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Radical difference trigonometry (Posted on 2023-04-16) Difficulty: 3 of 5
Simplify the expression:

√[sin4x+4cos2x] - √[cos4x+4sin2x]

Source: AMC12 2002

  Submitted by Jer    
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Solution: (Hide)
√[sin^4(x)+4cos^2(x)] - √[cos^4(x)+4sin^2(x)]
√[sin^4(x)+4(1-sin^2(x))] - √[cos^4(x)+4(1-cos^2(x))]
√[sin^4(x)-4sin^2(x)+4] - √[cos^4(x)-4cos^2(x)+4]
√[(sin^2(x)-2)^2] - √[(cos^2(x)-2)^2]

The trap is here: √[x^2] does not equal x, it equals |x|. In this case sin^2(x)-2 and cos^2(x)-2 are always negative, so we get

-(sin^2(x)-2) - -(cos^2(x)-2)
cos^2(x) - sin^2(x) which is a double angle formula for
cos(2x)

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolutionBrian Smith2023-04-22 15:29:31
Hints/Tipsre: simplified / two wrongs made a right?Jer2023-04-18 09:29:15
SolutionsimplifiedLarry2023-04-16 08:11:53
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