Given that x is an acute angle, and the square of SiNx - the square of sinxcosx-2cosx = 0, the first question: find the value of TaNx Second question: find the value of sin (x - π / 3)

Given that x is an acute angle, and the square of SiNx - the square of sinxcosx-2cosx = 0, the first question: find the value of TaNx Second question: find the value of sin (x - π / 3)

(1) The square of SiNx - the square of sinxcosx-2cosx = 0. Obviously, both sides of cosx ≠ 0 are divided by the square of cosx to get the square of TaNx - tanx-2 = 0, TaNx = 2 or - 1, and X is an acute angle, so TaNx = 2 (2) TaNx = 2 x is an acute angle, so SiNx = 2 / √ 5 cosx = 1 / √ 5sin (x - π / 3) = 1 / 2sinx - √ 3 / 2cosx =

Given TaNx = 2, then SiNx ^ 2 + sinxcosx-2cosx ^ 2=

sinx/cosx=tanx=2
sinx=2cosx
Substitute sin ² x+cos ² x=1
Then cos ² x=1/5,sin ² x=4/5
sinxcosx=(2cosx)cosx=2cos ² x=2/5
So the original formula = 4 / 5 + 2 / 5-2 / 5 = 4 / 5

Given that x is an acute angle and TaNx = 3, then 2cosx + SiNx / SiNx sinx-2cosx =?

Divide the numerator and denominator by cosx to get the original formula = (tanx-2) / (TaNx + 2) = 1 / 5

Given TaNx = 2, then SiNx square + sinxcosx-2cosx square =?

This question is the application of string cutting method. Original formula = [sin ² x＋sinxcosx－2cos ² x]/[sin ² x＋cos ² x] 【1=sin ² x＋cos ² x】=[tan ² x＋tanx－2]/[1＋tan ² x] [numerator and denominator are the same as division by cos ² x. And SiNx / cosx = TaNx

The minimum positive period of y = sin4x + cos4x is __

∵y=sin4x+cos4x=(1−cos2x
2)2+(1+cos2x
2)2
=2+2cos22x
4=cos4x+3
4=1
4cos4x+3
four
The minimum positive period of the function is 2 π
4＝π
two
So the answer is: π
two

Minimum positive period of function y = sin ^ 4x + cos ^ 4x

y=sin^4x+cos^4x
y=[(sinx)^2+(cosx)^2]-2(sinxcosx)^2
=1-(2sinxcosx)^2/2
=1-(sin2x)^2/2
=0.75+(1-2(sin2x)^2)/4
=0.75+(cos4x)/4
Minimum positive period pi / 2