*Note: I am assuming that "昨變" is a typo and should be "再變" for the second expression. To solve the equation: 2tanα/(1-tan2α) = -2√2 First, we can simplify the left side of the equation by using the trigonometric identity: tan2α + 1 = sec2α: 2tanα/(1-tan2α) = -2√2 2tanα/sec2α = -2√2 2sinα/cosα * cos2α/sin2α = -2√2 2sinα * cosα * cos2α / (cosα * sin2α) = -2√2 2cos3α / sinα = -2√2 Next, we can simplify further by dividing both sides of the equation by 2: cos3α / sinα = -√2 Now, we can rewrite the equation using the trigonometric identity: sin2α + cos2α = 1: (cosα/sinα) * cos2α = -√2 cos3α/sinα = -√2 Now, we can square both sides of the equation to eliminate the square root: (cos3α/sinα)2 = (-√2)2 cos?α / sin2α = 2 Using the identity: cos2α = 1 - sin2α (1 - sin2α)3 / sin2α = 2 Now, we can expand the numerator and simplify the equation: (1 - 3sin2α + 3sin?α - sin?α) / sin2α = 2 1 - 3sin2α + 3sin?α - sin?α = 2sin2α -sin?α + 3sin?α - 5sin2α + 1 = 0 This is now a sextic equation in terms of sinα. To solve for sinα, we would need to use numerical methods or approximation techniques.

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