Pertanyaan

Jika 0 ≤ ( α , B ) ≤ 3 π , cos ( α + β ) ⋅ cos ( α − β ) = 0. dan sin ( 2 π − α ) cos α = 4 3 , maka α 2 − β 2 = …

Jika $0 \leq (α, B) \leq \frac{π}{3}, cos (α + β) \cdot cos (α - β) = 0.$ dan $sin (\frac{π}{2} - α) cos α = \frac{3}{4},$ maka $α^{2} - β^{2} = \dots$

$- \frac{π ^{2}}{12}$
$- \frac{π ^{2}}{6}$
$0$
$\frac{π ^{2}}{12}$
$- \frac{π ^{2}}{6}$

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah A

Pembahasan

Ingat bahwa : Sudut berelasi sin ( 90 − α ) = cos α Rumus perkalian trigonometri cos A ⋅ cos B = 2 1 [ cos ( A + B ) + cos ( A − B ) ] dari soal diketahui sin ( 2 π − α ) cos α cos α ⋅ cos α cos 2 α cos α cos α cos α α = = = = = = = 4 3 4 3 4 3 4 3 2 1 3 cos 6 π 6 π cos ( α + β ) ⋅ cos ( α − β ) 2 1 [ cos ( α + β + α − β ) + cos ( α + β − α + β ) ] 2 1 [ cos 2 α + cos 2 β ] 2 1 [ 2 cos 2 α − 1 + 2 cos 2 β − 1 ] 2 1 [ 2 cos 2 α + 2 cos 2 β − 2 ] cos 2 α + cos 2 β − 1 4 3 + cos 2 β − 1 4 3 − 1 + cos 2 β cos 2 β cos β cos β β = = = = = = = = = = = = 0 0 0 0 0 0 0 0 4 1 2 1 cos 3 π 3 π Sehingga α 2 − β 2 = = = = ( 6 π + 3 π ) ( 6 π − 3 π ) ( 6 π + 2 π ) ( 6 π − 2 π ) − 36 3 π 2 − 12 1 π 2 Jadi, jawaban yang tepat adalah A

Ingat bahwa :

Sudut berelasi

$sin (90 - α) = cos α$

Rumus perkalian trigonometri

$cos A \cdot cos B = \frac{1}{2} [cos (A + B) + cos (A - B)]$

dari soal diketahui

$sin (\frac{π}{2} - α) cos α cos α \cdot cos α cos^{2} α cos α cos α cos α α = = = = = = = \frac{3}{4} \frac{3}{4} \frac{3}{4} \frac{3}{4} \frac{1}{2} 3 cos \frac{π}{6} \frac{π}{6}$

$cos (α + β) \cdot cos (α - β) \frac{1}{2} [cos (α + β + α - β) + cos (α + β - α + β)] \frac{1}{2} [cos 2 α + cos 2 β] \frac{1}{2} [2 cos^{2} α - 1 + 2 cos^{2} β - 1] \frac{1}{2} [2 cos^{2} α + 2 cos^{2} β - 2] cos^{2} α + cos^{2} β - 1 \frac{3}{4} + cos^{2} β - 1 \frac{3}{4} - 1 + cos^{2} β cos^{2} β cos β cos β β = = = = = = = = = = = = 00000000 \frac{1}{4} \frac{1}{2} cos \frac{π}{3} \frac{π}{3}$

Sehingga

$α^{2} - β^{2} = = = = (\frac{π}{6} + \frac{π}{3}) (\frac{π}{6} - \frac{π}{3}) (\frac{π + 2 π}{6}) (\frac{π - 2 π}{6}) - \frac{3}{36} π^{2} - \frac{1}{12} π^{2}$