Simplify f (x) = 2Sin (x + θ / 2) cos (x + θ / 2) + 2 √ 3cos ^ (x + θ / 2) -√ 3 How did the last step come about?

Simplify f (x) = 2Sin (x + θ / 2) cos (x + θ / 2) + 2 √ 3cos ^ (x + θ / 2) -√ 3 How did the last step come about?

(x) = sin (2x + θ) + s (2x + θ) + 3 * [2 * cos 2 (x + θ / 2) - 1] = sin (2x + θ) + and √ 3 * cos (2x + θ) = 2 * [sin (2x + θ) * (1 / 2) + (√ 3 / 2) * cos (2x + θ)] = 2 * [sin (2x + θ) * cos (π / 3) + sin (π / 3) + sin (π / 3) * cos (2x + θ)] = 2 * sin (2X + π + π + π + π + π + π + π + π + π + π + π + π + π + 2x + 2 * sin2 * sin2 * sin (2x get it this time!

sin(x+π/3)+2sin(x-π/3)-√3cos(2π/3 - x) Simplify, simplify, and

The original formula = sinxcos π / 3 + cosxsin π / 3 + 2sinxcos π / 3-2cosxsin π / 3 - √ 3cosxs2 π / 3 - √ 3sinxsin2 π / 3
=1/2*sinx+√3/2*cosx+sinx-√3cosx+√3/2*cosx-3/2*sinx
=0

sin20°sin70+sin10°cos50°=? RT

=sin20°cos20°+sin10°cos(60°-10°)
=1/2sin40°+sin10°(1/2cos10°+√3/2sin10°)
=sin20°cos20°+1/2sin10°cos10°+√3/2(sin10°^2)
=1/2sin40°+1/4sin20°+√3/4(2sin10°^2-1)+√3/4
=1/2sin40°+1/2(1/2sin20°-√3/2sin20°)+√3/4
=1/2sin(-40°)+1/2sin(40°)+√3/4
=√3/4
^Two is square, and the answer is three quarter root
I have done something complicated by myself. Excuse me

sin20°cos50°=1/2(sin70°-sin30°) This is for Mao

It is proved that sin20 ° cos50 ° = 1 / 2 (sin70 ° - sin30 °),
sin20°cos50°-1/2(sin70°-sin30°)= sin(30°-10°)cos(60°-10°)-1/2[sin(60°+10°)-sin30°]
=( sin30°cos10°- cos30° sin10°)( cos60°cos 10°+ sin60°sin10°)-1/2(sin60°cos10°+ cos60°sin10°-1/2)
=(1/2* cos10°-√3/2* sin10°)( 1/2* cos10°+√3/2* sin10°)- 1/2(√3/2*cos10°+ 1/2*sin10°-1/2)
=1/4* cos10°^2-3/4* sin10°^2-√3/4*cos10°-1/4*sin10°+1/4
=1/4*( cos10°^2-3 sin10°^2-√3*cos10°-sin10°+1)
=1/4*( cos10°^2-3 sin10°^2-√3*cos10°-sin10°+ cos10°^2+ sin10°^2)
=1/4*( 2cos10°^2-2 sin10°^2-√3*cos10°-sin10°)
=1/2*( cos10°^2- sin10°^2-√3/2*cos10°-1/2*sin10°)
=1/2*( cos20°-√3/2*cos10°-1/2*sin10°)
=1/2*( cos20°- cos30°*cos10°- sin30*sin10°)
=1/2*( cos20°- cos(30°-10°))
=1/2*( cos20°- cos20°)=0
Therefore, the above formula is established

Cos20 ° sin50 ° + sin20 ° sin50 ° is equal to?

