Find the minimum positive period of the function f (x) = 2cosxcos (Π / 6-x) - Radix 3sin ^ 2x + sinxcosx 1. The minimum positive period of F (x) 2. X belongs to [- π / 3, π / 2], find the range of F (x)

Find the minimum positive period of the function f (x) = 2cosxcos (Π / 6-x) - Radix 3sin ^ 2x + sinxcosx 1. The minimum positive period of F (x) 2. X belongs to [- π / 3, π / 2], find the range of F (x)

First of all, it is simplified that f (x) = 2cosxcos (x - π / 6) - √ 3sin ^ 2x + sinxcos x = 2cosxcos (x - π / 6) - 2sinx (x - π / 6) - 2sinx (√ 3 / 2sinx-1 / 2cosx) = 2cosxcos (x-π / 6) - 2sinx (sinxcos π / 6-cosxsin π / 6) = 2cosxcos (x - π / 6) - 2sinxsin (x - π / 6) = 2cos (x + X - π / 6) = 2cos (x + X - π / 6) = 2cos (x + X - π / 6) = 2cos (x + X + X / 6) = 2cos (2x - 2x - 2x - 2x - 2x - 2xπ / 6

When 0 degree

sin²θ-2sinθ+1=（1-sinθ）²
Because 0 degrees

Given 3 π / 2 < θ < 2 π, simplify 1 + sin θ - 1 - sin θ + 2 - sin θ under radical sign

Simplification of half angle formula 3 π / 4

Simplified root sign (1-2sin10 ° cos10 °) / [cos10 ° - root sign (1-sin ^ 280 °)]

Under root sign (1-2sin10 ° cos10 °) / [cos10 ° - root sign (1-sin ^ 280 °)]
=Under root sign (sin10 ° - cos10 °) ^ 2 / [cos10 ° - root sign (COS ^ 280 °)]
=|sin10°-cos10°|/[cos10°-cos80°] cos10°>sin10° cos80°=sin10°
=(cos10°-sin10°)/(cos10°-sin10°)
=1

2Sin ^ 2 α - (1 + radical 3) sin α + radical 3 / 2 = 0, calculate the degree of acute angle α

(2sina-√3)(sina-1/2)=0
sina=√3/2 sina=1/2
a=60° a=30°

Who knows how much simplified 1 + sin10 + 1-sin10 under radical

1+sin10
=sin²5+cos²5+2sin5cos5
=(sin5+cos5)²
Similarly, 1 -- sin10 = (cos5-sin5) 2
The original formula = sin5 + cos5 + cos5-sin5 = 2cos5

Simplification: 1 in 10 'minus 3 in COS 10';

=Sin10 '× cos10' (cos10 '- Radix 3 × sin10')
= 2 sin 10'cos 10 '(1 / 2 × cos 10' - 2 / 3 × sin 10 ')
{convert 1 / 2 to sin 30 'and 2 / 3 to Cos 30'}
＝sin20'/sin20'
＝1

How many units do you need to shift the cosine curve to the left (right) to get the sine curve? Find the value of 1 / sin10 degree - radical 3 / cos10 degree Monotone increasing interval of y = sin (2x + π / 3) It is known that Tan α ζ = 4 / 3 α belongs to [π, 3 / 2 π]. Find the values of sin α and cos α

The value of π / 2 Units 1 / sin10 degree root 3 / cos10 degree = (cos10 root 3sin10) / sin10cos10 = 1 / 2 (1 / 2cos10-3 / 2sin10) / sin10cos10 = 1 / 2 (sin30cos10-cos30sin10) / sin10cos10 = sin (30-10) / sin20 = 1 - π / 2 + 2K π

1 / sin 10 degrees minus (root 3 divided by cos 10 degrees) Simplification

1 / minus (root 3 divided by)
=(cos10 degree root 3 times sin10 degree) / (sin10 degree cos10 degree)
=4 * [(1 / 2cos10 degree - root 3 / 2sin10 degree) / sin20 degree]
=4 * [(sin30 degree cos10 degree sin10 degree cos30 degree) / sin20 degree]
=4 * sin (30-10) degrees / sin20 degrees
=4

1 / sin 10 ° minus 3 / cos 10 ° to find the process

1/sin10°-√3/cos10°=（cos10°-√3sin10°）/ sin10°cos10°=2[1/2cos10°-(√3/2)sin10°]/ sin10°cos10°=2(sin30°cos10°-cos30°sin10°)/ sin10°cos10°=2sin(30°- 10°)/ sin10°cos10°=2sin20°/ si...