The answer of 40.5 | level 2 | 2011-4-18: 54 mentioned the formula of vector point multiplication, and used the range of trigonometric function; But two basic problems are obviously ignored: first, the length of PA and PA is not radius, so it is not necessarily1; Secondly, when both PA and PB are tangent lines, the included angle between these two vectors obviously cannot be 180 degrees, that is, the cosine will not be-1. In addition, the dot multiplication symbol cannot be represented by "×".
Our solution is as follows, and we still use the point multiplication formula: (PA and PB both represent vectors in the solution process)
Let ∠APB=θ, because the length of the connecting line between the center and the tangent point is radius and perpendicular to the tangent, | ao | = | bo | =1;
∠APO=∠BPO=θ/2, and use tangent trigonometric functions in two right triangles respectively.
|PA|=|PB|= 1/tan(θ/2), which can be obtained by point multiplication formula: pa Pb = | pa ||| Pb | cos θ,
Bring in: = cos θ/[tan (θ/2) tan (θ/2)], and then get from the double angle formula or universal formula of trigonometric transformation:
Cos θ = [1-tan (θ/2) tan (θ/2)]/[1+tan (θ/2) tan (θ/2)], which brings simplification (let t=tan(θ/2)).
Papb = [1/T2-1]/[1/T2+1], this function is about the increasing function of1/T2 at (0, +∞), according to the inverse proportional function. PA PB & lt 1。
If there are no other additional conditions, this should be the result. If yes, please refer to Join.
I wish you good grades.