The three interior angles of △ ABC satisfy the equation (SINB Sina) x2 + (Sina sinc) x + (sinc SINB) = 0, and have two equal real roots, then: 1. Prove: ∠ B ≤ π / 3 2. If ∠ B takes the maximum value, try to judge the shape of △ ABC

The three interior angles of △ ABC satisfy the equation (SINB Sina) x2 + (Sina sinc) x + (sinc SINB) = 0, and have two equal real roots, then: 1. Prove: ∠ B ≤ π / 3 2. If ∠ B takes the maximum value, try to judge the shape of △ ABC

(1) The original formula = (sina-sinc) 2-4 (sinb-sina) (sinc-sinb) 2-4 (sinb-sinc) 2-4 (sinb-sinc) 2-4 SINB (sinb-sinc) 2-4 SINB (sina-sinc) 2-4 SINB (Sina + sinc) 2 (sinb-sinc) 2-4 (sinb-sinb) (sinb-sinb) (sinc-sinc-sinb) = sin2a-2-2-2 sin-2-2 sin-2 sin-2 sin-2 sin as a-2-2-2-2-4 (sinb-sinb-sinc) 2-4-4-4-4-4-4-4-4-4-4-4 sinc) 2-4-4-4-4-4-4 (sinc) 2-4-4 (sinb-sinb-sinc-sinc) 2-sinb (SINB (sinb-sinb-sinb-sinb 90o 〈 0 ＜ cos [(A-C) / 2] ≤ 1 〉 0 ＜ SINB / 2 ≤ 1 / 2, The shape of Δ ABC is equilateral triangle

In △ ABC, acab = cosbcosc. (I) prove that B = C: (II) if cosa = - 13, find the value of sin (4B + π 3)

In the triangle ABC, if the triangle area = the square of C - (a-b), a + B = 1, find COSC

S = C ^ 2 - (a-b) ^ 2 = C ^ 2 - (a ^ 2 + B ^ 2) + 2Ab (all items are disassembled) - S = a ^ 2 + B ^ 2-C ^ 2-2ab, s = 1 / 2absinc (1) cosine theorem: COSC = [a ^ 2 + B ^ 2-C ^ 2] / 2Ab (2) (1) substituting (2), AB reducing, 4cosc = sinc-4, (3) sinc ^ 2 + COSC ^ 2 = 1 (4) (3) (4) simultaneous, find COSC = - 1 (

（1）sin^4α-cos^4α=2sin^2α-1 （2）tan^2α-sin^2α=tan^2αsin^2α （3）cosα/1-sinα=1+sinα/cos
（1）sin^4α-cos^4α=2sin^2α-1
（2）tan^2α-sin^2α=tan^2αsin^2α
（3）cosα/1-sinα=1+sinα/cosα
Verification

(1) Let x = (sin α) ^ 2, then (COS α) ^ 2 = 1-x. so (sin α) ^ 4 - (COS α) ^ 4 = x ^ 2 - (1-x) ^ 2 = 2x - 1 = 2 (sin α) ^ 2 - 1