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

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7 votes

Answer:

6

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

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