Question

If tan3a/tana = k , prove that sin3a/sina = 2k/ k - 1 and hence deduce that either k > 3 or k < 1/3?

Answer 1

Tan3a/Tana=k => 3tana-tan^3a/tana-3tan^3a = k => 3-tan^2a/1-3tan^2a = k => tan^2a = a-3/3a-1 => tana = root of a-3/3a-1 Tana= opp/adj Hypotenuse = root of 4a-4 Sina = root of a-3/root of 4a-4 Now simplify sin3a/sina Sin3a/sina = 3sina - 4sin^3a/sina => 3-4sin^2a => 3-4(a-3/4a-4) => 3-(a-3/a-1) => 3a-3-a+3/a-1 => 2a/a-1

Answer 2

HELLO DEAR,

2k = 2tan3A/tanA

k - 1 = (tan3A/tanA) - 1 = (tan3A-tanA)/tanA

Hence,

2k/(k-1) = 2tan3A/(tan3A - tanA)

= 2tan3Acos3A/[cos3A (tan3A-tanA)]

= 2sin3A/[sin3A - tanA (4cos^3 A - 3cosA)]

= 2sin3A/(3sinA - 4sin^3 A - 4sinAcos^2 A + 3sinA)

=2sin3A/[sinA (3 - 4sin^2 A - 4cos^2 A + 3)]

= 2sin3A/[sinA {6-4 (sin^2 A + cos^2 A)}]

= 2sin3A/[sinA (6-4)]

= 2sin3A/2sinA= sin3A/sinA

Hence,

sin3A/sinA = 2k/(k-1)

Or,

(3sinA - 4sin^3 A)/sinA = 2k/(k-1)

Or,

3 - 4sin² A = 2k/(k-1)

Or, 4sin² A = 3 - 2k/(k-1)

= (k-3)/(k-1)

Or, sin^2 A = (k-3)/4 (k-1)

For 0 < (k-3)/4 (k-1), i.e., (k-3)/4 (k-1) > 0,

k-3> 0, k-1 > 0 or k-3 <0, k-1 <0Or, k> 3, k > 1 or k <3, k <1Or, k > 3 or k < 1.

(k-3)/4 (k-1) > 1

When k > 3,

(k-3) > 4 (k-1)Or, k-3 > 4k - 4Or, k - 4k > -4 + 3Or, -3k > -1Or, 3k < 1

Hence, k < 1/3

I HOPE ITS HELP YOU DEAR,
THANKS ♥️