Tolong jawab ya! Buktikan : 1. Sin A ÷ (1 - cos A) = (1 + sin A + cos A) ÷ (1 + sin A - cos A) 2. Sin 1/2 (A-B) + cos 1/2 C = 2 cos 1/2 B sin 1/2 A
Matematika
Anonyme
Pertanyaan
Tolong jawab ya!
Buktikan :
1. Sin A ÷ (1 - cos A) = (1 + sin A + cos A) ÷ (1 + sin A - cos A)
2. Sin 1/2 (A-B) + cos 1/2 C = 2 cos 1/2 B sin 1/2 A
Buktikan :
1. Sin A ÷ (1 - cos A) = (1 + sin A + cos A) ÷ (1 + sin A - cos A)
2. Sin 1/2 (A-B) + cos 1/2 C = 2 cos 1/2 B sin 1/2 A
1 Jawaban
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1. Jawaban Anonyme
Trigonometri.
Nomor 1.
[tex]\displaystyle \frac{1+\sin A+\cos A}{1+\sin A-\cos A}\\ =\frac{1+\sin A+\cos A}{\frac{1+\cos^2 A}{1+\cos A}+\sin A}\\ =\frac{1+\sin A+\cos A}{\frac{\sin^2 A}{1+\cos A}+\sin A}\\ =\frac{1+\sin A+\cos A}{\frac{\sin^2 A+(1+\cos A)\sin A}{1+\cos A}}\\ =\frac{(1+\sin A+\cos A)(1+\cos A)}{\sin^2 A+\sin A(1+\cos A)}\\ =\frac{(1+\sin A+\cos A)(1+\cos A)}{\sin A(\sin A+1+\cos A)}\\ =\frac{1+\cos A}{\sin A}\\[/tex]
[tex]\displaystyle =\frac{\sin A(1+\cos A)}{\sin^2 A}\\ =\frac{\sin A}{\frac{\sin^2 A}{1+\cos A}}\\ =\frac{\sin A}{\frac{1-\cos^2 A}{1+\cos A}}\\ =\frac{\sin A}{1-\cos A} \\ [/tex]
Terbukti.
Nomor 2.
2 cos α sin β = sin (α + β) - sin (α - β)
2 cos (1/2 B) sin (1/2 A)
= sin (1/2 B + 1/2 A) - sin (1/2 B - 1/2 A)
= sin [1/2 (A + B)] - sin [1/2 (B - A)]
Tidak terbukti.