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Prove the following: cos12°+ cos 84° + cos 156° + cos132° = -12 - Mathematics and Statistics

Sum

Prove the following:

cos12°+ cos 84° + cos 156° + cos132° = `-1/2`

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Solution

L.H.S. = cos12°+ cos84° + cos156° + cos132°

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

= cos12° + cos84° – cos24° – cos48°  ...[∵ cos(π – θ) = – cos θ]

= (cos12° – cos48°) – (cos24° – cos84°)

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

= 2sin30°·sin18° – 2sin54°·sin30°

= sin 18° – sin54°   ...`[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

∴ sin2θ = sin (90° – 3θ) = cos3θ

∴ 2sinθ cosθ = 4cos3θ – 3cosθ

∴ 2siθ = 4cos2θ – 3

∴ 2sinθ = 4(1 – sin2θ) - 3

∴ 2sinθ = 4 – 4sin2θ - 3

∴ 4sin2θ + 2sinθ – 1 = 0

∴ sinθ= `(-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 = sin18° = `(sqrt(5) - 1)/4`  ...(2)

Now, sin3θ = 3sinθ - 4sin3θ

Put θ = 18°, we get,

sin54 ° = 3sin18° - 4sin318°

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

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

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APPEARS IN

Balbharati Mathematics and Statistics 1 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 3 Trigonometry - 2
Miscellaneous Exercise 3 | Q II.(5) | Page 57
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