How do you use an addition or subtraction formula to write the expression as a trigonometric function of one number and then find the exact value given #cos (7pi/6) cos (-pi/ 2)-sin(7pi/6)sin(-pi/2) #?

1 Answer
May 18, 2015

Recall the formula for #cos(alpha+beta)#:
#cos(alpha+beta)=cos(alpha)*cos(beta)-sin(alpha)*sin(beta)#

Using this formula for #alpha=7pi/6# and #beta=-pi/2#, we get:

#cos(7pi/6)*cos(-pi/2)-sin(7pi/6)*sin(-pi/2)=#
#=cos(7pi/6+(-pi/2))=#
#=cos((7pi)/6-(3pi)/6)=#
#=cos((2pi)/3)=-cos(pi-(2pi)/3)=-cos(pi/3)=-1/2#

As a check, notice that
#cos(7pi/6)=-cos(pi/6)=-sqrt(3)/2#,
#cos(-pi/2)=0#,
#sin(7pi/6)=-sin(pi/6)=-1/2#,
#sin(-pi/2)=-1#.
Performing direct calculations, we get the same #-1/2#.
Check!