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vivado [14]
3 years ago
15

Find the derivative of y=8ln(sin t+9)

Mathematics
1 answer:
ira [324]3 years ago
6 0
\bf y=8ln[sin(t+9)]\\\\
-----------------------------\\\\
\cfrac{dy}{dx}=8\cdot \cfrac{cos(t+9)\cdot 1}{sin(t+9)}\implies \cfrac{dy}{dx}=8cot(t+9)
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Jika sin x = p, dengan x tumpul, maka cos(x + 60) =​
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Menjawab:

[(√1-p²) -3√p] / 2

Penjelasan langkah demi langkah:

Dari identitas trigonometri, pemuaiannya benar:

Cos (A + B) = cosAcosB-sinAsinB

Menerapkan ini dalam memperluas cos (x + 60).

cos (x + 60) = cosxcos60 - sinxsin60

Jika sinx = p = berlawanan / sisi miring

opp = p, hyp = 1

adj² = 1²-p²

adj = √1-p²

Cos (x) = adj / hyp = √1-p² / 1

Cos (x) = √1-p²

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Mengganti nilai-nilai ini ke dalam rumus

cos (x + 60) = cosxcos60 - sinxsin60

cos (x + 60) = √1-p² (1/2) - p (√3 / 2)

cos (x + 60) = (√1-p²) / 2 - √3p / 2

Temukan KPK tersebut

cos (x + 60) = [(√1-p²) -3√p] / 2

Oleh karena itu cos (x + 60) = [(√1-p²) -3√p] / 2

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