In △ ABC, if C4-2 (A2 + B2) C2 + A4 + a2b2 + B4 = 0, then ∠ C is equal to () A. 90 ° B. 120 ° C. 60 ° D. 120 ° or 60 ° C

In △ ABC, if C4-2 (A2 + B2) C2 + A4 + a2b2 + B4 = 0, then ∠ C is equal to () A. 90 ° B. 120 ° C. 60 ° D. 120 ° or 60 ° C

Let the edges corresponding to the inner angles a, B and C of △ ABC be a, B and C respectively, cos (A-C) + CoSb = 32, B2 = AC, then=______ ．

∵ B = π - (a + C), ∵ the known equation is transformed into cos (A-C) - cos (a + C) = 32, that is, cosacosc + sinasinc cosacosc + sinasinc = 2sinasinc = 32, ∵ sinasinc = 34, and B2 = AC is reduced to sin2b = sinasinc = 34, ∵ SINB = 32 or SINB = - 32 by using the sine theorem

In the triangle ABC, LGA -- LGC = lgsinb = - LG change sign 2 and B is an acute angle, the shape of the triangle is judged

Right triangle
Lgsinb = - LG radical 2
SINB = radical 2 / 2
B=45°
LGA LGC = - LG radical 2
A / C = radical 2
From the sine theorem Sina / sinc = radical 2 / 2
Sin (a + C) = sinacosc + sinccosa = sin135 ° = radical 2 / 2
Let Sina = x, then sinc = root 2 * X
Cosa = root (1-x ^ 2), COSC = root (1-2 * x ^ 2)
If we take it to the above formula, we can find that x = root 2 / 2
That is, a = 45 degrees (a cannot be equal to 135 degrees)
So C = 90 degrees
Is a right triangle
I suddenly found that the above equation is very troublesome to solve. Let's provide a simpler method
Since a / C = radical 2 / 2, let a = 1 and C = radical 2
Since B = 45 degrees, we can use the cosine theorem to calculate the value of B
b^2=a^2+c^2-2*a*c*cosB
=1+2-2=1
b=1
Therefore, a ^ 2 + B ^ 2 = C ^ 2
It is a right triangle
Let's ignore the previous method
Note: the ^ sign indicates the power

Simplification of Sin & sup2; a * Tana + cos & sup2; acota + 2sinacosa

The original formula = Sin & sup3; a / cosa + cos & sup3; a / Sina + 2sinacosa = [(Sin & sup2; a) & sup2; + (COS & sup2; a) & sup2;] / sinacosa + 2sinacosa = [(Sin & sup2; a) & sup2; + (COS & sup2; a) & sup2; + 2Sin & sup2; ACOS & sup2; a] / sinacosa = (Sin & sup2; a + co