Question

please help!

cos(pi/3+b)

1/2(cosb-root3sinb)
1/2(cosb+root3sinb)
root3(cosb-1/2sinb)
root3(cosb+1/2sinb)

Answer 1

This?\[\cos\left(\dfrac{\pi}{3} + b\right)\]

Answer 2

If so, in that case, use\[\cos(a + b) = \cos (a)\cos(b) - \sin(a)\sin(b)\]

Answer 3

yes i got it till that part!

Answer 4

Also remember that \(\cos\left(\pi/3\right) = 1/2\) and \(\sin(\pi/3) = \sqrt3 /2\). That is all you need!

Answer 5

yes i remember! but the problem that i have is the last part of solving ): the calculation part; im kinda confused!

Answer 6

OK. Let's do this!

Answer 7

\[\cos\left(\dfrac{\pi}{3} + b\right) = \cos \left(\dfrac{\pi}{3} \right)\cos(b) - \sin\left(\dfrac{\pi}{3}\right)\sin(b)\]

Answer 8

It really can't be hard from here. Can it?

Answer 9

then it would be
(1/2) cos(b) - (root3) sin (b)??

Answer 10

Yup, except that is \(\dfrac{\sqrt3}{2}\)

Answer 11

@ParthKohli thank you so much!!

Answer 12

You're welcome!