(sinπ/6)cosθ+sinθ(cosπ/6)=? Well, I worked out the answer of passers-by- Can you keep driving?

(sinπ/6)cosθ+sinθ(cosπ/6)=? Well, I worked out the answer of passers-by- Can you keep driving?

(sinπ/6)cosθ+sinθ(cosπ/6)=sin[(π/6)+θ]

cos(6+a)cos(a-54)+sin(6+a)sin(a-54)=

The original formula = 1 / 2 {cos (2a-45) + cos60} - 1 / 2 {cos (2a-45) - cos60} = cos60 = 0.5

∫（sin∧3θ／cos∧6θ）dθ

∫ (sin θ) ^ 3 / (COS θ) ^ 6 d θ = - ∫ (sin θ) ^ 2 / (COS θ) ^ 6 DCOS θ = ∫ [(COS θ) ^ 2 - 1] / (COS θ) ^ 6 DCOS θ = ∫ (COS θ) ^ (- 4) + (COS θ) ^ (- 6) DCOS θ = - 1 / 3 (COS θ) ^ 3 - 1 / 5 (COS θ) ^ 5 + C, C is a constant

It is known that the square of the parabola y = 4x, the straight line AB passes (4,0), and intersects the parabola at two points a and B. It is proved that OA is perpendicular to ob

Let y = KX (∵ passing point 4,0)
From: y ^ 2 = 4x
y=kx
That is: K ^ 2x ^ 2-4x = 0
Delta = 0 (because there are two intersections)
Find K
Let's take the linear AB equation and let's not say it