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3 years ago

PLEASE HELP ME If 0 < z ≤ 90 and sin(9z − 1) = cos(6z + 1), what is the value of z? z = 3 z = 4 z = 5 z = 6

Mathematics

1 answer:

Burka [1]3 years ago

6 0

Answer:

z = 6

Step-by-step explanation:

We know that ...

sin(x) = cos(90 -x)

Substituting (9z-1) for x, this is ...

sin(9z -1) = cos(90 -(9z -1))

But we also are given ...

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

Equating the arguments of the cosine function, we have ...

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

90 = 15z . . . . . . . . . add (9z-1) to both sides

6 = z . . . . . . . . . . . . divide by 15

_____

<em>Comment on the graph</em>

The attached graph shows 5 solutions in the domain of interest. These come from the fact that the relation we used is actually ...

sin(x) = cos(90 +360k -x) . . . . . for any integer k

Then the above equation becomes ...

90 +360k = 15z

6 +24k = z . . . . . . . . . for any integer k

The sine and cosine functions also enjoy the relation ...

sin(x) = cos(x -90)

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

3z = 92 . . . . . equating arguments of cos( ) and adding 91-6z

z = 30 2/3

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