Problem (13) : Trigonometri


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buktikan sin A + sin B + sin C = cos \frac{1}{2} A + cos \frac{1}{2} B + cos \frac{1}{2} C jika A + B + C = 1800

Bukti :

A + B = 1800 – C … (i)

sin A + sin B = 2 sin \frac{1}{2} (A + B) cos \frac{1}{2} (A – B) … (ii) [Sifat Dasar Trigonometri]

sin C = sin (\frac{1}{2} C + \frac{1}{2} C)

= 2 sin \frac{1}{2} C cos \frac{1}{2} C … (iii)

sin \frac{1}{2} (A + B) = sin \frac{1}{2} (1800 – C)

= sin (900\frac{1}{2} C)

= sin 900 cos \frac{1}{2} C – cos 900 sin \frac{1}{2} C

= 1 . cos \frac{1}{2} C – 0 . sin \frac{1}{2} C

= cos \frac{1}{2} C … (iv)

sin \frac{1}{2} C = cos (1800\frac{1}{2} C) … (v)

sin A + sin B + sin C = 2 sin \frac{1}{2} (A + B) cos \frac{1}{2} (A – B) + 2 sin \frac{1}{2} C cos \frac{1}{2} C [berdasarkan (ii) dan (iii)]

= 2 cos \frac{1}{2} C cos \frac{1}{2} (A – B) + 2 sin \frac{1}{2} C cos \frac{1}{2} C [berdasarkan (iv)]

= 2 cos \frac{1}{2} C (cos \frac{1}{2} (A – B) + sin \frac{1}{2} C) [sifat distributif]

= 2 cos \frac{1}{2} C (cos \frac{1}{2} (A – B) + cos (1800\frac{1}{2} C)) [berdasarkan (v)]

= 2 cos \frac{1}{2} C (2 cos \frac{1}{4} (A – B + 1800 – C) cos \frac{1}{4} (A – B – 1800 + C)) [berdasarkan Sifat Dasar Trigonometri]

= 2 cos \frac{1}{2} C (2 cos \frac{1}{4}) (A – B + A + B) cos \frac{1}{4} (A – B – A – B)) [berdasarkan (i)]

= 2 cos \frac{1}{2} C (2 cos \frac{1}{2} A cos \frac{1}{2} (-B))

= 4 cos \frac{1}{2} C cos \frac{1}{2} A cos \frac{1}{2} B [INGAT : cos (-\alpha) = cos \alpha)

= 4 cos \frac{1}{2} A cos \frac{1}{2} B cos \frac{1}{2} C

Jadi menurut saya, soalnya ada kesalahan, seharusnya sin A + sin B + sin C = 4 cos \frac{1}{2} A cos \frac{1}{2} B cos \frac{1}{2} C.

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