Question

Question 1 Prove that: 2cos (π/13) cos (9π/13) + cos (3π/13) + cos (5π/13) = 0

Class X1 - Maths -Trigonometric Functions Page 81

Answer 1

2cos(π/13).cos(9π/13)+cos(3π/13).cos(5π/13) = 0
use the formula,
2cosC.cosD = cos(C+D)+cos(C-D)

LHS = cos(π/13+9π/13)+cos(9π/13-π/13) + cos(π/13).cos(5π/13)
= cos(10π/13) + cos(8π/13) + cos(3π/13).cos(5π/13)
= {cos(10π/13)+cos(3π/13)}+{cos(8π/13) + cos(5π/13)}

use formula,
cosC + cosD = 2cos(C+D)/2.cos(C-D)/2

= {2cos(π/2).cos(7π/26)}+ {2cos(π/2).cos(3π/26)}

we know,
cos(π/2) = 0

= 0 + 0 = RHS

Answer 2

Answer:

Your answer is given on the above attachment. Hope it helps!!