Plz help I need help
2 answers:
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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
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<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
