Question

In triangle ABC,if a+b-3c=0,then what is the value of cosA+cosB?​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

cos A + cos B = 6 sin² $\frac{C}{2}$

Step-by-step explanation:

Given that a + b - 3c = 0

=> a + b = 3c

According to sine rule,

$\frac{a}{sin A}$ = $\frac{b}{sinB}$ = $\frac{c}{sin C}$ = x(Let)

=> a = x sin A

    b = x sin B

    c = x sin C

We also know that

A + B + C = 180°           (sum of ∠s of a Δ = 180°)

=> A + B = 180° - C

=> $\frac{A+B}{2}$ = 90° - $\frac{C}{2}$                                                     --(i)

On substitution,

=> x sin A + x sin B = 3 * x sin C

=> x(sin A + sin B) = 3x sin C

=> sin A + sin B = 3 sin C

=> 2 sin($\frac{A+B}{2}$) cos ($\frac{A-B}{2}$) = 3 sin C                     (by formula)

=> 2 sin (90° - $\frac{C}{2}$)cos ($\frac{A-B}{2}$) = 3 sin C                   (from equation i)

=> 2 cos  $\frac{C}{2}$ cos ($\frac{A-B}{2}$) = 3 *2*sin $\frac{C}{2}$ cos $\frac{C}{2}$        (sin(90 - x) = cos x)

=>  cos ($\frac{A-B}{2}$) = 3 *sin $\frac{C}{2}$

=> cos ($\frac{A-B}{2}$) = 3 *sin (90° -  $\frac{A+B}{2}$ )                   (from equation i)

=> cos ($\frac{A-B}{2}$) = 3 *cos ( $\frac{A+B}{2}$ )

We need to find cos A + cos B

We know that cos A + cos B = 2cos($\frac{A+B}{2}$) cos ($\frac{A-B}{2}$)

                                               = 2 cos($\frac{A+B}{2}$) * [3 *cos ( $\frac{A+B}{2}$ ) ]

                                               = 6 cos²($\frac{A+B}{2}$)

                                               = 6 cos²(90° - $\frac{C}{2}$ )                  (from equation i)

                                               = 6 sin² $\frac{C}{2}$

Therefore, cos A + cos B = 6 sin² $\frac{C}{2}$