Question

Question 21 Prove that [cos 4x + cos 3x + cos 2x] / [sin 4x + sin 3x + sin 2x] = cot 3x

Class X1 - Maths -Trigonometric Functions Page 73

Answer 1

LHS = (cos4x + cos3x + cos2x)/(sin4x + sin3x +sin2x)
use the formula,
sinC + sinD = 2sin(C+D)/2.cos(C-D)/2
and, cosC+cosD = 2cos(C + D)/2.cos(C-D)/2

= {(cos4x + cos2x)+cos3x}/{(sin4x+sin2x)+sin3x}
= {2cos(4x+2x)/2.cos(4x-2x)/2 + cos3x}/{2sin(4x+2x)/2.cos(4x-2x)/2+sin3x}
=(2cos3x.cosx + cos3x)/(2sin3x.cosx+sin3x)
= cos3x(2cosx+1)/sin3x(2cosx+1)
= cos3x/sin3x
= cot3x = RHS

Answer 2

$\frac{cos4x + cos2x + cos3x}{sin4x + sin2x + sin3x} \\ \\ = \frac{2cos \frac{6x}{2}cos \frac{2x}{2} + cos3x}{2sin \frac{6x}{2} cos \frac{2x}{2} + sin3x } \\ \\ = \frac{2cos3xcosx + cos3x}{2sin3xcosx + sin3x} \\ \\ = \frac{cos3x(2cosx + 1)}{sin3x(2cosx + 1)} \\ \\ = cot3x$