Given the ellipse X / 4 + y = 1, a straight line at point a (- 3,0 under the root sign) crosses the ellipse at the origin and BC, the maximum area of triangle ABC is obtained

Given the ellipse X / 4 + y = 1, a straight line at point a (- 3,0 under the root sign) crosses the ellipse at the origin and BC, the maximum area of triangle ABC is obtained

SABC = Sabo + SACO | Ao | = 3 under the root sign, triangle ABO and triangle ACO have the same bottom and the same height. When the area of triangle ABO is the largest, the area of ABC is the largest. The bottom edge of triangle ABO | Ao | = 3 under the root sign, and the height is the absolute value of the ordinate of point B. when the absolute value of the ordinate of point B is the largest, the surface product of triangle ABO is the largest
Root 3, what's the process for you? Leave a Q or something?
Who can give the specific process of using calculator to calculate trigonometric function and inverse trigonometric function
The best answer format is x + X + x = answer
Take sin (30) as an example
Step 1: Press（
Step 2: press sin to build
Part 3: press 3
Step 4: press 0
Step 5: Press)
Press last=
If it's an inverse triangle
Step 1, then press shift
The rest is the same
All calculators are different. If you look at the manual of all calculators, maybe it will help you.
It is known that the sequence (an) satisfies the first n terms of A1 = 1 an + 1 = 3an sequence (BN) and Sn = n * n + 2n + 1
Finding the general term formula of sequence (an) (BN)
Let CN = anbn to find the first n terms and TN of (CN)
Is there anyone who would,
(1) An is an equal ratio sequence
an=3^(n-1)
Sn=n*n+2n+1
When n = 1
b1=4
n> 1:00
bn= Sn-S（n-1）=2n+1
(2) When n = 1
Tn=4
n> 1:00
tn=4+3^2*5+3^3*7+…… +3^(n-1)*(2n+1)
TN * 3-tn = the number of words is not enough, the difference can be subtracted
tn=3^n+4
The inverse function of F (x) = 2 ^ X / 1 + 2 ^ x,
y=(2^x)/(1＋2^x)
y＋y(2^x)=2^x
y=(1－y)2^x
2^x=y/(1－y)
x=log(2)[y/(1－y)]
Then the inverse function is y = log (2) [x / (1-x)] (0
The eccentricity e = (radical 3) / 2 of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is known, and the line L: y = 2x-3 intersects ellipse C and points a and B
If a circle with diameter AB passes through the origin, the equation of ellipse C is obtained
E = (radical 3) / 2,
∴c^2/a^2=3/4,
∴b^2=a^2-c^2=a^2/4,
∴C:x^2+4y^2=4b^2,
Substituting y = 2x-3 into the above formula,
x^2+4(4x^2-12x+9)=4b^2,
17x^2-48x+36-4b^2=0,
△=48^2-68（36-4b^2)=272b^2-144,
x1+x2=48/17,
The midpoint of AB is m (24 / 17, - 3 / 17),
OM^2=585/289=(AB/2)^2=[(1/2)√（5△)/17]^2=5(68b^2-36)/289,
117=68b^2-36,
b^2=9/4,a^2=9,
The equation of ellipse C is x ^ 2 / 9 + 4Y ^ 2 / 9 = 1
1. Find the equation of ellipse
2. Find the equation of line L
3. Find the area of triangle PAB
(1) C / a = √ 6 / 3, C = 2 √ 2, a = 2 √ 3, a & # 178; = 12, B & # 178; = 4
So the elliptic equation is X & # 178 / 12 + Y & # 178 / 4 = 1
(2) Let the midpoint of AB be d (x0, Y0), and the linear AB equation be y = x + B
