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The minimum value of the function y = cosx + cos2x is______ .
y=cosx+cos2x=cosx+2cos2x-1，
Let cosx = t, then - 1 ≤ t ≤ 1,
Function f (T) min = f (- 1)
4）=1
2-1
4-1=-5
4，
So the answer is: - 5
4．
The value range of the function y = cos2x-cosx-1 is
y=cos2x-cosx-1
=2cos^2x-1-cosx-1
Let cosx = X
Then y = 2x ^ 2-x-2
Opening up
Axis of symmetry x = - B / (2a) = 1 / 4
The minimum value is: y = (4ac-b ^ 2) / (4a) = - 17 / 8
So the range is [- 17 / 8, positive infinity)
It is known that TaNx = 2, the value of cos2x / (sinx-cosx) ^ 2 is?
cos2x=(cosx)^2-(sinx)^2
The original formula is (cosx + SiNx) / (cosx SiNx)
Then the denominator is divided by cosx to (1 + TaNx) / (1-tanx) = - 3
It is proved that 1-sin2x / cos2x = 1-tanx / 1 + TaNx How does cosx SiNx / cosx + SiNx become 1-tanx / 1 + TaNx? Where does the 1 come from? Isn't it cosx?
【1】
∵tanx=(sinx)/(cosx)
∴sinx=cosxtanx.
right
=(cosx cosxtanx) / (cosx + cosxtanx)
=(cosx-sinx)/(cosx+sinx).
=(cosx-sinx)²/[(cosx+sinx)(cosx-sinx)]
=(cos²x-2sinxcosx+sin²x)/(cos²x-sin²x)
=(1-sin2x)/cos(2x).
=On the left
Proof of formula TaNx = (1-cos2x) / sin2x
(1-cos2x)/sin2x
=[1-(1-2sin²x)]/(2sinxcosx)
=(2sin²x)/(2sinxcosx)
=sinx/cosx
=tanx
Simplify TaNx + tan2x + TaNx * tan2x * tan3x
Because Tan (x + 2x) = (TaNx + tan2x) / (1-tanx * tan2x), TaNx + tan2x = (TaNx + tan2x) * (1-tanx * tan2x) = tan3x * (1-tanx * tan2x) = tan3x-tan3x * TaNx * tan2x
Confirmation: 1 / sin2x = 1 / tanx-1 / tan2x
1/sin2x
=(1+cos2x)/sin2x-cos2x/sin2x
=(1+cos2x)/sin2x-1/tan2x
=2cosxcosx/(2sinxcosx)-1/tan2x
=1/tanx-1/tan2x
It is proved that tanx-1 / TaNx = - 2 / tan2x
tanx-1/tanx
=sinx/cosx-cosx/sinx
=2(sinx^2-cosx^2)/sin2x
=-2/tanx
Solution set of tan3x = TaNx Write down the process, thank you!
TaNx = (TaNx + tan2x) / (1-tanxtan2x) ν (TaNx + tan2x) / (1-tanxtan2x) = TaNx divisor TaNx + tan2x = (1-tanxtan2x) * tanxtanx + tan2x = tanx-tan-xtan2x
It is proved that cos2x = cos4x-sin4x
cos²x+sin²x=1
So cos2x = cos? X-sin? X = (COS? X-sin? X) (COS? X + sin? X) = cos4x-sin4x
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