Given sin2a = a, cos2a = B, find Tan (π / 4 + a) RT

Given sin2a = a, cos2a = B, find Tan (π / 4 + a) RT

Because cos2a = B, that is cos ^ 2a-sin ^ 2A = B (^ 2 represents Square), so 1-2sin ^ 2A = B is reduced to Sina = (| 2-2b) / 2 (| represents root sign) (1) from sin2a = a to Sina = A / 2cosa (2) (1) to (2) to cosa = A / (| 2-2b) (3) Tan (π / 4 + a) = - Tana, so (2) / (3) * - 1) is reduced to

If 1 + Tan A / 1-tan a = 3 + 2 radical 2, then sin2a =?
How to calculate?

(1 + Tan a) / (1-tan a) = 3 + 2 √ 2 → (Tan π / 4 + Tan a) / (1-tan π / 4 · Tan a) = Tan (π / 4 + a), that is, Tan (π / 4 + a) = 3 + 2 √ 2. According to the universal formula, cos (π / 2 + 2a) [1-tan ^ 2 (π / 2 + 2a)] / [1 + Tan ^ 2 (π / 2 + 2a)] = [1 - (3 + 2 √ 2) ^ 2] / [1

The number of intersections between the image of function y = x ^ 2 + 3 | x | + 7 and the image of function y = x ^ 2-3x + | x ^ 2-3x | + 6 is
There are four

The graph of function y = x2-3 | x | + 7 is drawn, and the value of x2-3x is discussed. It can be divided into three cases: greater than 0, less than 0 and equal to 0, and the corresponding image solution can be made. For example, the graph of function y = x2-3 | x | + 7 is y = 2x2-6x + 6 when x2-3x > 0, and there are two intersections between the graph and the graph of function y = x2-3 | x | + 7; when x2-3x ≤ 0, 0 |

It is known that sin (45 ° + a) sin (45 ° - a) = - 1 / 4,0 °

1, sin (45 + a) sin (45-a) = sin (45 + a) cos (45 + a) = 1 / 2Sin (90 + 2a) = 1 / 2cos (2a) = - 1 / 4, the solution is cos (2a) = - 1 / 2; a = 602, sin (a + 10) (1 - (radical 3) Tan (A-10)) = sin70 * (1 - (radical 3) * tan50) = cos20 / cos50 * (cos50 - (radical 3) * sin50) = 2cos20 / sin40 * (1 / 2 * Co