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Serggg [28]
3 years ago
6

How do you do... cos2x-sin2x=1-2sin2x

Mathematics
1 answer:
Bess [88]3 years ago
3 0
sin (2 \alpha) =2\cdot sin  \alpha \cdot cos  \alpha \\cos(2 \alpha )=1-2sin^2  \alpha \\-------------------- \\cos2x-sin2x=1-2sin2x\\cos2x+sin2x=1\\1-2sin^2x+2sinx\cdot cosx=1\\-2sinx(sin x-cosx)=0\\ -2sinx=0\ \ \ \ or\ \ \ \ sinx-cosx=0\\\\1)\ -2sinx=0\ \ \Leftrightarrow\ \ sinx=0\ \ \Leftrightarrow\ \ x_1=k \pi \ \ and\ \ k\in I\\2)\ \ sinx-cosx=0\ \ \Leftrightarrow\ \ sinx=cosx\ /:cosx\ \ \ \ \ \wedge\ \ \ \  cos x \neq 0\\tanx=1\ \ \Leftrightarrow\ \ x_2= \frac{ \pi }{4} +k \pi \ \ and\ \ k\in I
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Answer:

Step-by-step explanation:

Given f_{XY} (x,y) = c(4x + 2y +1) ; 0 < x < 40\,and\, 0 < y

a)

we know that \int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty} {f(x,y)} \, dxdy=1

therefore \int\limits^{40}_{-0}\int\limits^2_{0} {c(4x+2y+1)} \, dxdy=1

on integrating we get

c=(1/6640)

b)

P(X>20, Y>=1)=\int\limits^{40}_{20}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy

on doing the integration we get

                        =0.37349

c)

marginal density of X is

f(x)=\int\limits^2_{0} {\frca{1}{6640}(4x+2y+1)} \, dy

on doing integration we get

f(x)=(4x+3)/3320 ; 0<x<40

marginal density of Y is

f(y)=\int\limits^{40}_{0} {\frca{1}{6640}(4x+2y+1)} \, dx

on doing integration we get

f(y)=\frac{(y+40.5)}{83}

d)

P(01)=\int\limits^{40}_{0}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy

solve the above integration we get the answer

e)

P(X>20, 0

solve the above integration we get the answer

f)

Two variables are said to be independent if there jointprobability density function is equal to the product of theirmarginal density functions.

we know f(x,y)

In the (c) bit we got f(x) and f(y)

f(x,y)cramster-equation-2006112927536330036287f(x).f(y)

therefore X and Y are not independent

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