Given the function y = radical 3sin2x - cos2x + 1. Find the minimum positive period of the function The formula of auxiliary angle should be used to calculate the original formula = 2 ((root 3 / 2) SiNx - (1 / 2) cosx) + 1 = 2six (2x - π / 6) + 1. I wonder how to get the step of π / 6?

Given the function y = radical 3sin2x - cos2x + 1. Find the minimum positive period of the function The formula of auxiliary angle should be used to calculate the original formula = 2 ((root 3 / 2) SiNx - (1 / 2) cosx) + 1 = 2six (2x - π / 6) + 1. I wonder how to get the step of π / 6?

The original formula = 2 ((root 3 / 2) SiNx - (1 / 2) cosx) + 1
=2[sinxcosπ/6-cosxsinπ/6]+1
=2six﹙2x－π／6﹚＋1

Let f (x) = radical 3sin2x + cos2x + A + 1,1. If the maximum value is 2, find the value of A

f(x)=√3sin2x+cos2x+a+1
=2[sin2x*(√3/2)+cos2x*(1/2)]+a+1
=2[sin2x*cos(π/6)+cos2x*sin(π/6)]+a+1
=2sin(2x+π/6)+a+1
The range of sine function is [- 1,1]
So the maximum value of F (x) is 2 + A + 1 = 3 + a = 2
So a = - 1,

Given that f (x) = cos2x + root sign 3sin2x-1, X belongs to R 1, find the value of F (π / 6), find the minimum positive period, maximum value and minimum value of the function

F (x) = cos2x + radical 3sin2x-1 = 2Sin (2x + Pai / 6) - 1
f(Pai/6)=2sin(Pai/3+Pai/6)-1=2sinPai/2-1=2-1=1
Minimum positive period T = 2pai / 2 = Pai
The maximum value is: 2-1 = 1
The minimum value is: - 2-1 = - 3

Given that f (x) = cos2x + root sign 3sin2x-1, the minimum positive period, maximum value and minimum value of the function are obtained

F (x) = 2 (cos2x / 2 + Radix 3sin2x / 2) - 1
=2(cos2x*cos60°+sin2x*sin60°)-1
=2cos(2x-60°）-1
So t = 2 π / 2 = π
Fmax = 2-1 = 1
Fmin = - 2-1 = - 3

If y = f (x) SiNx is an odd function with period π, then f (x) can be a SiNx B cosx C sin2x D cos2x reason

B cosx
because
y=cosxsinx = 1/2 * sin(2x)
It happens to be an odd function with period = 2 π / 2 = π
To be familiar with sin's double angle formula, the problem is very simple
For a, y = SiNx ^ 2 = (cos2x-1) / 2
The period is π
But it's not an odd function
CD is not a periodic function

What are the periods of sin ^ 2x, sin2x * SiNx, SiNx * cos2x? The question is how many, how many, how many, how many, how many

(SiNx) ^ 2 = (1-cos2x) / 2, because the period of cosx is 2 π, the period of SiNx ^ 2 is π;
Sin2x * SiNx = 2 (SiNx) ^ 2cosx = 2 [cosx - (cosx) ^ 3], because the period of cosx is 2 π, so the period of the original formula is 2 π;
SiNx * cos2x = SiNx (1-2 (SiNx) ^ 2) = sinx-2 (SiNx) ^ 3, because the period of SiNx is 2 π, so the period of the original formula is 2 π

If sin ^ x-sinxcosx-6cos ^ x = 0, then [(cos2x-sin2x) / (1-cos2x) (1-tan2x)] =?

Here's the answer

SiNx = sin2x and cosx = cos2x

analysis
sinx=2sinxcosx
Qi
cosx=1/2
cosx=cos2x
Then cosx = 2cos? X-1
It is similar to the quadratic equation of one variable

The known function f (x) = 3sin ^ 2x + 2 √ 3 sinxcosx + 5cos ^ 2x 1. Find the period and maximum of function f (x) 2. When f (a) = 5, find the value of Tan a

1、f(x)=3sin��x+2√3 sinxcosx+5cos��x=3(1-cos2x)/2+√3sin2x+5(1+cos2x)/2=3/2-3/2cos2x+√3sin2x+5/2+5/2cos2x=5/2cos2x-3/2cos2x+√3sin2x+3/2+5/2=cos2x+√3sin2x+4=2(1/2cos2x+√3/2sin2x)+4=2sin(2x+π/...

The known function f (x) = - √ 3sin ^ 2x + sinxcosx Find the minimum positive period and maximum value of function f (x)

f(x)=-√3sin²x+sinxcosx=(1/2)*sin2x+(√3/2)*cos2x-√3/2=sin(2x+π/3)-√3/2
So the minimum positive period of the function f (x) is t = 2 π / 2 = π
The maximum value is 1 - √ 3 / 2
The minimum value is - 1 - √ 3 / 2
If you do not understand, please hi me, I wish you a happy study!