Question

:cos alpha cos(60degree-alpha)cos(60degree+alpha)=1/4cos 3 alpha

Answer 1

are they being multiplied together?

Answer 2

yes.....

Answer 3

angle sum and difference formulas for cos
\[\cos(a+b) = \cos a*\cos b - \sin a*\sin b\]
\[\cos(a-b) = \cos a*\cos b +\sin a*\sin b\]

cos(60) = 1/2

Answer 4

and aftr dat??

Answer 5

\[\cos(a+60) = (1/2)\cos a - (\sqrt{3}/2)\sin a\]
\[\cos(a-60) = (1/2)\cos a + (\sqrt{3}/2)\sin a\]
\[\cos(a+60)*\cos(a-60) = (1/4)\cos^{2} a - (3/4)\sin^{2} a\]

Answer 6

multiply by cos(a)
\[\rightarrow (1/4)\cos^{3} a - (3/4)\cos a*\sin^{2} a\]
sin^2 = 1 - cos^2
\[\rightarrow (1/4)\cos^{3} a - (3/4)\cos a*(1-\cos^{2} a) = (1/4)\cos^{3} a +(3/4)\cos^{3} a -(3/4)\cos a\]
\[= \cos^{3} a - (3/4)\cos a\]

Answer 7

ok yeah this equals (1/4)cos(3a)
\[\cos 3a = \cos 2a*\cos a - \sin 2a*\sin a\]
\[= \cos a(2\cos^{2} a -1) - 2\cos a*\sin^{2} a\]
\[= (2\cos^{3} a-\cos a) - 2\cos a(1-\cos^{2} a)\]
\[= 4\cos^{3} a - 3\cos a\]