cos20°cos50°+sin20°sin50°
=cos(50°-20°)
=cos30°
=√3/2

Evaluate sin 10 degrees, cos 20 degrees, sin 30 degrees, cos 40 degrees, sin 50 degrees, cos 60 degrees, sin 70 degrees, cos 80 degrees, sin 90 degrees Evaluation, it's best to write down the process

cos20cos40cos80
=sin20cos20cos40cos80/sin20
=(1/2)sin40cos40cos80/sin20
=(1/4)sin80cos80/sin20
=(1/8)sin160/sin20
=1/8
sin70*sin50*sin10=-1/2(cos120-cos20)*sin10=1/2(sin10cos20+1/2sin10)=1/2[1/2(sin30-sin10)+1/2sin10]=1/4*sin3=1/8
[[[or there is a triple angle formula that can be calculated directly
sin3x=4sinx*sin(60+x)*sin(60-x) ]]]
sin30*cos60*sin90=1/4
So the original formula = 1 / 8 * 1 / 8 * 1 / 4 = 1 / 256

What is cos 20 ° cos 40 ° cos 60 ° cos 80

cos20°cos40°cos60°cos80°
=(sin20cos20cos40cos60cos80) / sin20
=(sin 40 ° cos 40 ° cos 60 ° cos 80 °) / 2 sin 20 ° (2-fold angle formula)
=(sin 80 ° cos 60 ° cos 80 °) / 4sin 20 ° (2-fold angle formula)
=(sin 160 ° cos 60 °) / 8sin 20 ° (2-fold angle formula)
=sin(180°-160°)cos60°/8sin20°.(sin（π－α）＝sinα)
(Note: here α = 160)
=(sin20 ° cos60 °) / 8sin20 °
=1/16
Among them:
=(sin 40 ° cos 40 ° cos 60 ° cos 80 °) / 2 sin 20 ° (2-fold angle formula)
=(sin 80 ° cos 60 ° cos 80 °) / 4sin 20 ° (2-fold angle formula)
=(sin 160 ° cos 60 °) / 8sin 20 ° (2-fold angle formula)
=sin(180°-160°)cos60°/8sin20°.(sin（π－α）＝sinα)

cos20°cos40°cos60°cos80° =(sin20cos20cos40cos60cos80) / sin20 =(sin 40 ° cos 40 ° cos 60 ° cos 80 °) / 2 sin 20 ° (2-fold angle formula) =(sin 80 ° cos 60 ° cos 80 °) / 4sin 20 ° (2-fold angle formula) =(sin 160 ° cos 60 °) / 8sin 20 ° (2-fold angle formula) =sin(180°-160°)cos60°/8sin20°.(sin（π－α）＝sinα) (Note: here α = 160) =(sin20 ° cos60 °) / 8sin20 ° =1/16 Among them: =(sin 40 ° cos 40 ° cos 60 ° cos 80 °) / 2 sin 20 ° (2-fold angle formula) =(sin 80 ° cos 60 ° cos 80 °) / 4sin 20 ° (2-fold angle formula) =(sin 160 ° cos 60 °) / 8sin 20 ° (2-fold angle formula) =sin(180°-160°)cos60°/8sin20°.(sin（π－α）＝sinα) How do you get these steps, (multiply or divide by what number), please write your ideas

∵sinαcosα=1/2sin2α∴sin20°cos20°=1/2sin40°sin40°cos40°=1/2sin80°sin80°cos80°=1/2sin160°∴=(sin40°cos40°cos60°cos80°)/2sin20°=（sin80°cos60°cos80°）/4sin20°=(sin160°cos60°)/8sin...

cos20°•cos40°•cos60°•cos80°=（　　） A. 1 Four B. 1 Eight C. 1 Sixteen D. 1 Thirty-two

∵cos20°•cos40°•cos80°=2sin20°cos20°cos40°cos80°2sin20°=2sin40°cos40°cos80°22sin20°=2sin80°cos80°23sin20°∴cos20°•cos40°•cos80°=18×sin160°sin20°=18×sin(180°−20°)sin20°=18...

Find cos20 × cos40 × cos60 × cos80

cos20×cos40×cos60×cos80
=sin20×cos20×cos40×cos60×cos80/sin20
=sin40×cos40×cos60×cos80/2sin20
=sin80×cos60×cos80/4sin20
=sin160×cos60/8sin20
=sin20×cos60/8sin20
=cos60/8=1/16
Because cos60 = 1 / 2, the double angle formula is used gradually