The simultaneous equations are 4x & # 178; + 6bx + 3B & # 178; - 12 = 0
X... Unfold
1. Find the equation of ellipse
2. Find the equation of line L
3. Find the area of triangle PAB
(1) C / a = √ 6 / 3, C = 2 √ 2, a = 2 √ 3, a & # 178; = 12, B & # 178; = 4
So the elliptic equation is X & # 178 / 12 + Y & # 178 / 4 = 1
(2) Let the midpoint of AB be d (x0, Y0), and the linear AB equation be y = x + B
The simultaneous equations are 4x & # 178; + 6bx + 3B & # 178; - 12 = 0
x1+x2=-3b/2，y1+y2=x1+x2+2b=b/2
So x0 = - 3B / 4, Y0 = B / 4
PD is perpendicular to AB, so the slope of PD is - 1
[(1 / 4) B-2] / [(- 3 / 4) B + 3] = - 1, B = 2
So the linear AB equation is y = x + 2
(3) From (2), we can see that | ab | = (√ 2) √ [(x1 + x2) & # 178; - 4x1x2]
If x1x2 = (3b & # 178; - 12) / 4 = 0, X1 + x2 = - 3B / 4 = - 3 / 2, then | ab | = (3 / 2) √ 2
|PD|=|-3-2+2|/√2=（3/2）√2
S △ PAB = (1 / 2) | PD | * | ab | = 9 / 4
Trigonometric value help press calculator
73 degrees 69. Thank you
It's torture. Tan 73 ° 69 '= Tan 74 ° 09' equals Tan 75 ° approximately
Do you know how to use trigonometric Tan internal angle and formula?
Calculate Tan [45 + 30] = [4 + 2 * radical 3] / 2
It is known that in the sequence an, A1 = 1 / 2, an + 1 = 3an / an + 3, the sum of the first n terms of BN is SN
And for any positive integer n, BN · n (3-4an) / an = 1 holds, 1 / 2 ≤ Sn < 1
A (n + 1) = 3A (n) / [a (n) + 3], if a (n + 1) = 0, then a (n) = 0,..., a (1) = 0, is contradictory to a (1) = 1 / 2. Therefore, a (n) is not 0.1/a (n + 1) = (1 / 3) [a (n) + 3] / a (n) = 1 / a (n) + 1 / 3, {1 / a (n)} is an arithmetic sequence with the first term of 1 / a (1) = 2 and the tolerance of 1 / 3
F (x) = (2 ^ x-1) / 2 ^ + 1) inverse function process
y=(2^x-1)/(2^x+1)=(2^x+1-2)/(2^x+1)=1-2/(2^x+1)
1-y=2/(2^x+1) 2^x+1=2/(1-y)
2^x=2/(1-y) -1=(1+y)/(1-y)
X = log (lower 2) (1 + y) / (1-y)
y=(2^x -1)/(2^x +1)
(2^x +1)y=2^x -1
(2^x)y+y -(2^x)=-1
(2^x)(y-1）=-1-y
2^x=-1-y/y-1
x=log2 (1+y)/(1-y)
As shown in the figure, F1 and F2 are the left and right focal points of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, and P is on the ellipse, △ pof2 is an equilateral triangle with an area of root 3, and B ^ 2 is?
B ^ 2 = 4 because △ pof2 is a triangle. According to s = 1 / 2Ab * sinc, we can calculate the side length as 2. A = 2. Because P is on the vertical line of of2, the abscissa of P is 1. According to the area, the absolute value of the ordinate of point P is root sign 3. So x = 1, y = root sign 3. Bring this coordinate into the elliptic equation
Where is the picture? ~ ~ ask: according to the known drawing, I focus on the x-axis and P in the first quadrant
Use a calculator to find the values of the following formulas
(1)cos76°39′+sin17°52′
（2）sin57°18′-tan22°30′
（3）tan83°6′-cos4°59′
（4）tan12°30′-sin15°
I can help you. You just need the answer, right?
What's the exact number of decimals
Forget it, I'll take two decimal places if you don't say it
zero point five four
zero point four three
seven point two seven
-0.04