How to simplify Sina + SINB

How to simplify Sina + SINB

Let (a + b) / 2 = a, (a-b) / 2 = B
The original formula = sin (a + b) + sin (a-b) = sin a * cos B + cos a * sin B + sin a * cos b-cos a * sin B
=2sina*cosb=2sin[(A+B)/2]cos[(A-B)/2]

Simplify [sin (60 + a) + cos120 * Sina] / cosa =?

[sin(60+a)+cos120*sina]/cosa
=[sin60cosa+cos60sina+cos120sina]/cosa
=[√3/2cosa+1/2sina-1/2sina]/cosa
=√3/2

How to simplify Cosa × cos (a + b) + Sina times sin (a + b)
This is the original topic 2cos (a + b) cosa cos (2a + b) simplification!

cos(a+b-a)=cosa×cos(a+b)+sina*sin(a+b) =cosb
2cos(a+b)cosa-cos(2a+b)=2cos(a+b)cosa-(cos(a+b)cosa-sin(a+b)sina)=cosa×cos(a+b)+sina*sin(a+b) =cosb

Simplification: sin (3a TT) / Sina + cos (3a TT) / cosa

sin(3a-TT)/sina+cos(3a-TT)/cosa
=(sin(3a-TT)cosa+cos(3a-TT)sina)/(sinacosa)
=sin(3a-TT+a)/sinacosa
=2sin(4a-TT)/sin2a

Known 0 degree

sinA·sin(120°-A)=0.5cos[A-(120°-A)]-0.5cos[A+(120°-A)]
=0.5cos(2A-120°)-0.5cos120°
=0.5cos(2A-120°)+0.25
Because 0 degree

Evaluate Sina + sin (120 + a) - sin (120-a)

sinA+sin(120°+A)-sin(120°-A)
=sinA+sin120°cosA+cos120°sinA-(sin120°cosA-cos120°sinA)
=sinA+ 2cos120°sinA
=sinA- 2cos60°sinA
=sinA-sinA
=0

Can 3 / 4 = sin (120-a) Sina solve a?

3/4=sin(120-A)sinA
3/4=-1/2[cos(120-A+A)-cos(120-A-A)]
-3/2=cos120-cos(120-2A)
-3/2=-1/2-cos(120-2A)
cos(120-2A)=1
120-2A=2kπ
2A=120-2kπ
A=60-kπ

Sina + sin (120-a) = 3 / 2 to find degree a

sinA+sin(120-A)=3/2
A=30°
I can see it

Simplification: 1 + sin * (α - 2 faction) * sin (faction + α) - 2cos & # 710; 2 * (- α)

1 + sin * (α - 2 faction) * sin (faction + α) - 2cos & # 710; 2 * (- α)
=1+sinα(-sinα)-2cos²α
=1-sin²α-2cos²α
=-3cos²α

F (x) = 2cos & # 178; ω x + 2 √ 3cos ω x sin ω x how to get Cos2 ω x + √ 3sin2 ω x + 1
f(x)=2cos²ωx+2√3cosωx sinωx
How to get Cos2 ω x + √ 3sin2 ω x + 1

Cos2 ω x = cos & # 178; ω x-sin & # 178; ω x = cos & # 178; ω X - (1-cos & # 178; ω x) = 2cos & # 178; ω X-1 double angle formula
∴2cos²ωx=cos2ωx+1
2Sin ω xcos ω x = sin2 ω x double angle formula