Given TaNx = 2, find the value of (cosx + SiNx) / (cosx SiNx) 2) sinxcosx-1 3) 2 + sinxcosx cos? X

Given TaNx = 2, find the value of (cosx + SiNx) / (cosx SiNx) 2) sinxcosx-1 3) 2 + sinxcosx cos? X

(1)(cosx+sinx)/(cosx-sinx)=(1+tanx)/(1-tanx)=(1+2)/(1-2)=-3
(2)sinxcosx-1=sinxcosx/[(sinx)^2+(cosx)^2]-1=tanx/[1+(tanx)^2]-1=2/5-1=-3/5
(3)2+sinxcosx-(cosx)^2=2+[sinxcosx-(cosx)^2]/[(sinx)^2+(cosx)^2]
=2+[tanx-1]/[(tanx)^2+1]=2+(2-1)/(4+1)=2+1/5=11/5

It is known that SiNx + cosx = 1 / 5.0

sinx+cosx=1/5
The square is 1 + 2sinxcosx = 1 / 25
So sinxcosx = - 12 / 25 < 0
Therefore, cosx < 0, SiNx > 0, SiNx cosx > 0
(sinx-cosx)^2=(sinx+cosx)^2-4sinxcosx=49/25
sinx-cosx=7/5
(sinx)^2-(cosx)^2=(sinx+cosx)(sinx-cosx)=7/25

Given the vector a = (SiNx, 1), B = (cosx, negative half), find the monotone increasing interval of the function f (x) = A. (2b-a) + cos quadratic X seek

f(x)=2ab-a²+cos²x=2(sinxcosx-1/2)+(cos2x+1)/2=sin2x+cos2x-1/2
=2Sin (2x + π / 4) - 1 / 2 under radical
The monotone increasing interval is 2K π - π / 2 ≤ 2x + π / 4 ≤ 2K π + π / 2
kπ-3π/8≤x≤kπ+π/8
The increasing interval is [K π - 3 π / 8, K π + π / 8]

Let x ∈ (0, π), try to compare the size of COS (SiNx) and sin (cosx)

When x ∈ [0, π / 2] SiNx is an increasing function, cosx is a decreasing function, when x ∈ [π / 2, π] SiNx is a decreasing function and COS (SiNx) = sin (π / 2-sinx) x > π / 2, cosxsin (cosx) x0sinx (π / 2-sinx) > sin (cosx) is always cos (SiNx) > sin (cosx)

X is an acute angle, try to compare the size of cosx, sin (cosx), cos (SiNx)

x∈（0,π/2）
Comparing cosx with cos (SiNx), cosx decreases monotonically in the range of 0 to π / 2
Let f (x) = x-sinx, then f (x) is an odd function, f '(x) = 1-cos (x) > 0, f (x) increases monotonically
Because f (0) = 0, when x > 0, f (x) > 0, that is, x > SiNx
In the range of 0 to π / 2, cos (SiNx) > cosx
Compare cosx and sin (cosx)
cosx∈（0,1）
Let cosx = y, f (y) = y-siny, f '(y) = 1-cos (y) > 0, then cosx is in (0,1), cosx > sin (cosx)
The conclusion is cos (SiNx) > cosx > sin (cosx)

Let the plane vector a = (cosx, SiNx), B = (cosx + 2 root 3, SiNx), C = (Sina, COSA), X belongs to R, (1) if a is perpendicular to C, find C If a is perpendicular to C, find the value of COS (2x + 2a)

a=(cosx,sinx), b=(cosx+2√3,sinx), c=(sina,cosa)
if a⊥c
=> a.c =0
(cosx,sinx).(sina,cosa) =0
cosxsina+sinxcosa=0
sin(a+x) =0
a+x = 0
a = -x
c=(sina,cosa)=(sin(-x), cos(-x)) = (-sinx, cosx)
--------------
cos（2x＋2a）
=cos2(x+a)
= 1- 2[sin(x+a)]^2
= 1- 2(0) =1

Let a = (cosx, SiNx), B = (cosx + 2, radical 3, SiNx), C = (Sina, COSA), X ∈ R If a ⊥ B, find the value of COS (2x + 2a)

Whether a is perpendicular to C
If a is perpendicular to C, then a * C = cosxsina + sinxcosa = 0
That is, sin (a + x) = 0
cos(2x+2a)=1-2(sin(x+a))^2=1-0=1

If SiNx + Sina = root 2 / 2, find the value range of cosx + cosa

Let cosa + cosx = KCOs? A + cos? X + 2cosxcosk = k? 2, because sin? 2 + cos? 2 = 1, then 2 + 2 (cosacosx + sinasinx = k? 2 + 1 / 2cos (A-X) = (k? - 3 / 2) / 2 is - 1

F (x) = SiNx / 2 * cosx / 2 + cos ^ 2 (x / 2) - 2 find the maximum and minimum values of function f (x) on [π, 17 / 12 π]

f(x)=sinx/2*cosx/2+cos^2(x/2)-2
=1/2sinx+1/2(cosx+1)-2
=1/2(sinx+cosx)
=√2/2sin(x+π/4)
Y = SiNx is a decreasing function at (π, 3 π / 2)
PI

The range options of the function f (x) = 1 / 2 (SiNx + cosx) + 1 / 2 | SiNx cos | A. [- 1,1] B. [negative root two, 1] C. [- 1 / 2,1 / 2] D. [- 1, root two]

1) If SiNx > cosx, then SiNx > - √ 2 / 2, f (x) = SiNx, - 2 / 22) SiNx < = cosx, then cosx > - √ 2 / 2, f (x) = cosx, - 2 / 2 < = f (x) < = 1
Therefore, the value range of F (x) is [- √ 2 / 2,1], select B