Simplification: 1 − (sin4x − sin2xcos2x + cos4x) sin2x + 3sin2x

Simplification: 1 − (sin4x − sin2xcos2x + cos4x) sin2x + 3sin2x

The original formula = 1 − [(sin2x + cos2x) 2 − 3sin2xcos2x] sin2x + 3sin2x = 3sin2xcos2xsin2x + 3sin2x = 3cos2x + 3sin2x = 3 (sin2x + cos2x) = 3

Simplification (1 + sin4x-cos4x) / (1 + sin4x + cos4x)

（1+sin4X-cos4X)=(sin2x)^2+(cos2x)^2+2sin2xcos2x-(cos2x)^2+(sin2x)^2=2(sin2x)^2+2sin2xcos2x=2sin2x(sin2x+cos2x)（1+ sin4X+cos4X）=(sin2x)^2+(cos2x)^2+2sin2xcos2x+(cos2x)^2-(sin2x)^2=2(cos2x)^2+2sin2xc...

Simplify 1 + cos4x + sin4x (1-cos4x + sin4x)

I'll do it later

Simplification: cos4x-2sinxcosx-sin4x

f(x)=(cos²x+sin²x)(cos²x-sin²x)-2sinxcosx
=1×cos2x-sin2x
=√2(cos2x×√2/2-sin2x×√2/2)
=√2(cos2xcosπ/4-sin2xsinπ/4)
=√2cos(2x+π/4)

If Tan α = m, then sin (- 5 π - α) cos (3x + α)=

sin(-5π-α)cos(3π+α）
=sina*(-cosa)
=-sinacosa
=-The denominator of sinacosa / (sin ^ 2A + cos ^ 2a) is also divided by cos ^ 2A
=-tana/(1+tan^2a) tanα=m
=-m/(1+m^2)

Limx→o(x-sinx)/tan^3x

This paper is to find the limit of X \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\x) / (6sinxcos

Finding the minimum positive period of Tan (x / 3) - cos (3x)

The minimum positive period of Tan (x / 3) is π / (1 / 3) = 3 π,
The minimum positive period of COS (3x) is 2 π / 3,
Therefore, the minimum positive period of the function is 6 π. (2 * 3 π = 6 π, 9 * 2 π / 3 = 6 π)

Given sin2x = 3 / 5, find cos2x and cos2x
The other is cosx

sin2x=3/5
cos2x =±√{1-（3/5)^2}= ±4/5
cosx =±√{1+cos2x}/2= ±2 √5/5