Pertanyaan

Jika α , β , γ sudut di dalam segitiga, buktikan bahwa: cos α + cos β + cos γ = 1 + 4 sin 2 α sin 2 β sin 2 γ

Jika $α, β, γ$ sudut di dalam segitiga, buktikan bahwa:

$cos α + cos β + cos γ = 1 + 4 sin \frac{α}{2} sin \frac{β}{2} sin \frac{γ}{2}$

S. Nur

Master Teacher

Jawaban terverifikasi

Pembahasan

Ingat : Jumlah sudut dalam segitiga adalah 18 0 ∘ cos A + cos B = 2 cos 2 1 ( A + B ) cos 2 1 ( A − B ) cos ( 180 − A ) = − cos A cos 2 A = 1 − 2 sin 2 A cos A − cos B = − 2 sin 2 1 ( A − B ) sin 2 1 ( A − B ) Perhatikan perhitungan berikut A + B + C C = = 18 0 ∘ 18 0 ∘ − ( A + B ) Shingga cos α + cos β + cos γ = ( cos α + cos β ) + cos γ = 2 cos 2 1 ( α + β ) cos 2 1 ( α − β ) + ( 1 − 2 sin 2 2 γ ) = 2 cos 2 1 ( 180 − γ ) cos 2 1 ( α − β ) + 1 − 2 sin 2 2 γ = 2 cos ( 90 − 2 γ ) cos 2 1 ( α − β ) − 2 sin 2 2 γ + 1 = 2 sin ( 2 γ ) cos 2 1 ( α − β ) − 2 sin 2 2 γ + 1 = 2 sin ( 2 γ ) { cos 2 1 ( α − β ) − sin 2 1 ( 18 0 ∘ − ( α + β ) ) } + 1 = 2 sin ( 2 γ ) { cos 2 1 ( α − β ) − sin ( 9 0 ∘ − 2 1 ( α + β ) ) } + 1 = 2 sin ( 2 γ ) { cos 2 1 ( α − β ) − cos ( 2 1 ( α + β ) ) } + 1 = 2 sin ( 2 γ ) { cos ( 2 1 α − 2 1 β ) − cos ( 2 1 α + 2 1 β ) } + 1 = 2 sin ( 2 γ ) { − 2 sin 2 1 ( ( 2 1 α − 2 1 β ) + ( 2 1 α + 2 1 β ) ) sin ( 2 1 ( 2 1 α − 2 1 β ) − ( 2 1 α + 2 1 β ) ) } + 1 = 2 sin ( 2 γ ) { − 2 sin 2 1 ( α ) sin 2 1 ( − β ) } + 1 = 2 sin ( 2 γ ) { 2 sin 2 1 ( α ) sin 2 1 ( β ) } + 1 = 4 sin ( 2 γ ) sin ( 2 α ) sin ( 2 β ) + 1 = 1 + 4 sin ( 2 α ) sin ( 2 β ) sin ( 2 γ )

Ingat :

Jumlah sudut dalam segitiga adalah $18 0^{\circ}$
$cos A + cos B = 2 cos \frac{1}{2} (A + B) cos \frac{1}{2} (A - B)$
$cos (180 - A) = - cos A$
$cos 2 A = 1 - 2 sin^{2} A$
$cos A - cos B = - 2 sin \frac{1}{2} (A - B) sin \frac{1}{2} (A - B)$

Perhatikan perhitungan berikut

$A + B + C C = = 18 0^{\circ} 18 0^{\circ} - (A + B)$

Shingga

$cos α + cos β + cos γ = (cos α + cos β) + cos γ = 2 cos \frac{1}{2} (α + β) cos \frac{1}{2} (α - β) + (1 - 2 sin^{2} \frac{γ}{2}) = 2 cos \frac{1}{2} (180 - γ) cos \frac{1}{2} (α - β) + 1 - 2 sin^{2} \frac{γ}{2} = 2 cos (90 - \frac{γ}{2}) cos \frac{1}{2} (α - β) - 2 sin^{2} \frac{γ}{2} + 1 = 2 sin (\frac{γ}{2}) cos \frac{1}{2} (α - β) - 2 sin^{2} \frac{γ}{2} + 1 = 2 sin (\frac{γ}{2}) {cos \frac{1}{2} (α - β) - sin \frac{1}{2} (18 0^{\circ} - (α + β))} + 1 = 2 sin (\frac{γ}{2}) {cos \frac{1}{2} (α - β) - sin (9 0^{\circ} - \frac{1}{2} (α + β))} + 1 = 2 sin (\frac{γ}{2}) {cos \frac{1}{2} (α - β) - cos (\frac{1}{2} (α + β))} + 1 = 2 sin (\frac{γ}{2}) {cos (\frac{1}{2} α - \frac{1}{2} β) - cos (\frac{1}{2} α + \frac{1}{2} β)} + 1 = 2 sin (\frac{γ}{2}) {- 2 sin \frac{1}{2} ((\frac{1}{2} α - \frac{1}{2} β) + (\frac{1}{2} α + \frac{1}{2} β)) sin (\frac{1}{2} (\frac{1}{2} α - \frac{1}{2} β) - (\frac{1}{2} α + \frac{1}{2} β))} + 1 = 2 sin (\frac{γ}{2}) {- 2 sin \frac{1}{2} (α) sin \frac{1}{2} (- β)} + 1 = 2 sin (\frac{γ}{2}) {2 sin \frac{1}{2} (α) sin \frac{1}{2} (β)} + 1 = 4 sin (\frac{γ}{2}) sin (\frac{α}{2}) sin (\frac{β}{2}) + 1 = 1 + 4 sin (\frac{α}{2}) sin (\frac{β}{2}) sin (\frac{γ}{2})$