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Simplify completely into an expression with sin(A) or cos(A) only: 1−(sin(A)+cos(A))^2=
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expand the square first
is it 1-(sin^2A + sinAcosA + sinAcosA + cos^2A) ?
yup now use the indentity \[\sin^2A+\cos^2A=1\]
do you get 1-1+2sinAcosA ? so just 2sinAcosA
yes and \[2sinA.cosA=\sin2A\]
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hmmm it says its wrong
this one too
there will be minus ,look we missed it
now what it says?
okay yeah that was the problem, thank you!
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wlcm :)
wait how did you know there was supposed to be negative? @sidsiddhartha
\[1-[sinA+cosA]^2=1-[\sin^2A+\cos^2a+2sinA.cosA]\\=1-[1+2sinA.cosA]=1-1-2sinA.cosA=-2sinA.cosA=-\sin2A\]
make sense?
it like\[1-(a+b)=1-a-b\]
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yes it does, thanks again!
np :)
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