cos^2(pi/6)/sec^2(pi/4)sin(pi/3)
How would I approach this?
How would I approach this?
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cos(pi/6) = sqrt(3)/2 (look at an equilateral triangle)
so cos^2(pi/6)=3/4
sec^2(pi/4) = 1 / cos^2(pi/4) = 1 / (1/2) = 2
sin(pi/3) = cos(pi/6) = sqrt(3)/2 so
cos^2(pi/6)/sec^2(pi/4)sin(pi/3) = (3/4) / (2sqrt(3)/2) = (3/2) / (sqrt(3)) = sqrt(3)/2
so cos^2(pi/6)=3/4
sec^2(pi/4) = 1 / cos^2(pi/4) = 1 / (1/2) = 2
sin(pi/3) = cos(pi/6) = sqrt(3)/2 so
cos^2(pi/6)/sec^2(pi/4)sin(pi/3) = (3/4) / (2sqrt(3)/2) = (3/2) / (sqrt(3)) = sqrt(3)/2