartha):

make sense?

OpenStudy (sidsiddhartha):

it like\[1-(a+b)=1-a-b\]

OpenStudy (anonymous):

yes it does, thanks again!

OpenStudy (sidsiddhartha):

np :)

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