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sattari [20]
3 years ago
11

Find sin 2x, cos 2x, and tan 2x from the given information. cscx=4,tanx<0.

Mathematics
1 answer:
lutik1710 [3]3 years ago
5 0

Answer:

Step-by-step explanation:

Given the expression cosec (x) = 4 and tan(x)< 0

since cosec x = 1/sinx

1/sinx = 4

sinx = 1/4

From SOH, CAH TOA

sinθ = opposite/hypotenuse

from sinx = 1/4

opposite = 1 and hypotenuse = 4

to get the adjacent, we will use the Pythagoras theorem

adj² = 4²-1²

adj² = 16-1

adj ²= 15

adj = √15

cosx = adj/hyp = √15/4

tanx = opposite/adjacent = 1/√15

since tan < 0, then tanx = -1/√15

From double angle formula;

sin2x = 2sinxcosx

sin2x = 2(1/4)(√15/4)

sin2x = 2√15/16

sin2x = √15/8

for cos2x;

cos2x = 1-2sin²x

cos2x = 1-2(1/4)²

cos2x = 1-2(1/16)

cos2x= 1-1/8

cos2x = 7/8

for tan2x;

tan2x  = tanx + tanx/1-tan²x

tan2x = 2tanx/1-tan²x

tan2x = 2(-1/√15)/1-(-1/√15)²

tan2x = (-2/√15)/(1-1/15)

tan2x = (-2/√15)/(14/15)

tan2x = -2/√15 * 15/14

tan2x = -30/14√15

tan2x = -30/7√15

rationalize

tan2x =  -30/7√15 * √15/√15

tan2x =  -30√15/7*15

tan2x =  -2√15/7

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x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }

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p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}

Next, we determine the rate at which the quantity demanded changes with respect to time.

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\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}

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\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}

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Recall that at t=25, p(25)=\dfrac { 5800 } {13} \approx 446.15

Therefore:

\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009

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