Question

两道三角函数题求详解

详细过程，答的好另加分，谢谢！

Answer 1

sintheta >= 1/2, so pi/6<= theta <= 5pi/6
cos theta <= 1/2 so pi/3 <= theta <= pi
A belongs to [pi/3, 5pi/6]
cos^2 B + sin B = 1-sin^2 B + sin B =5/4 - (sinB - 1/2)^2
B = A-7pi/12
B belongs to [-3pi/12,3pi/12] => [-pi/4,pi/4]
in [-pi/4,pi/4], min of |sinB - 1/2| = 0, maximum = [sqrt(2)+1]/2
5/4 - (sinB - 1/2)^2 maximum is 5/4, minimum is 5/4 - (sqrt(2)+1)^2 /4 =[ 1-sqrt(2)]/2
所以值域为[1-sqrt(2)]/2, 5/4

追问

22题呢？求解，谢谢！

追答

22. f(x) = 2sqrt(3)sin(3wx+pi/3)
f(x+theta) = 2sqrt(3) sin(3w(x+theta)+pi/3) = 2sqrtsin(3wx + 3wtheta + pi/3)
3w = 1, w = 1/3
3wtheta + pi/3 = pi/2, theta = pi/6

(2)
3w*pi/3 + pi/3 <= pi/2
w <= 1/6