Let SiNx + cosx = root 2, then the value of the fourth power of SiNx + the fourth power of cosx is

Let SiNx + cosx = root 2, then the value of the fourth power of SiNx + the fourth power of cosx is

sinx+cosx=√2
square
sin ² x+cos ² x+2sinxcosx=2
1+2sinxcosx=2
sinxcosx=1/2
Original formula = (SIN) ² x+cos ² x) ²- 2(sinxcosx) ²
=1 ²- two × (12) ²
=1/2

Given SiNx + cosx = m, find the value of the third power of SiNx + the third power of cosx?

SiNx + cosx = m square sin ²+ cos ² x+2sinxcosx=1+2sinxcosx=m ² sinxcosx=(m ²- 1) / 2 original formula = (SiNx + cosx) (SIN) ² x-cosxsinx+cos ² x)=m[1-(m ²- 1)/2]=(2m-m ³+ 1)/2

Simplification: [1 - (cosx) 4th power - (SiNx) 4th power] / [1 - (cosx) 6th power - (SiNx) 6th power]

Tip: Set 1 = sin ²+ cos ² Substitute, extract the common factor and sort it out. The result is 2 / 3

Function f (x) = the fourth power of cosx - the fourth power of 2sinxcosx SiNx, find the minimum positive period of F (x)? If x belongs to zero to 2 π, find the maximum and small value

f(x)=(cosx)^4-2sinxcosx-(sinx)^4
=[(cosx)^2+(sinx)^2]*[(cosx)^2-(sinx)^2]-sin2x
=cos2x-sin2x
=√2cos(2x+π/4)
So t = 2 π / 2 = π
x∈[0,2π]
2x+π/4∈[π/4,17π/4]
Therefore, the maximum value of F (x) is √ 2 and the minimum value is - √ 2
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Function f (x) = (cosx) 4th power - (SiNx) 4th power, find the period

f(x)=cos⁴x-sin⁴x
=(cos ² x+sin ² x)(cos ² x-sin ² x)
=1 × cos(2x)
=cos(2x)
Minimum positive period Tmin = 2 π / 2 = π

Let the function y = y (x) be determined by the square of equation x + 2 (square of Y) = 4, and find dy / DX

Both sides have the same differential, 2x * DX + 2Y * dy = 0, so dy / DX = - X / y = - X / radical 4-x ^ 2