If 0 < z ≤ 90 and sin(9z − 1) = cos(6z + 1), what is the value of z?

Step-by-step explanation:

Given the expression sin(9z − 1) = cos(6z + 1), for 0 < z ≤ 90, we are to find z

Note that sin theta = cos(90-thetsa)

sin(9z − 1) = cos(6z + 1)

cos (90 - (9z-1)) = cos(6z + 1)

cos (90 - 9z+1)) = cos(6z + 1)

cos (91-9z) = cos(6z + 1)

Cos will cancel out to have;

91-9z = 6z+1

-9z - 6z = 1-91

-15z = -90

z = 90/15

z = 6

Hence the value of z is 6

22

step-by-step explanation:

ab will be equal to