(1) If SiNx = 2cosx, then sin? X = (2) if Tan α = cos α is known, then sin α =? Ask brothers to write the detailed process It's mainly about how to write the first question, detailed process steps

(1) If SiNx = 2cosx, then sin? X = (2) if Tan α = cos α is known, then sin α =? Ask brothers to write the detailed process It's mainly about how to write the first question, detailed process steps

(1)
sinx=2cosx
tanx=2
(sinx)^2=(sinx)^2/[(sinx)^2+(cosx)^2]=(tanx)^2/[1+(tanx)^2]
(sinx)^2=4/5
（2）
(sinx)^2+(cosx)^2=1
sinα/cosα=cosα
sinα=(cosα)^2
(sinα)^2+sinα=1
sinα=(√5 -1)/2

1. If Tan α = cos α, then sin α is equal to? 2. SiNx = 2cosx, then sin? 2x =? 2?

（1）
(sinx)^2+(cosx)^2=1
sinα/cosα=cosα
sinα=(cosα)^2
(sinα)^2+sinα=1
sinα=(√5 -1)/2
(2)
sinx=2cosx
tanx=2
(sinx)^2=(sinx)^2/[(sinx)^2+(cosx)^2]=(tanx)^2/[1+(tanx)^2]
(sinx)^2=4/5

SiNx = 2cosx, the square of SiNx + 1?

sinx=2cosx
Square on both sides
sin²=4cos²x
Because sin? X + cos? X = 1
So sin? X + (1 / 4) sin? X = 1
sin²x=4/5
So sin? X + 1 = 9 / 5

If sinx-2cosx = O, find sin squared X

sinx-2cosx=0.
sinx=2cosx.
sinx=4cosx=4（1-sinx）=4-4sinx.
5sinx=4.
sinx=4/5.

Simplify cosx ^ 4 + SiNx ^ 4 + SiNx ^ 2cosx ^ 2 / (SiNx ^ 6 + cosx ^ 6 + 2sinx ^ 2cosx ^ 2)

You can multiply the above formula by 1, that is (OSX ^ 4 + SiNx ^ 4 + SiNx ^ 2cosx ^ 2) * 1
It is a multiplication of ((cosx) ^ 2 + (SiNx) ^ 2), i.e. (OSX ^ 4 + SiNx ^ 4 + SiNx ^ 2cosx ^ 2) * (cosx) ^ 2 + (SiNx) ^ 2)
Then the simplification is equal to ((SiNx) ^ 6 + (cosx) ^ 6 + 2 (sinxcosx) ^ 2)
It's equal to the following formula
The result is equal to 1

Simplification (cosx ^ 4 + SiNx ^ 4 + SiNx ^ 2cosx ^ 2) / (SiNx ^ 6 + cosx ^ 6 + 2sinx ^ 2cosx ^ 2)

(cosx^4+sinx^4+sinx^2cosx^2)/(sinx^6+cosx^6+2sinx^2cosx^2)＝[﹙cos2x﹚2＋﹙sin2x﹚2＋sin2xcos2x]／[﹙sin2x﹚3＋﹙cos2x﹚3＋2sin2xcos2x]＝[﹙sin2x＋cos2x﹚2－sin2xcos2x]／﹛﹙sin2x＋cos2x﹚[﹙sin2x﹚2...

Let (2cosx SiNx) (SiNx + cosx + 3) = 0, then what is the value of 2cosx + SiNx / 2sinx + cosx

Analysis:
Because SiNx ≥ - 1, cosx ≥ - 1, so: SiNx + cosx + 3 > 0
In order for (2cosx SiNx) (SiNx + cosx + 3) = 0 to hold, it is necessary to make (2cosx SiNx) (SiNx + cosx + 3) = 0
2cosx SiNx = 0 means SiNx = 2cosx
So:
(2cosx+sinx)/(2sinx+cosx)
=(2cosx+2cosx)/(4cosx+cosx)
=4/5

I want to know how √ 2 (√ 2 / 2sinx - √ 2 / 2cosx) ≠ 0 and √ 2 (x - π / 4) ≠ 0 is what I want to know

sin(x-y)=sinxcosx-cosxsinx
sinx-cosx=√2sin(x-π/4)\x05
\x05\x05
\x05
\X05 analytical geometry test paper (a) of Jinan University in the first semester of the first academic year
\x05
Professional class 2006 class name student number
Sealing wire
\x05
Score / X05 reviewer
\x05
\(12 minutes) rotate the straight line around the axis, and find the
In this paper, we discuss what surfaces they represent in terms of the possible values of
If the point is any point of a straight line, then the equation of latitude circle passing through the point is
……………………………… . 4 points
And . 5 points
The equation of the surface of revolution is obtained by eliminating the parameters
………………………………………………… 8 points
\X05 discussion: at that time, it means axis;
\When X05, it means that the generatrix is parallel to the cylindrical surface of the axis;
At that time, it represents the conic surface whose vertex is at the origin;
At that time, it means hyperboloid of one sheet 12 points
Score / X05 reviewer
\x05
7、 (13 points) taking the main diameter of conic as the new coordinate axis, the following equation of conic is simplified
\x05
And make its figure
The matrix of conic is
The characteristic equation is
The characteristic root is 2 points
The non asymptotic principal direction corresponding to is
\The non asymptotic principal direction corresponding to X05 is 4 points
\The main diameter equation of X05 corresponding to the non asymptotic principal direction is
\X05 is
\The main diameter equation of X05 corresponding to the non asymptotic principal direction is
\X05 means 6 points
\X05 score
Score / X05 reviewer
\x05
5、 (10 points) solve the plane equation of two straight generatrix of points on hyperboloid of one sheet
\x05
Two families of straight generatrix equations of hyperboloid of one sheet
\X05 family: clan . 2 points
By substituting the point into the above two equations, we can get . 4 points
\Therefore, the equation of two straight generatrix passing through the point is
\X05 and
\The direction vectors of X05 are respectively 8 points
Therefore, the plane equation obtained is
I.e . 10 points
\x05
\X05 analytical geometry test paper (a) of Jinan University in the first semester of the first academic year
\x05
Professional class 2006 class name student number
Sealing line
\X05 takes the main diameter as the coordinate axis for coordinate transformation
………………………………………… . 8 points
The solution
By substituting it into the original equation, we can get the following result 10 points
The standard equation is
The figure is shown in the figure 13 points

Two senior one compulsory four simplification questions: Question 1 3 √ 15sinx + 3 √ 5cosx; (simplification) question 2 3 / 2cosx - √ 3 / 2sinx (simplification) There are steps to solve the problem

1.3√15sinx+3√5cosx=3√5（√3sinx+cosx)
=6√5(sinx√3/2+cosx1/2)
=6√5sin(x+π/6）
Two
3/2cosx-√3/2sinx=/3（/3/2cosx-1/2sinx)
=/3(sin60°cosx-cos60°sinx)
=/3sin(60°-x)

How to simplify 3sinx + cos (π / 3 + x) = 3sinx + 1 / 2cosx-v3 / 2sinx = (3-v3 / 2) SiNx + 1 / 2cosx

cos(a+b)=cosacosb-sinasinb
cos(a-b)=cosacosb+sinasinb
.
Sincerely hope to get your adoption!