TaNx = SiNx / Cox = 2 SiNx = 2cosx sin ^ 2x + cos ^ 2x = 1 Why does the book say "so cos ^ 2x = 1 / 5?

TaNx = SiNx / Cox = 2 SiNx = 2cosx sin ^ 2x + cos ^ 2x = 1 Why does the book say "so cos ^ 2x = 1 / 5?

Replace SiNx = 2cosx into sin ^ 2x + cos ^ 2x = 1
5cos ^ 2x = 1, so cos ^ 2x = 1 / 5

The known function f (x) = 3cos ^ 2x + 2cosx * SiNx + sin ^ 2x (1) Finding the period of a function (2) Write the monotone interval of F (x) (3) Find the maximum value of F (x), and find the value of X at this time

f(x)=3(cosx)^3+2sinxcosx+(sinx)^2
=sin2x+2(cosx)^2+1
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
(1) The minimum positive period is t = 2 π / 2 = π, the period is k π, K is an integer which is not 0
(2) If 2K π - π / 2 < 2x + π / 4 < 2K π + π / 2, then K π - 3 π / 8 2K π + π / 2 < 2x + π / 4 < 2K π + 3 π / 2, then K π + π / 8
(3) When 2x + π / 4 = 2K π + π / 2, that is, x = k π + π / 8, the maximum value of F (x) is √ 2 + 2

sin^2x+2sinxcosx+3cos^x Write down the specific process, summer vacation has not reviewed, the previous study has forgotten

The last item in the title should be squared
If we use the formula of decreasing power, that is, the formula of cosine double angle is reversed. (SiNx) ^ 2 = (1-cos2x) / 2. (cosx) ^ 2 = (1 + cos2x) / 2
There is also the sine double angle formula: sin2x = 2sinxcosx
The original formula is simplified as follows:
sin^2x+2sinxcosx+3cos^2x=(1-cos2x)/2 +sin2x+3(1+cos2x)/2
=sin2x+cos2x+2
=√2[√2/2sin2x+√2/4cos2x]+2
=√2[cosπ/4sin2x+sinπ/4cos2x]+2
=√2sin(2x+π/4)+2

The known function y = sin2x + 2sinxcosx + 3cos2x, X ∈ R (1) The minimum positive period of function y; (2) The increasing interval of function y

（1）y=sin2x+2sinxcosx+3cos2x
=（sin2x+cos2x）+sin2x+2cos2x
=1+sin2x+（1+cos2x）
=sin2x+cos2x+2
=
2sin(2x+π
4)+2，
The minimum positive period of the function T = 2 π
2=π．
(2) By 2K π - π
2≤2x+π
4≤2kπ+π
2, K π - 3 π
8≤x≤kπ+π
8（k∈Z），
The increasing interval of the function is [K π − 3 π
8，kπ+π
8]（k∈Z）．

If sinx-3cosx = 0, then 2sinxcosx / cos ^ 2x-sin ^ 2x= Just give the result

sinx=3cosx
tanx=3
Easy to know
2sinxcosx/(cos^2x-sin^2x)
=2tanx/(1-tan^2x)
=6/(1-9)=-6/8=-3/4

Is sin-x equal to - SiNx? Why The main reason is why

sin(-x)=-sinx
The values of angular functions with the same final edge are equal
The sign of sine function in 1234 quadrant is + + --

If SiNx = 2cosx, then sin2x + 1 = 2___ .

∵sinx=2cosx，∴tanx=2．
Then sin2x + 1 = sin2x
sin2x+cos2x+1=tan2x
tan2x+1+1=22
22+1+1=9
5．
So the answer is: 9
5．

SiNx = 2cosx find sin ^ 2 + 1

sinx=2cosx
tanx=2
sin^2x+1
=sin^2x/(sin^2x+cos^2x)+1
=tan^2x/(1+tan^2x)+1
=9/5

Given the vector M = (2cos? X, SiNx), n = (1,2cosx). 1, if M is perpendicular to N and 0 is less than x less than PI, find X.2, Let f (x) = m · n, find f (x) symmetric axis equation, symmetric center, monotone increasing interval

(1) If M is perpendicular to N, then m · n = 2 (cosx) ^ 2 * 1 + SiNx * 2cosx = 0
cosx*(cosx+sinx)=0
Cosx = 0 or cosx + SiNx = 0, that is TaNx = - 1
And 0 (2) f (x) = m · n = 2 (cosx) ^ 2 + 2sinxcosx = cos2x + 1 + sin2x = root 2Sin (2x + Pai / 4) + 1
Axis of symmetry: 2x + Pai / 4 = KPAI + Pai / 2
I.e. Pai + 2 / 8
Symmetry center: 2x + Pai / 4 = KPAI, i.e. x = KPAI / 2-pai / 8
The center of symmetry is (kPa I / 2-pai / 8,0)
Monotonic increasing range: 2kpai Pai / 2 < = 2x + Pai / 4 < = 2kpai + Pai / 2
That is to say, the increasing range is: [kpai-3pai / 8, KPAI + Pai / 8]

It is proved that [SiNx (1 + SiNx) + cosx (1 + cosx)] [SiNx (1-sinx) + cos (1-cosx)] = sin2x

[sinx(1+sinx)+cosx(1+cosx)][sinx(1-sinx)+cos(1-cosx)]=(sinx+sin²x+cosx+cos²x)(sinx-sin²x+cosx-cos²x)=(sinx+cosx+1)(sinx+cosx-1)=(sinx+cosx)²-1=sin²x+2sinxcosx+cos²x-...