Question

in ∆ABC cos(B+2C+3A/2)+cos(A-B/2)=?​

Answer 1

Step-by-step explanation:

so this is the correct answer

Answer 2

Concept:

We need to first recall the concept of trigonometric ratios and triangle to answer this question.

• The sum of all angles of a ΔABC = $180^{0}$.
• cos($180^{0}$ +α) = - cos α                      (2)

To find:

We have to find the value of cos$(\frac{B+2C+3A}{2})$ + cos $(\frac{A-B}{2})$ in triangle ABC.

Solution:

In ΔABC,

       A + B + C = $180^{0}$                                   (1)

cos$(\frac{B+2C+3A}{2})$ = cos $(\frac{(A+B+C)+2A+C}{2})$

By using (1)

                        = cos$(\frac{180^{0}+2A+C}{2})$

                        = cos $(\frac{180^{0}A+180^{0}-B} {2})$                     [  using (1)   A+C = $180^{0}$- B ]

                        = cos $(180^{0}+\frac{A-B} {2})$

By using (2)

                        = - cos$(\frac{A-B} {2})$

Hence, cos$(\frac{B+2C+3A}{2})$ + cos $(\frac{A-B}{2})$ =  - cos$(\frac{A-B} {2})$ + cos$(\frac{A-B} {2})$

                                                         = 0