Question

cos(x+pi/3)-cos(x-pi/3)=-sqrt(3)
the possible answers are:
x= pi/2
x=2pi/3 +2pin
x= pi/2 +2pin
none of the above

I attempted to use cos(u) - cos (v)= 2*sin((u+v/)2)*sin((u-v)/2)
2*sin(((x+pi/3)+(x-pi/3)/)2)*sin((x+pi/3)-(x-pi/3)/2)
2* sin(x) * sin( pi/3)
2* sqrt(3)/2* sin(x)
sqrt(3)*sin(x) = -sqrt(3)
sin(x)= -sqrt(3)/ sqrt(3)
x= arcsin(-1)
That would make x = -pi/2 correct?? Why is it not an option??

Answer 1

Is it none of the above

Answer 2

i ran -pi/2 though the equation and it isnt a solution, where did i mess up?

Answer 3

Ah you need the formula \[\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\]

Answer 4

Keith... where would i use that formla

Answer 5

From that you can break up \(\cos(x+\pi/3)\) and \(\cos(x-\pi/3)\)

Answer 6

cos(x)cos(pi/3) - sin(x)sin(pi/3) - cos(x)cos(-pi/3) - sin(x)sin(-pi/3)
like that?

Answer 7

So you would have:
\[\cos(x+\pi/3)-\cos(x-\pi/3) \\
=\cos(x)\cos(\pi/3) - \sin(x)\sin(\pi/3) - \cos(x)\cos(\pi/3) - \sin(x)\sin(\pi/3) \\
=-2\sin(x)\sin(\pi/3)=-\sqrt{3}\]

Answer 8

And so with algebra you result in \[\sin(x)=\frac{\sqrt{3}}{2\sin(\pi/3)}\]

Answer 9

shouldn't pi/3 be negative in the second half of the equation?

Answer 10

No because the equation is
\[cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\]
So:
\[cos(a-b)=cos(a+(-b))=\cos(a)\cos(-b)-\sin(a)\sin(-b)\]
Since \(\cos(b)=\cos(-b)\) and \(\sin(-b)=-\sin(b)\) then:
\[\cos(a-b)=\cos(a)\cos(b)-\sin(a)\times-\sin(b)=\cos(a)\cos(b)+\sin(a)\sin(b)\]

Answer 11

I see, thanks :)

Answer 12

Yeah so \(\pi/2\) is the correct answer