Asked by zaki
cos 4x=7/18. 4x is on quadrant 4. sin x?
Answers
Answered by
Steve
3pi/2 < 4x < 2pi, so
3pi/8 < x < pi/2
sin2x = √((1+cos4x)/2)
sinx = √((1+cos2x)/2)
sin2x = √((1+7/18)/2) = √(25/36) = 5/6
so, cos2x = √11/6
sinx = √((1+√11/6)/2) = 1/2 √(2+√11/3)
3pi/8 < x < pi/2
sin2x = √((1+cos4x)/2)
sinx = √((1+cos2x)/2)
sin2x = √((1+7/18)/2) = √(25/36) = 5/6
so, cos2x = √11/6
sinx = √((1+√11/6)/2) = 1/2 √(2+√11/3)
Answered by
Steve
oops -- got my formulas mangled:
cos2x = √((1+cos4x)/2)
sinx = √(1-cos2x)/2)
cos2x = 5/6, as above
sinx = √((1 - 5/6)/2) = √(1/12) = 1/(2√3) = √3/6
Now, that's a lot less work!
cos2x = √((1+cos4x)/2)
sinx = √(1-cos2x)/2)
cos2x = 5/6, as above
sinx = √((1 - 5/6)/2) = √(1/12) = 1/(2√3) = √3/6
Now, that's a lot less work!
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