Draw an isosceles triangle ABC, where AB = AC, ∠ a ≠ 90 ° and find a point P on the line where AC is located, so that PA = Pb

Draw an isosceles triangle ABC, where AB = AC, ∠ a ≠ 90 ° and find a point P on the line where AC is located, so that PA = Pb

As shown in the figure: point P is the desired point

In the isosceles triangle ABC, ab = AC = 6, P is a point on BC, and PA = 4, then what is Pb * PC equal to?

Let ad be perpendicular to BC and BC be perpendicular to D PA ^ 2 = ad ^ 2 + PD ^ 2 (Pythagorean theorem) BD = CD (three lines in one) Pb * PC = bd-pd (CD + PD) = (bd-pd) (BD + PD) = BD ^ 2-PD ^ 2 = AB ^ 2-ad ^ 2 - (AP ^ 2-ad ^ 2) = AB ^ 2-AP ^ 2 = 36-16 = 20

If P is a point on the bottom edge BC of isosceles triangle ABC, ab = AC = 5, find the square of PA = Pb times PC

Let ad ⊥ BC, then BD = CD (set to x), let p be on BD, PD = y, ad = Z
Then PA & # 178; + Pb · PC = y & # 178; + Z & # 178; + (X-Y) (x + y) = y & # 178; + Z & # 178; + X & # 178; - Y & # 178; = Z & # 178; + X & # 178; = AB & # 178; = 25

The moving point P starts from the vertex of the square ABCD with side length of 1, passes through B, C, D in turn, and then returns to A. let x denote the journey of point P, and Y denote the length of PA, and find the analytic function of Y with respect to X

When p is on AB, i.e. 0 ≤ x ≤ 1, y = PA = x; when p is on BC, i.e. 1 ≤ x ≤ 2, y = PA = 1 + (x − 1) 2; when p is on CD, i.e. 2 ≤ x ≤ 3, y = PA = 1 + (3 − x) 2; when p is on Da, i.e. 3 ≤ x ≤ 4, y = PA = 4-x

Given the function f (x) = asin (Wx + a) (W is greater than 0, the absolute value of a is less than π / 2), the image intersects the y-axis at the point (0,3 / 2), which is (x0,3), (XO + 2 π, - 3) with the first maximum and minimum point on the right side of the y-axis to find the analytic expression of F (x)

The first maximum point and the minimum point are (x0,3), (XO + 2 π, - 3) respectively, which can give π / ω = 2 π
So ω = 1 / 2
At the maximum, sin (Wx + a) = 1, so a = 3
Then (0,3 / 2) is brought in to get Sina = 1 / 2
a=π/6
f(x）=3sin（π/2+π/6)

Given the function f (x) = asin (π x + α) + bcos (π x + β), where a, B, α and β are all non-zero real numbers. If f (2005) = - 1, what is f (2006) equal to

f(x)=asin(πx+α)+bcos(πx+β)
f(2005)=asin(2005π+α)+bcos(2005π+β)=asin(π+α)+bcos(π+β)=-1
asin(π+α)+bcos(π+β)=-1
asin(α)-bcos(β)=-1
bcos(β)=asin(α)+1
f(2006)=asin(2006π+α)+bcos(2006π+β)=asin(α)+bcos(β)=asin(α)+asin(α)+1
=2asin(α)+1

∫(0,1) x·(1/e^x)·arctan(e^x) dx

The original function of arctan (e ^ x) is unknown

Who can make ∫ arctan x DX (upper and lower limits are 0 to 1)=

x*arctan x-(ln(1+x^2))/2+C

For definite integral, the upper limit of integral is 4, the lower limit of integral is 0, and the integral part is arctan (x / 4) DX,

The original formula = xarctan (x / 4) | (0 ~ 4) -∫ xdarctan (x / 4)
=π-∫x/[1+(x/4)^2]dx
=π-8∫dx^2/(16+x^2)
=π-8*ln|16+x^2||(0~4)
=π-8ln2

Arctan X / (1 + x ^ 2) DX the upper limit is positive infinity and the lower limit is 1

Original = ∫ arctanxdarctanx
=(arctanx)² (1,+∞)
=(π/2)²-(π/4)²
=3π²/16