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OpenStudy (anonymous):
cos 3x is cos (2x + x) so you apply cos (A + B) and cos(2x) here
OpenStudy (anonymous):
do you know what those are?
OpenStudy (anonymous):
Yes those are familiar. cos a cos b + sin a sin b = something right?
OpenStudy (anonymous):
wait it equals the cos(a+b) yeah
OpenStudy (anonymous):
cos (A+B) is cosAcosB - sinAsinB
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OpenStudy (anonymous):
so cos(2x+x) = cos 2x cos x - sin 2x sin x
?
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
then cos 2x is cos^2 x - sin^2 x
OpenStudy (anonymous):
if you want to see why
cos (2x) = cos (x + x) = cosx cosx - sin x sin x
thus cos^2 x - sin ^2 x
OpenStudy (anonymous):
do you know the value of sin(2x)?
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OpenStudy (anonymous):
sin(2x) would be sinx sinx - cosx cosx
?
OpenStudy (anonymous):
nope. sin(A+B) is sinAcosB + sinBcosA
so sin(2x) = sin(x+x) = sinxcosx + sinxcosx
OpenStudy (anonymous):
that gives you 2sinxcosx
OpenStudy (anonymous):
a)cos x - 4 cos x sin2x
b) -sin3x + 2 sin x cos x
c) -sin2x + 2 sin x cos x
d) 2 sin2x cos x - 2 sin x cos x
Those are the answers. I think all we need to do now is simplify