Use the expansions of cos ( 2 x + x ) and cos ( 2 x - x ) , to prove that: cos 3 x + cos x ≡ 2 cos 2 x cos x Solve cos 3 x + cos 2 x + cos x > 0 for 0 ° < x < 360 ° .

Take cos 3 x = cos ( 2 x + x ) = cos 2 x cos x - sin 2 x sin x ...(i) cos x = cos ( 2 x - x ) = cos 2 x cos x + sin 2 x sin x ..............(ii) (i) + (ii) cos 3 x + cos x = 2 cos 2 x cos x Hence, proved. Using cos 3 x + cos x = 2 cos 2 x cos x ⇒ cos 3 x + cos 2 x + cos x > 0 ⇒ ( cos 3 x + cos x ) + cos 2 x > 0 ⇒ 2 cos 2 x cos x + cos 2 x > 0 ⇒ cos 2 x ( 2 cos x + 1 ) > 0 ⇒ cos 2 x > 0 and cosx > - 1 2 ....(iii) or cos 2 x < 0 and cosx < - 1 2 ....(iv) Now, cos x > - 1 2 when x < 120 ° or x > 240 ° cos x < - 1 2 when 120 ° < x < 240 ° cos 2 x > 0 when 0 ° < 2 x < 90 ° or 270 ° < 2 x < 360 ° i.e., when 0 ° < x < 45 ° or 135 ° < x < 180 ° cos 2 x < 0 when 45 ° < x < 135 ° Here, (iii) requires x < 120 ° and 0 ° < x < 45 ° so 0 ° < x < 45 ° Here, (iv) requires 120 ° < x < 240 ° and 45 ° < x < 135 ° , so 120 ° < x < 135 °

Prove each of these identities. tan A + co t A ≡ 2 cosec 2 A

To prove: tan A + co t A ≡ 2 cosec 2 A LHS : sinA cosA + cosA sinA = sin 2 A + cos 2 A sin A cos A = 1 sin A cos A = 2 2 sin A cos A = 2 sin 2 A = 2 cosec 2 A = RHS Hence, proved.