Can someone please help me understand how
sin(4Θ + 2°) csc(3Θ + 5°) = 1, has an answer of 3°
Our professor really didn't go over identities in depth and I have a loose understanding of them but I really can't fathom this one.
sin(4Θ + 2°) csc(3Θ + 5°) = 1, has an answer of 3°
Our professor really didn't go over identities in depth and I have a loose understanding of them but I really can't fathom this one.
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This is not an identity. It is algebra.
Since csc = 1/sin you have and equation that says sin/sin = 1
This can only be true when the 2 sin values are the same, therefore 4theta + 2 must equal 3theta +5
Set those 2 values equal to one another in an equation and solve for theta.
4x + 2 = 3x + 5...Solve for x
Since csc = 1/sin you have and equation that says sin/sin = 1
This can only be true when the 2 sin values are the same, therefore 4theta + 2 must equal 3theta +5
Set those 2 values equal to one another in an equation and solve for theta.
4x + 2 = 3x + 5...Solve for x
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3° is just one of the solutions though!!