In the triangle ABC, BAC = 90 °, P is outside the triangle, PA = Pb = PC, and the relationship between face PBC and face ABC is?

In the triangle ABC, BAC = 90 °, P is outside the triangle, PA = Pb = PC, and the relationship between face PBC and face ABC is?

If angle a = 90 degrees, then on the circle where points a, B and C are located, BC is the diameter, and the distance from P to points a, B and C is equal, then point P is the center of the circle. Face PBC is actually the diameter BC, a line segment and the chord of face ABC
I hope I can give you some help

The maximum value of function y = 2Sin ^ 2x-2cosx-3

y=2sin^2x-2cosx-3
=2-2cos^2x-2cosx-3
=-2(cosx+1/2)^2-1/2
When cosx = - 1 / 2, the maximum value is - 1 / 2

The minimum value of function y = 2Sin ^ 2x + 2cosx-3, X ∈ [- π / 3,2 π / 3]

y=2sin^2x+2cosx-3
=2(1-cos²x)+2cosx-3
=-2cos²x+2cosx-1
=-2(cosx-1/2)²-1/2
x∈[-π/3,2π/3]
∴cosx∈[-1/2,1]
When cosx = - 1 / 2
y=-2*（-1）²-1/2
=-2-1/2
=-5/2
Is the minimum value
If you still have doubts, please ask. I wish you progress in your study!

Find the range of function y = (2Sin α - 1) / (1 + sin α)

Original formula = 2-3 / (1 + sin α)
The range of 1 + sin α is [0,2]
So the range of - 3 / (1 + sin α) is [- OO, - 3 / 2]
The range of the original formula is [- OO, 1 / 2]

Given the function f (x) = sin ^ 2 (x - π / 6) + sin ^ 2 (x + π / 6) (1) find the minimum positive period of function f (x) (2) x ∈ [- π / 3, π / 6], find the function range

f(x)=(1/2)[1-cos(2x-π/3)]+(1/2)[1-cos(2x+π/3)]
=1-[cos(2x-π/3)+cos(2x+π/3)]
=1-2cos2xcos(π/3)
=1-cos2x
(1) The period is t = 2 π / 2 = π;
(2) F (x) is a decreasing function on [- π / 3,0] and an increasing function on [0, π / 6],
The maximum value is f (- π / 3) = 3 / 2, and the minimum value is f (0) = 0

Given function y = 2Sin (2x + π / 3)
Find: (1) its amplitude, period, initial phase angle (2) its monotone increasing interval symmetric axis equation

The amplitude is 2, the period is π, and the initial phase is π / 3
Single increasing interval: K π - 5 π / 12 ≤ x ≤ K π + π / 12
Axis of symmetry: x = (1 / 2) k π + (1 / 12) π

Given the vector OP = (COS θ, sin θ) and the vector OQ = (1 + sin θ, 1 + cos θ) (θ∈ [0, π]), then the value range of PQ is____ .
(1 + sin θ - cos θ) ^ 2 + (1-sin θ + cos θ) ^ 2

(1+sinθ-cosθ)^2+(1-sinθ+cosθ)^2 T=sinθ-cosθ
=(1+T)^2+(1-T)^2
=2+2T^2
=2+2(sinθ-cosθ)^2
=2+4sin（θ-45°）^2
The maximum is six and the minimum is two
The value range of PQ is 2 ^ (1 / 2) to 6 ^ (1 / 2)

Let a = (COS (θ - π / 6), sin (θ - π / 6)), B = (2cos (θ + π / 6), 2Sin (θ + π / 6))
When t changes in the interval (0.1), we find the minimum value of the module of vector 2T, vector B + m / T, vector a (M is a constant, M > 0)

Module length is root sign (16t ^ 2 + m ^ 2 / T ^ 2 + 8mcos 2 θ)
When m > = 4, the minimum module length is the root sign (16 + m ^ 2 + 8mcos 2 θ)
When 0

Simplifying sin (a + b) / (Sina + SINB)
Can we use the sum difference product formula to make this formula look the best
What about half angle?
Go back to the second floor. I'm the same as you. Can I simplify it?

The original formula = 2Sin [(a + b) / 2] cos [(a + b) / 2] / {2Sin [(a + b) / 2] cos [(a-b) / 2]}
=cos[(a+b)/2]/cos[(a-b)/2]}

How can (Sina + SINB) (Sina SINB) = 1 / 2 sinc ^ 2 be reduced to sin (a + b) sin (a-b) = 1 / 2 sin (a + b) ^ 2

(sinA+sinB)(sinA-sinB)=(sinA)^2-(sinB)^2=(sinA)^2-(sinAsinB)^2+(sinAsinB)^2-(sinB)^2=(sinA)^2*[1-(sinB)^2]-(sinB)^2*[1-(sinA)^2]=(sinAcosB)^2-(sinBcosA)^2=(sinAcosB+cosAsinB)*(sinAcosB-cosAsinB)=sin(A+B)sin(A-B)=1/2(sinC)^2=1/2[sin(A+B)]^2
Sinc ^ 2 and sin (a + b) ^ 2 in your topic should mean (sinc) ^ 2 and [sin (a + b)] ^ 2
Sorry, it's a bit cumbersome, but it's relatively simple and easy to understand,