Question

sin(A+B) + sin(A+B) = 2 sinAcosB how it will get​

Answer 1

sin(a+b)+sin(a+b)=2sinAcosB

EXPLAINATION:WRITE SIN(A+B) FORMULA THAT IS sinAcosB+cosAsinB

and add it again with sinAcosB+cosAsinB

and simplify it you will get 2sinAcosB

Answer 2

$\sf{Answer}$

Given to prove :-

sin(A+B) + sin(A-B) = 2sinAcosB

Formulaes Implemented :-

sin( A +B) = sinAcosB+ sinBcosA

sin (A - B) = sinAcosB - sinB cosA

Lets do !

sin(A+B) + sin(A- B) = sinAcosB+ sinBcosA + sinAcosB - sinB cosA

Keep like terms together

sinAcosB + sinAcosB + sinBcosA - sinBcosA

2sinBcosA get cancelled

Then ,

sinAcosB + sinA cosB

2sinAcosB

So, sin(A+B) + sin(A+B) = 2 sinAcosB

Hence proved !

Know more :-

cos(A + B) = cosAcosB - sinAsinB

cos ( A - B) = cosA cosB + sinAsinB

tan ( A +B ) = tanA + tanB / 1 - tanAtanB

tan( A-B) = tanA - tanB/1 + tanAtanB

cot ( A + B ) = cotBcotA -1 / cotB + cotA

cot ( A - B ) = cotB cotA + 1/ cotB - cotA

tan(45+ A) = 1+tanA/1 - tanA

tan (45 - A ) = 1-tanA/1 + tanA

Trignometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trignometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trignometric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj