In the triangle ABC, if sin (2 π - a) = - √ 2Sin (π - b), √ 3cos (2 π - a) = - √ 2cos (π - b), find the three angles of the triangle

In the triangle ABC, if sin (2 π - a) = - √ 2Sin (π - b), √ 3cos (2 π - a) = - √ 2cos (π - b), find the three angles of the triangle

Sin (2 π - a) = - Sina = - 2 SINB (Sina) 2 = 2 (SINB) 2 = 3 cos (2 π - a) = 3 cosa = 2 (CoSb) 2 (Sina) 2 + (COSA) 2 = 2 sin 2 B + (2 cos 2 b) / 3 = 14 sin

Given that sin α = Radix 2Sin β, Radix 3cos α = Radix 2cos β, and 0 < α, β < π, calculate the values of α and β

Because sin α = root 2Sin β, sin α ^ 2 = 2Sin β ^ 2
Because the root 3cos α = root 2cos β, 3cos α ^ 2 = 2cos β ^ 2
By adding the two formulas, 2cos α ^ 2 + 1 = 2
So cos α = radical 2 / 2
We can get α = 45 ° and β = 30 °

If sin (3 π - 2) = 2 sin (2 π + β), 3cos (- α) = - 2cos (π + β) Zero

Sin (3 π - a) = root 2Sin (2 π + β), root 3cos (- α) = - root 2cos (π + β) Ψ Sina = √ 2Sin β ① √ 3cosa = √ 2cos β ② ① J2 + ② sin? A + 3cos? A = 2 = 2Sin? A + 2cos? A ﹤ sin? A = cos? A ﹣ Tana = 1 or - 1 (1

A tan α = radical 3, find the value of (2Sin α - 3cos α) / (sin α + cos α) Given Tan α = radical 3, find the value of (2Sin α - 3cos α) / (sin α + cos α)

(2sinα-3cosα)/(sinα+cosα)
=(2tan α - 3) / (Tan α + 1) in this step, the denominator is divided by cos α
=(2√3 -3)/(√3+1)
=(2√3-3)(√3-1)/[(√3+1)(√3-1)]
=(9-5√3)/2

Simplification: sin (x + 60 °) + 2Sin (X-60 °) - √ 3cos (120 ° - x)=

Sin(x+60)+2sin(x-60)-√3cos(120-x)
=(sinxcos60+cosxsin60)+2(sinxcos60-cosxsin60)-√3(cos120cosx+sin120sinx)
=1/2sinx+√3/2cosx+sinx-√3cosx+√3/2cosx-3/2sinx
=0

Simplify y = sin ^ 2 (x) + 2Sin (x) cos (x) + 3cos ^ 2 (x)

y=sin²x+2sinxcosx+3cos²x
y=(sin²x+cos²x)+2sinxcosx+(2cos²x-1)+1
=1+sin2x+cos2x+1
=√2sin(2x+π/4)+2

Simplification: sin (x + π / 3) -√ 3cos (2 π / 3-x) + 2Sin (x - π / 3)=

Expand to get
simple form
=sinx*cosπ/3 +cosx *sinπ/3 - √3 *cos2π/3 *cosx -√3 *sin2π/3*sinx +2sinx*cosπ/3 -2cosx *sinπ/3
=sinx* 1/2 +cosx *√3/2 -√3 *(-1/2) *cosx -√3 *√3/2 *sinx +2sinx *1/2 -2cosx *√3/2
=(1/2 -3/2+1)*sinx +(√3/2 +√3/2 -√3)*cosx
=0

Calculation: sin20 ° cos40 ° + sin70 ° cos50 °

sin20=sin(90-70)
cos40=cos(90-50)
If you open it, you can delete two items and multiply them to sin50cos70
The sum is sin120

(evaluation) sin20 degrees cos50 degrees - sin70 degrees cos40 degrees (main process)

Sin 20 degrees cos 50 degrees - Sin 70 degrees cos 40 degrees
=sin20cos50-cos20sin50
=sin(20-50)
=-sin30
=-1/2

The values of sin20 ° cos50 ° - sin70 ° cos40 ° This is the content of trigonometric identity transformation in Chapter 3 of senior high school mathematics compulsory course 4. Please explain the process of using sum angle formula, and I will adopt it immediately,

sin20°cos50°-sin70°cos40°=sin20°cos50°-cos20sin50=-sin30=-0.5