Question

How do you evaluate `cos((3pi)/5)cos((4pi)/15)+sin((3pi)/5)sin((4pi)/15)`?

Answer 1

Use the standard trig. formula:
`cos( a - b ) = cos(a)*cos(b) + sin(a)*sin(b)`

`cos( (3Pi)/5) cos( (4Pi)/15) + sin( (3Pi)/5) sin( (4Pi)/15)` becomes

`cos ( (3Pi)/5 - (4Pi)/15)`

`= cos( (5Pi)/15) = cos( Pi/3)`

`Pi/3` (`60^o`) is one of the standard triangles

#cos(Pi/3) = 1/2