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Nikolay [14]
3 years ago
8

Plz prove this for me.....

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
3 0

I've answered your other question as well.

Step-by-step explanation:

Since the identity is true whether the angle x is measured in degrees, radians, gradians (indeed, anything else you care to concoct), I’ll omit the ‘degrees’ sign.

Using the binomial theorem, (a+b)3=a3+3a2b+3ab2+b3

⇒a3+b3=(a+b)3−3a2b−3ab2=(a+b)3−3(a+b)ab

Substituting a=sin2(x) and b=cos2(x), we have:

sin6(x)+cos6(x)=(sin2(x)+cos2(x))3−3(sin2(x)+cos2(x))sin2(x)cos2(x)

Using the trigonometric identity cos2(x)+sin2(x)=1, your expression simplifies to:

sin6(x)+cos6(x)=1−3sin2(x)cos2(x)

From the double angle formula for the sine function, sin(2x)=2sin(x)cos(x)⇒sin(x)cos(x)=0.5sin(2x)

Meaning the expression can be rewritten as:

sin6(x)+cos6(x)=1−0.75sin2(2x)=1−34sin2(2x)

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Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software,
vichka [17]

Answer:

A(-1,0) is a local maximum point.

B(-1,0)  is a saddle point

C(3,0)  is a saddle point

D(3,2) is a local minimum point.

Step-by-step explanation:

The given function is  

f(x,y)=x^3+y^3-3x^2-3y^2-9x

The first partial derivative with respect to x is  

f_x=3x^2-6x-9

The first partial derivative with respect to y is  

f_y=3y^2-6y

We now set each equation to zero to obtain the system of equations;

3x^2-6x-9=0

3y^2-6y=0

Solving the two equations simultaneously, gives;

x=-1,x=3  and y=0,y=2

The critical points are

A(-1,0), B(-1,2),C(3,0),and D(3,2).

Now, we need to calculate the discriminant,

D=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2

But, we would have to calculate the second partial derivatives first.

f_{xx}=6x-6

f_{yy}=6y-6

f_{xy}=0

\Rightarrow D=(6x-6)(6y-6)-0^2

\Rightarrow D=(6x-6)(6y-6)

At A(-1,0),

D=(6(-1)-6)(6(0)-6)=72\:>\:0 and f_{xx}=6(-1)-6=-18\:

Hence A(-1,0) is a local maximum point.

See graph

At B(-1,2);

D=(6(-1)-6)(6(2)-6)=-72\:

Hence, B(-1,0) is neither a local maximum or a local minimum point.

This is a saddle point.

At C(3,0)

D=(6(3)-6)(6(0)-6)=-72\:

Hence, C(3,0) is neither a local minimum or maximum point. It is a saddle point.

At D(3,2),

D=(6(3)-6)(6(2)-6)=72\:>\:0 and f_{xx}=6(3)-6=12\:>\:0

Hence D(3,2) is a local minimum point.

See graph in attachment.

3 0
3 years ago
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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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