since cos(pi/2) = 0, you have
sin(pi/4) cos(pi/12) = sin(pi/4) cos ((pi/6)/2)
= 1/√2 * (1+√3)/(2√2) = (1+√3)/4
on the off chance you had a typo, review your sum/difference formulas to see that
cosx cosy - sinx siny = cos(x+y)
use an appropriate compound angle formula to express as a single trig function, and determine the exact value for cos(pi/2)cos(pi/12) - sin(pi/4)sin(pi/12)
3 answers
cos(pi/2)cos(pi/12) - sin(pi/4)sin(pi/12)
I suspect a typo , perhaps you meant
cos(pi/2)cos(pi/12) - sin(pi/2)sin(pi/12)
If so, then
you should recognize the expansion:
cos(A + B) = cosAcosB - sinAsinB
so we get:
cos(π/2 + π/12)
= cos (7π/12)
If you meant it the way you typed it, all except the cos π/12 should
be in your list of common trig ratios.
for the cos π/12 or cos 15°
you could use the half angle formula
cos 30° = 2cos^2 (15) - 1
√3/2 + 1 = 2cos^2 (15)
cos 15 = √[(√3/2 + 1)/2]
I suspect a typo , perhaps you meant
cos(pi/2)cos(pi/12) - sin(pi/2)sin(pi/12)
If so, then
you should recognize the expansion:
cos(A + B) = cosAcosB - sinAsinB
so we get:
cos(π/2 + π/12)
= cos (7π/12)
If you meant it the way you typed it, all except the cos π/12 should
be in your list of common trig ratios.
for the cos π/12 or cos 15°
you could use the half angle formula
cos 30° = 2cos^2 (15) - 1
√3/2 + 1 = 2cos^2 (15)
cos 15 = √[(√3/2 + 1)/2]
I’m sad I almost got AB honor roll but I have a 70 in math :< all my other grades are good :(