a) sin^2(-π/3) + cos^2(-π/6)
First, we can find the values of sin(-π/3) and cos(-π/6):sin(-π/3) = -√3/2cos(-π/6) = √3/2
Now we can substitute these values into the equation:sin^2(-π/3) = (-√3/2)^2 = 3/4cos^2(-π/6) = (√3/2)^2 = 3/4
Therefore:3/4 + 3/4 = 6/4 = 3/2
So, sin^2(-π/3) + cos^2(-π/6) = 3/2
b) 4sin(π/6) sin^4(π/4) tan^2(π/3)
First, we can find the values of sin(π/6), sin(π/4), and tan(π/3):sin(π/6) = 1/2sin(π/4) = √2/2tan(π/3) = √3
Now we can substitute these values into the equation:4sin(π/6) sin^4(π/4) tan^2(π/3) = 4(1/2)(√2/2)^4(√3)^2= 2(1/2)(2/4)(3)= 1(1/2)(2)*(3)= 3
Therefore, 4sin(π/6) sin^4(π/4) tan^2(π/3) = 3
a) sin^2(-π/3) + cos^2(-π/6)
First, we can find the values of sin(-π/3) and cos(-π/6):
sin(-π/3) = -√3/2
cos(-π/6) = √3/2
Now we can substitute these values into the equation:
sin^2(-π/3) = (-√3/2)^2 = 3/4
cos^2(-π/6) = (√3/2)^2 = 3/4
Therefore:
3/4 + 3/4 = 6/4 = 3/2
So, sin^2(-π/3) + cos^2(-π/6) = 3/2
b) 4sin(π/6) sin^4(π/4) tan^2(π/3)
First, we can find the values of sin(π/6), sin(π/4), and tan(π/3):
sin(π/6) = 1/2
sin(π/4) = √2/2
tan(π/3) = √3
Now we can substitute these values into the equation:
4sin(π/6) sin^4(π/4) tan^2(π/3) = 4(1/2)(√2/2)^4(√3)^2
= 2(1/2)(2/4)(3)
= 1(1/2)(2)*(3)
= 3
Therefore, 4sin(π/6) sin^4(π/4) tan^2(π/3) = 3