Jika tan θ = a untuk θ sudut lancip, nilai cos 3θ sama dengan . . . .

<p>Jawaban : D</p><p>&nbsp;</p><p>Ingat!</p><p>sin 2a = 2 sin a cos a</p><p>cos (a+b) = cos a cos b - sin a sin b</p><p>Perbandingan trigonometri:&nbsp;</p><p>sin a = de/mi&nbsp;</p><p>cos a = sa/mi&nbsp;</p><p>tan a = de/sa</p><p>dengan mi = √(de²+sa²)&nbsp;</p><p>de : sisi depan sudut a</p><p>sa : sisi samping sudut a</p><p>mi : sisi miring&nbsp;</p><p>&nbsp;</p><p>Diketahui tan θ = a untuk θ sudut lancip</p><p>&nbsp;</p><p>tan θ = a</p><p>de/sa = a/1</p><p>&nbsp;</p><p>Diperoleh&nbsp;</p><p>de = a</p><p>sa = 1</p><p>&nbsp;</p><p>mi = √(de²+sa²)&nbsp;</p><p>mi = √(a²+1²)&nbsp;</p><p>mi = √(a²+1)&nbsp;</p><p>&nbsp;</p><p>Sehingga</p><p>sin θ = de/mi = a/√(a²+1)</p><p>cos θ = sa/mi = 1/√(a²+1)</p><p>&nbsp;</p><p>Sehingga&nbsp;</p><p>cos (a+b) = cos a cos b - sin a sin b</p><p>cos (θ+θ) = cos θ cos θ - sin θ sin θ</p><p>cos (2θ) = 1/√(a²+1) . 1/√(a²+1) - a/√(a²+1) . a/√(a²+1)</p><p>cos (2θ) = 1/(a²+1) - a²/(a²+1)</p><p>cos (2θ) = (1-a²)/(a²+1)</p><p>&nbsp;</p><p>sin 2θ = 2 sin θ cos θ</p><p>sin 2θ = 2 . a/√(a²+1) . 1/√(a²+1)</p><p>sin 2θ = 2a/(a²+1)</p><p>&nbsp;</p><p>cos (a+b) = cos a cos b - sin a sin b</p><p>cos (2θ+θ) = cos 2θ cos θ - sin 2θ sin θ</p><p>cos (3θ) = (1-a²)/(a²+1) . 1/√(a²+1) - 2a/(a²+1) . a/√(a²+1)</p><p>cos (3θ) = (1-a²)/((a²+1)√(a²+1)) - 2a²/((a²+1)√(a²+1))</p><p>cos (3θ) = (1-a²-2a²)/((a²+1)√(a²+1))</p><p>cos (3θ) = (1-3a²)/((a²+1)√(a²+1))</p><p>cos (3θ) = (1-3a²)/((a²+1)√(a²+1)) . √(a²+1)/√(a²+1)</p><p>cos (3θ) = ((1-3a²)√(a²+1))/((a²+1)²)</p><p>cos (3θ) = ((1-3a²)/((a²+1)²)) √(a²+1)</p><p>cos (3θ) = ((1-3a²)/((1+a²)²)) √(1+a²)</p><p>&nbsp;</p><p>Jadi, cos (3θ) = ((1-3a²)/((1+a²)²)) √(1+a²)</p><p>Pilihan jawaban yang benar adalah D</p>