How do you find the minimum of the equation sin^2(x) + 4sin(x)?
The first step to solving this problem is to treat it as a normal quadratic equation; if you are struggling with comparing our equation to a normal quadratic, try substituting sin(x) = y into the equation as shown:
sin2(x) + 4sin(x) becomes y2 + 4y
Even though our equation does not equal 0 we can still use the 'complete the square' method to help us find the minimum, after applying this method our equation becomes:
(y + 2)2 - 4
From this we can substitue y for sin(x), giving:
(sin(x) + 2)2 - 4
To find the minimum of our equation we have to take in to account the fact that sin(x) has a range of -1 to 1, which limits (sin(x) + 2) to a range of 1 to 3.
From this you should be able to deduct that the smallest value of (sin(x) + 2)2 - 4 is -3. This occurs when sin(x) = -1.
(-1 + 2)2 - 4 = 12 - 4 = -3
Hence the minimum of sin2(x) + 4sin(x) is -3.
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