If 0 < b ≤ 90 and cos(11b + 2) = sin(12b − 4), what is the value of b?

The value of b from the given equation is 4

Given the equation cos(11b + 2) = sin(12b − 4)

This can also be expressed as:

cos(11b + 2) = cos [90-(12b − 4)]

cos(11b + 2) = cos[90-12b+4]

cos(11b + 2) = cos(94-12b)

11b + 2 = 94 - 12b

11b + 12b = 94 - 2

23b = 92

b = 4

Hence the value of b from the given equation is 4

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b = 4

Step-by-step explanation:

The given expression is

Cos (11b + 2) = Sin (12b - 4)

Since, Sin(θ) = Cos(90 - θ)

Therefore, Cos (11b + 2) = Cos [90 - (12b - 4)]

Where 0 < b ≤ 90°

11b + 2 = 90 - (12b - 4)

11b + 2 = 94 - 12b

11b + 12b + 2 = 94 - 12b + 12b

11b + 12b + 2 - 2 = 94 - 2

23b = 92

b = 4

Therefore, b = 4 will be the answer.