The minimum value of the function y = 4 / cos ^ 2 + 9 / SiNx ^ 2 y=4/cosx^2+9/sinx^2

y=4/cos^2+9/sinx^2
=4(cos^2x+sin^2x)/cos^2x+9(cos^2x+sin^2x)/sinx^2
=4+4tan^2x+9+9/tan^2x
≥13+2√36
=25
If and only if 4tan ^ 2x = 9 / Tan ^ 2x, that is, Tan ^ 2x = 3 / 2

The function f (x) = cos2x + SiNx in the interval [- π 4，π 4] The minimum value on is () A. 2−1 Two B. -1+ Two Two C. -1 D. 1− Two Two

f（x）=cos2x+sinx=1-sin2x+sinx=-（sinx-1
2）2+5
4．
∵x∈[-π
4，π
4] So SiNx ∈ [−
Two
2，
Two
2]
Therefore, when SiNx = -
Two
When 2, the minimum value of the function is Ymin = 1 −
Two
2．
That is, if x = - π
At 4, Ymin = 1 −
Two
2．
Therefore, D

The value range of the function y = ︱ SiNx CoS / ︱ cosx is Detailed process

Function y = 2A cos square X-2 root 3A SiNx cosx + A + B (a

y=a[(1+cos2x)-(√3)sin2x]+a+b=a[cos2x-(√3)sin2x]+2a+b
=2a[(1/2)cos2x-(√3/2)sin2x]+2a+b=2a[cos2xcos(π/3)-sin2xsin(π/3)]+2a+b
=2acos(2x+π/3)+2a+b (a

The known function f (x) = cos quartic x-2sinxosx sin quartic X (1) Find the minimum positive period of F (x) (2) When x ∈ [0, π / 2], find the minimum value of F (x) and the set when the minimum value is obtained

F (x) = cos quartic x-2sinxosx sin quartic x
=(cos^2x+sin^2x)(cos^2x-sin^2x)-sin2x
=1*cos2x-sin2x
=Root 2 cos (2x + Pai / 4)
The minimum positive period T = 2pai / 2 = Pai
When x ∈ [0, π / 2], 2x + Pai / 4 belongs to [Pai / 4,5pai / 4]
So - radical 2 / 2=

The known function f (x) = cos quartic x-2sinxcosx sin quartic x (2) When x ∈ [0, tt / 2], find the minimum value of F (x) and the corresponding set of X when the minimum value is obtained Why is the monotone decrease of = - √ 2Sin (2x - π / 4) to be [- π / 8,3 π / 8] Instead of [π / 8,3 π / 8]

f(x)=(cosx)^4-2sinxcosx-(sinx)^4
=[(cosx)^2+(sinx)^2][(cosx)^2-(sinx)^2]-2sinxcosx
=(cosx)^2-(sinx)^2-2sinxcosx
=cos(2x)-sin(2x)
= √2*[cos(2x)*√2/2-sin(2x)*√2/2]
= √2*[cos(2x)cos(π/4)-sin(2x)sin(π/4)]
= √2cos(2x+π/4) .
(2) Because 0

It is known that the function f (x) = the fourth power of COS x + 2sinxcosx -- the fourth power of SiN x (1). Find the minimum positive period of F (x): (2). Question: the function f (x) is a function of How is the function f (x) obtained from the function y = SiNx

(cosx) ^ 4 + 2sinxcosx - (SiNx) ^ 4 = (COS? X + sin? X) (COS? - sin? 2x) + sin2x = cos2x + sin2x = root 2Sin (2x + π / 4)
(1) Minimum positive period: T = 2 π / 2 = π
(2) I don't know what to ask

Given the function f (x) = the fourth power of SiNx + the second power of cosx, if FX = a has a solution, find the range of real number a

First of all, we simplify f (x) = the fourth power of SiNx + the second power of cosx, f (x) = (SiNx) ^ 4 + (cosx) ^ 2 = [(SiNx) ^ 2] ^ 2 + 1 - (SiNx) ^ 2 = [(SiNx) ^ 2-1 / 2] ^ 2 + 3 / 4 = 1 / 4 [2 * (SiNx) ^ 2-1-1] ^ 2 + 3 / 4 = 1 / 4 (cos2x) ^ 2 + 3 / 4 = 1 / 4 (cos2x) ^ 2 + 3 / 4 = 1 / 1 / 8 + 1 / 8 + 3 / 4 = 1 / 8cos4x + 8 + 3 / 4 = 1 / 8cos4x + 8 + 3 / 4 = 1 / 8cos4x + 8 + 3 / 4 = 1 / 8cos4x + 8 + 3 / 4 = the interval of 7 / 8 cos4x, [- 1

Let f (x) = the fourth power of SiNx - the fourth power of cosx + sin2x-3

What do you want to do, the problem is not complete! However, I hope my simplification can help you! F (x) = (SiNx) ^ 4 - (cosx) ^ 4 + sin2x-3f (x) = ((SiNx) ^ 2 + (cosx) ^ 2) ((SiNx) ^ 2 - (cosx) ^ 2) + sin2x-3f (x) = (SiNx) ^ 2 - (cosx) ^ 2 + sin2x-3 and (SiNx) ^ 2 + (cosx) ^ 2 = 1

How to expand cos (SiNx), cos (cosx), sin (cosx), sin (SiNx) to x ^ 3 with Taylor expansion? And the above four forms are rewritten as Taylor type with Lagrange remainder. Thank you for your thanks This is the problem of mathematical analysis in mathematics department!

Original Taylor formula:
SiNx = x minus one sixth of X to the third power
Cosx = one minus one half x square
Replace x with what you need
Lagrangian remainder sin; R2N (x)
cos;Rn(x)
Will you

The minimum value of the function y = 4 / cos ^ 2 + 9 / SiNx ^ 2 y=4/cosx^2+9/sinx^2