The price for hardcover books is 1.50 and the price for paperback is .50
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π → C = π - (A + B)
→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))
→ sin C = sin (A + B) cos C = - cos(A + B)
Use the following Sum to Product Identity:
sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]
cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]
Use the following Double Angle Identity:
sin 2A = 2 sin A · cos A
<u>Proof LHS → RHS</u>
LHS: (sin 2A + sin 2B) + sin 2C




![\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cqquad%20%5Cqquad%20%5Cqquad%202%5Csin%20C%5Ccdot%20%5B%5Ccos%20%28A-B%29%2B%5Ccos%20%28A%2BB%29%5D)


LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C 
f(x)= 3x³ - 18x +9
Algebraic identities are algebraic equations that are true regardless of the value of each variable. Additionally, they are employed in the factorization of polynomials. Algebraic identities are employed in this manner for the computation of algebraic expressions and the solution of various polynomials.
Identity I: (a + b)² = a² + 2ab + b²
Identity II: (a – b)² = a² – 2ab + b²
Identity III: a² – b²= (a + b)(a – b)
Identity IV: (x + a)(x + b) = x² + (a + b) x + ab
Identity V: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Identity VI: (a + b)³ = a³ + b³ + 3ab (a + b)
Identity VII: (a – b)³ = a³ – b³ – 3ab (a – b)
Identity VIII: a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
f(x) = (3x + 6) (x - 3)²
= ( 3x + 6) ( x - 3 )²
= ( 3x + 6)( x² - 6x + 9)
= 3x( x² - 6x + 9) + 6( x² - 6x + 9)
= 3x³ - 6x² + 18x + 6x² - 36x +9
= 3x³ - 18x +9
To learn more about algebraic expansions, refer to brainly.com/question/4344214
#SPJ9
<h3><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u>-</u></h3>
Lets Solve
