If | x | ≤ π 4, then the minimum value of the function f (x) = cos2x + SiNx is___ .

If | x | ≤ π 4, then the minimum value of the function f (x) = cos2x + SiNx is___ .

The function f (x) = cos2x + SiNx = 1-sin2x + SiNx = - (sinx-1)
2）2+5
4，
Because | x | π
So SiNx ∈[-
Two
2，
Two
2]，
When SiNx=-
Two
2, the function gets the minimum value: 1-
Two
2．
So the answer is: 1-
Two
2．

The known function f (x) = asinx * cosx - √ 3A (COS ^ 2) x + ((√ 3) / 2 * a) + B (1) Write the monotone decreasing interval of a function (2) Let x ∈ [0, ΠΠΠΠΠΠ], the minimum value of F (x) is - 2, and the maximum value is √ 3

The function f (x) = asinx · cosx - radical 3acos? X + (radical 3) / 2A + B (a > 0)
=a/2*sin2x-a*√3/2*cos2x+b
=asin(2x-∏/3)+b.
∏/2+2k∏≤2x-∏/3)≤2k∏+3∏/2,
k∏+5∏/12≤x≤k∏+11∏/12.
In other words, the monotone decreasing interval of the function is: {x|k Π + 5 Π / 12 ≤ x ≤ K Π + 11 Π / 12, K ∈ Z}
2) Let x ∈ [0, π / 2], then there is
-∏/3≤(2X-∏/3)≤2∏/3.
f(x)=asin(2x-∏/3)+b.
If the minimum value of F (x) is - 2 and the maximum value is root 3, then there is
-√3/2*a+b=-2,
a+b=√3,
A = 2, B = √ 3-2

The following relation is correct: a cos (π / 2-x) = cosx B Tan (2 π + x) = - TaNx C cos (π + x) = cosx D sin (x + 2 π) = SiNx

D. Because no matter which trigonometric function angle changes 360 degrees, the function value will not change. Secondly, if a B = 90 degrees, Sina = CoSb, Tana = cotb,

It is known that SiNx cosx = 1 / 5 and 0 It is suggested that SiNx cosx = 1 / 5 (sinX-cosX)^2=1/25 -2sinXcosX=1/25-1=-24/25 2sinXcosX=24/25 sinXcosX=12/25

One
sinX-cosX=1/5
1-2sinxcosx=1/25
sinxcosx=12/25
Two
(sinx+cosx)²=1+2sinxcosx=1+24/25=49/25
Because SiNx cosx = 1 / 5,0

Given that SiNx + cosx = 1 / 5, X belongs to (0, π), find the value of COS ^ 3 x-sin ^ 3 X

sinx+cosx=1/5
(sinx)^2+(cosx)^2=1
(sinx+cosx)^2
=(sinx)^2+(cosx)^2+2sinxcosx
=1/25
2sinxcosx=-24/25
sinxcosx=-12/25
simultaneous equation:
sinx+cosx=1/5
sinxcosx=-12/25
cos^3 X-sin^3 X
=(sinx-cosx)((sinx+cosx)^2-sinx+cosx)
=The root sign ((SiNx + cosx) ^ 2-4sinxcosx) * ((SiNx + cosx) ^ 2-sinx + cosx) can be carried in

Let a and B be the solutions of the equations cos (SiNx) = x, sin (cosx) = x on the interval (0, PI / 2), respectively Please be more specific

When cos (SiNx) = x, the deformation can be obtained as follows:
Sin (Π / 2-sinx) = x, now compare it with
Sin (cosx) = the size of X, i.e. need to compare
The size of Π / 2-sinx and cosx
The hypotheses are as follows:
∏/2-sinx- cosx〉0
We can get the following results:
Π / 2 〉 SiNx + cosx > = (2 radix), both sides are multiplied by 2,
Π〉 3 > (8 radix)
So the hypothesis holds,
So:
A>B

When the range of X is (0, π / 2), let a = sin (cosx), B = cos (SiNx), and find the size relationship between a and B

X ∈ ((0, π / 2), then SiNx ∈ (0,1), cosx ∈ (0,1)
One radian is about 50 degrees, in the first quadrant,
In (0, π / 4] sinxb

The acute angle x =? Satisfying sin (x + SiNx) = cos (x-cosx)?

∵0

SiNx + sin γ = sin β cos β + cos γ = cosx, the three of them are acute angles to find β - X=

sinx+sinγ=sinβ cosβ+cosγ=cosx
sinβ-sinx=siny cosβ-cosx=cosy
After the two formulas are squared, add up
2-2 cos (β - x) = 1
cos(β-x)=1/2
β - x = 2K π ± π / 6 (k is natural number)

2011 supplementary question of the fifteenth question of the second mock exam in Nantong, Jiangsu: set up the plane vector a= (cosx, SiNx), b= (cosx+2 root 3, SiNx), c= (siny, cosy), X belongs to R. Let a = (cosx, SiNx), B = (cosx + 2 root 3, SiNx), C = (siny, cosy), X belongs to R 1. If a is perpendicular to C, the value of COS (2x + 2Y) 2. If x belongs to (0,90 °), it is proved that a and B cannot be parallel 3. If y = 0, find the maximum value of the function f (x) = a times (b-2c), and find the corresponding x value

(1)
A vertical C
=> a.c =0
(cosx,sinx).(siny,cosy)=0
cosxsiny+ sinxcosy =0
sin(x+y) =0
x+y = k(180°) k =0,1,2,..
2(x+y) = k(360°)
cos(2x+2y) = cosk(360°) = 1
(2)
if a // b
=>cosx/sinx=(cosx+2√3)/sinx
cosx =cosx+2√3
0 = 2√3 ( contradiction)
=>A and B can't be parallel
(3)
F(x) = a.(b-2c)
= ( cosx.sinx ).(cosx+2√3 ,sinx-2)
= (cosx)^2+2√3 cosx + (sinx)^2-2sinx
= 4(√(3/2)cosx- (1/2)sinx) +1
= 4sin(60°-x) +1
max F(x) at x = -30°
max F(x) = 5