![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
F (x) = the maximum of SiNx cosx
(friendly note: auxiliary angle formula: asinx + bcosx = (√ (a ^ 2 + B ^ 2)) sin (x + α), where Tan α = B / a)
f(x)=(√2)sin(x-π/4)
Therefore, when X - π / 4 = π / 2 + 2K π, that is, x = 3 π / 4 + 2K π, the function f (x) has the maximum value √ 2
How to calculate the maximum value of F (x) = SiNx cosx?
How to transform? How to extract the common cause? How to extract
In the form of F (x) = asinx + (or -) bcosx, we can extract a square + b square under the root sign
This topic a, B are equal to 1, so we can extract the root sign 2, the original formula is equivalent to
If f (x) = 2 times of the root (2 / 2 root 2sinx-2 / 2 root 2cosx), then we can use the sine formula of the sum of two angles as the inverse
F (x) = the - sin (x - π / 4) of radical 2
The maximum value is the root sign 2, where x is 7 π / 4
The maximum of F (x) = 1 + cosx + SiNx
y=cosx+sinx +1
=√2[(√2/2)cosx+(√2/2)sinx] +1
=√2[sin(π/4)cosx+cos(π/4)sinx] +1
=√2sin(π/4＋x)+1
Maximum = 1 + √ 2
Minimum = 1 - √ 2
The maximum value of F (x) = SiNx + cosx is
A. 1 B. radical 2 C. radical 3 d. 2
f(x)=sinx+cosx
=√2sin(x+π/4)
Sine function is bounded and belongs to [- 1,1]
So the maximum value of F (x) is √ 2
Choose B
Given SiNx + cosx = 1 / 5 (0 is less than or equal to X is less than or equal to pie), find TG =?
If you know how to do it, please tell me quickly! Thank you
(sinx+cosx)^2=1/25
1+2sinxcosx=1/25
2sinxcosx=-24/25
Then (SiNx cosx) ^ 2 = 49 / 25
sinx-cosx=±7/5
∵0=
It is known that sinxcosx = 60 / 169 0
The result is wrong
sinx=5/13
cosx=12/13
In fact, the result of this problem can be guessed
169=13*13
13*13=12*12+5*5
5*12=60
Given that SiNx + cosx = 60 / 169 and tt / 4 "x 〈 tt / 2, find the value of SiNx and cosx
The original title should be: SiNx &; cosx = 60 / 169,
∵π/40.
So (SiNx + cosx) & sup2; = 1 + 2sinxcosx = 289 / 169,
sinx+cosx=17/13.…… ①
（sinx-cosx）²=1-2sinxcosx=49/169,
sinx-cosx=7/13.…… ②
① + 2: SiNx = 12 / 13
① (2) COSIX = 5 / 13
Given sinxcosx = 1 / 2, X ∈ [0, π / 2], it is necessary to find the value of 1 at 1 + SiNx + 1 at 1 + cosx
∵sinxcosx=1/2,∴(sinx+cosx)²=sin²x+cos²x+2sinxcosx=1+2×1/2=2∵x∈[0,π/2],∴sinx+cosx>0∴sinx+cosx=√2∴1/(1+sinx)+1/(1+cosx)=[(1+cosx)+(1+sinx)]/[(1+sinx)(1+cosx)]=(2+sinx+cosx)/[1+(...
If 0 ≤ x ≤ π 2, sinxcosx = 12, then 11 + SiNx + 11 + cosx=______ ．
∵ 0 ≤ x ≤ π 2, sinxcosx = 12, ∵ (SiNx + cosx) 2 = sin2x + cos2x + 2sinxcosx = 1 + 2sinxcosx = 2, ∵ SiNx + cosx = 2, then the original formula = 1 + cosx + 1 + sinx1 + cosx + SiNx + sinxcosx = 2 + 232 + 2 = 4-22
When x=______ SiNx + cosxsinx − cosx has no significance (0 ° < x < 90 ° & nbsp;)
∵ SiNx + cosxsinx − cosx is meaningless ∵ SiNx cosx = 0, ∵ SiNx = cosx, ∵ 0 ° < x < 90 °, ∵ x = 45 °. So the answer is: 45 °
RELATED INFORMATIONS
- 1. Given that - 90 degree is less than X and less than 0 degree, SiNx + cosx = 0.2, find SiNx cosx =?
