Question

Prove the following:

cos 12°+ cos 84° + cos 156° + cos 132° = `-1/2`

Answer 1

L.H.S. = cos 12°+ cos 84° + cos 156° + cos 132°

= cos12°+ cos84° + cos(180° – 24°) + cos(180° – 48°)

= cos 12° + cos 84° – cos 24° – cos 48°  ...[∵ cos(π – θ) = – cos θ]

= (cos 12° – cos 48°) – (cos 24° – cos 84°)

= `2sin((12^circ + 48^circ)/2)*sin((48^circ - 12^circ)/2) - 2sin((24^circ + 84^circ)/2)*sin((84^circ - 24^circ)/2)`

= 2 sin 30° · sin 18° – 2 sin 54° · sin 30°

= sin 18° – sin 54°  ....`[because sin30^circ = 1/2]` ...(1)

Let θ = 18°

Then 2θ = 36° and 3θ = 54°

We have 36° + 54° = 90°

i.e., 2θ = 90° – 3θ

∴ sin 2θ = sin (90° – 3θ) = cos 3θ

∴ 2 sin θ cos θ = 4 cos³θ – 3 cos θ

∴ 2 sin θ = 4 cos²θ – 3

∴ 2 sin θ = 4(1 – sin²θ) – 3

∴ 2 sin θ = 4 – 4 sin²θ – 3

∴ 4 sin²θ + 2 sin θ – 1 = 0

∴ sin θ= `(-2 ± sqrt(4 - 4(4)(-1)))/(2 xx 4)`

= `(-2 ± 2sqrt(5))/8`

= `(-1 ± sqrt(5))/4`

Since θ = 18° is acute, sin θ is positive.

∴ sin θ = sin 18° = `(sqrt(5) - 1)/4`  ...(2)

Now, sin 3θ = 3 sin θ – 4 sin³θ

Put θ = 18°, we get,

sin 54 ° = 3 sin 18° - 4 sin³18°

= `3((sqrt(5) - 1)/4) - 4((sqrt(5) - 1)/4)^3`

= `(3(sqrt(5) - 1))/4 - 1/16 (5sqrt(5) - 15 + 3sqrt(5) - 1)`

= `(12sqrt(5) - 12 - 8sqrt(15) + 16)/16`

= `(4sqrt(5) +  4)/16`

= `(sqrt(5) + 1)/4`  ...(3)

From (1), (2) and (3), we get,

L.H.S. = `((sqrt(5) - 1)/4) - ((sqrt(5) + 1)/4)`

= `1/4(sqrt(5) - 1 - sqrt(5) - 1)`

= `-2/4`

= `-1/2`

= R.H.S.