- 2. It is known that the focus of the ellipse x ^ 2 / 4 + y ^ 2 = 1 is F1, F2, and the intersection of the parabola y ^ 2 = PX (P > 0) and the ellipse in the first quadrant is Q, if ∠ f1qf2 = 60 °, (1) Calculate the area of △ f1qf2; (2) Find the equation of this parabola
- 3. Point P is a point on the hyperbola x2 / 25-y2 / 9 with F1 and F2 as the focus, and the absolute value of Pf1 = 12, then the absolute value of PF2=
- 4. The left and right focus of the hyperbola x2a2 − y2b2 = 1 is F1, F2, P is a point on the hyperbola, satisfying | PF2 | = | F1F2 |, and the straight line Pf1 is tangent to the circle x2 + y2 = A2, then the eccentricity e of the hyperbola is () A. 3B. 233C. 53D. 54
- 5. How to solve the equation (LNX) ^ 3-1 + x = 0?
- 6. The left and right fixed points of ellipse C1: x ^ 2 / A ^ 2 + y ^ 2 / 4 = 1 (a is greater than 2) are a and B respectively, and point P is the image of hyperbola C2: x ^ 2 / A ^ 2-y ^ 2 / 4 = 1 in the first quadrant If the area of triangle ACD and triangle PCD is equal, (1) calculate the coordinates of point p; (2) whether the line CD can pass the right focus of ellipse C1, if it can, calculate the value of A. if not, please explain the reason
- 7. If there are two different intersections between the line y = KX + 2 and the circle (X-2) + (Y-3) = 1, find the value range of K
- 8. If the line y = KX (k > 0) and hyperbola y = 4 / X intersect at two points a (x1, Y1) and B (X2, Y2), then the value of 2x1y2-7x2y2
- 9. Prove: no matter what the value of K is, the straight line L: kx-y-4k + 3 = 0 and the curve C: x ^ 2 + y ^ 2-6x-8y + 21 = 0 always have several intersections 1. To make the line intersect the circle, the distance from the center of the circle to the line is | 3k-4-4k + 3 | / (radical K ^ 2 + 1) Radical (1 + 2K / (1 + K ^ 2))
- 10. The position relationship between the square of X + the square of Y - 6x = 0 and the square of circle x + the square of Y - 8y = 0. If they intersect, the common chord length can be obtained,
- 11. Are there overlaps between infinite discontinuities and oscillatory discontinuities, such as piecewise function x < 0, f (x) = 1 / x, x > 0, f (x) = sin Is there any overlap between infinite discontinuity and oscillation discontinuity For example, when the piecewise function x < 0, f (x) = 1 / x, x > 0, f (x) = sin (1 / x), what is the breakpoint of x = 0?
- 12. F (x) = sin (ω x + φ) (ω is greater than 0, φ is greater than or equal to 0 and less than or equal to π) is even function, and the distance between two adjacent highest points on the image is 2 π It is known that the function f (x) = sin (ω x + φ) (ω greater than 0,0 ≤ φ ≤ π) is even function, and the distance between two adjacent highest points on the image is 2 π 1. Find the analytic expression of F (x) 2. If a belongs to (- π / 3, π / 2), f (a + (π / 3)) = 1 / 3, find the value of sin (2a + (5 π / 3))
- 13. If the sum of the first n terms of the equal ratio sequence {an} is 2N-1, then the sum of the first n terms of the sequence {an ^ 2} is?
- 14. Let the sum of the first n terms of the equal ratio sequence {an} be Sn, S4 = 1, S8 = 17, and find the general term formula an
- 15. Let the sum of the first n terms of the sequence an be Sn, and Sn = n ^ 2-4n + 4, BN = an / 2 ^ n. can we use dislocation subtraction to find the sum of the first n terms of BN and TN?
- 16. What is dislocation subtraction
- 17. There is an arithmetic sequence with tolerance of 4 and item number of 7. The total number is 147
- 18. If the arithmetic sequence {an} satisfies A1 = 1, and A1, A2 and A4 are equal proportion sequence, find an
- 19. Observe the equation 1x2 = 1 / 2 minus 1 / 2, 2x3 = 1 / 2 minus 1 / 3, 3x4 = 1 / 3 minus 1 / 4 (1) Fill in the blanks according to the above rules (2) According to the above rules, the calculation is: 1x2 / 1 + 2x3 / 1 + 3x4 / 1 +... + 2012x2013 / 1 (3) If the value of a is the smallest positive integer and B is 3, find one part of AB + (a + 2) (B + 2) + (a + 4) (B + 4) +... + (a + 200) (B + 200)
- 20. Please write down the process and way of thinking Given an = 1 / N √ (n + 1) + (n + 1)) √ n, find the sum of the first n terms of the sequence If the deformation method of an can not be written in detail, the method can be written directly, but the important steps should be given It's an = (1 / N √ (n + 1)) + (n + 1) √ n. part of the plus sign in front of it is a fraction, but not after it~