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Posted on Aug 07, 2010

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Posted on Jan 02, 2017

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Square each side

Sin ^2 (x) + cos ^2 (x) +2 sin (x)cos (x) = 49/25

1 + 2sin (x)cos (x) = 1.960

sin (2x) = 0.960

2x = 73.74 deg

x = 36.87 deg

Sin ^2 (x) + cos ^2 (x) +2 sin (x)cos (x) = 49/25

1 + 2sin (x)cos (x) = 1.960

sin (2x) = 0.960

2x = 73.74 deg

x = 36.87 deg

Sep 07, 2014 | Computers & Internet

Use the identity cos(2x)=2(cos(x))^2-1

cos(2x)+3=5cos(x) becomes 2(cos(x))^2-1+3=5cos(x)

Arrange a bit: 2(cos(x))^2-5cos(x)+2=0

Get rid of the 2-factor

(cos(x))^2-(5/2) cos(x)+1=0

This is a quadratic equation for the unknown U=cos(x)

U^2-(5/2)U+1=0

Solve it by factoring or with the quadratic equation formula. The solutions are** U=2 or U=1/2.**

Since U=cos(x), the root U=cox(x)=2 must be rejected.

What is left is cos(x)=(1/2). The solutions are x=60 or x=-60 plus or minus 360 degrees.

cos(2x)+3=5cos(x) becomes 2(cos(x))^2-1+3=5cos(x)

Arrange a bit: 2(cos(x))^2-5cos(x)+2=0

Get rid of the 2-factor

(cos(x))^2-(5/2) cos(x)+1=0

This is a quadratic equation for the unknown U=cos(x)

U^2-(5/2)U+1=0

Solve it by factoring or with the quadratic equation formula. The solutions are

Since U=cos(x), the root U=cox(x)=2 must be rejected.

What is left is cos(x)=(1/2). The solutions are x=60 or x=-60 plus or minus 360 degrees.

Apr 01, 2014 | SoftMath Algebrator - Algebra Homework...

**Dimension
mismatch** error occurs when arguments are not with same dimensions. Two or more arguments must be of the same
dimension. For example, [1,2]+[1,2,3] is a dimension mismatch because the matrices
contain a different number of elements. By the way, it it very easy to resolving
your problem with differentiation. See captured image

As a result of differentiation you had to obtained:

(3b+2c)cos2x+(3c-2b)(sin2x)e^(3x)+2ae^(2x)

Feb 23, 2012 | Texas Instruments TI-89 Calculator

Your angle unit is set to degrees. If you do not change the Xmin and Xmax window dimensions to say -180 to -90 for Xmin, and 90 to 180 for Xmax you will not see much of the features of the graph; In the default interval Xmin, Xmax the curve is a staright line.

Sep 17, 2011 | Texas Instruments TI-83 Plus Calculator

Use the rule for differentiating products of functions: ()' signifies derivative

(29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* (sin(2X))'*sin(X) +29*sin(2X)*(sin(X))'

But

(29*sin(2X)*sin(X))'= 29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)

You could also have cast your formula in the form

sin(2X)*sin(X)= 1/2[ cos(2X-X)-cos(2X+X)]=1/2[cos(X)-cos(3X)]

then calculated the derivative of

29/2*[cos(X)-cos(3X)]

which is

29/2*[-si(X) +3*sin(3X)]

The challenge for you is to prove that the two forms are equivalent

29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)=29/2*[-si(X) +3*sin(3X)]

(29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* (sin(2X))'*sin(X) +29*sin(2X)*(sin(X))'

But

- (29)'=0 derivative of a constant is zero
- (sin(2X))'=cos(2X)*(2X)'=2*cos(2X)
- (sin(X))'=cos(X)

(29*sin(2X)*sin(X))'= 29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)

You could also have cast your formula in the form

sin(2X)*sin(X)= 1/2[ cos(2X-X)-cos(2X+X)]=1/2[cos(X)-cos(3X)]

then calculated the derivative of

29/2*[cos(X)-cos(3X)]

which is

29/2*[-si(X) +3*sin(3X)]

The challenge for you is to prove that the two forms are equivalent

29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)=29/2*[-si(X) +3*sin(3X)]

Jun 21, 2010 | Vivendi Excel@ Mathematics Study Skills...

sin x cos x = -1/2

=> 2sinx cosx = -1

=> sin(2x) = -1

=> 2x = (3pi)/2 OR 2x = 270°

=> x = 3pi/4 OR x = 135°

=> 2sinx cosx = -1

=> sin(2x) = -1

=> 2x = (3pi)/2 OR 2x = 270°

=> x = 3pi/4 OR x = 135°

Jan 04, 2010 | SoftMath Algebrator - Algebra Homework...

X=13.2825(APPROX.)

Nov 08, 2009 | ValuSoft Bible Collection (10281) for PC

sec^4X- sec^2X = 1/cot^4X + 1/cot^2X

RHS

1/cot^4X + 1/cot^2X

=1/(Cos^4X/Sin^4X) + 1/(Cos^2X/Sin^2X)

=Sin^4X/Cos^4X + Sin^2X/Cos^2X

=Sin^4X/Cos^4X + Cos^2X.Sin^2X/Cos^4X

=Sin^2X/Cos^4(Sin^2X + Cos^2X)

=Sin^2X/Cos^4X

=(1-Cos^2X)/Cos^4X

=1/Cos^4X - Cos^2X/Cos^4X

=1/Cos^4X - 1/Cos^2X

=Sec^4X - Sec^2X

=LHS

RHS

1/cot^4X + 1/cot^2X

=1/(Cos^4X/Sin^4X) + 1/(Cos^2X/Sin^2X)

=Sin^4X/Cos^4X + Sin^2X/Cos^2X

=Sin^4X/Cos^4X + Cos^2X.Sin^2X/Cos^4X

=Sin^2X/Cos^4(Sin^2X + Cos^2X)

=Sin^2X/Cos^4X

=(1-Cos^2X)/Cos^4X

=1/Cos^4X - Cos^2X/Cos^4X

=1/Cos^4X - 1/Cos^2X

=Sec^4X - Sec^2X

=LHS

Feb 02, 2009 | Super Tutor Trigonometry (ESDTRIG) for PC

(sinX-cosX)(sinX+cosX)
=(sin^2 - Cos^2X)

=Sin^2X - (1-Sin^2X)

=Sin^2X -1 + Sin^2X

=2Sin^2X -1

Zulfikar Ali

ali_zulfikar@yahoo.com

9899780221

=Sin^2X - (1-Sin^2X)

=Sin^2X -1 + Sin^2X

=2Sin^2X -1

Zulfikar Ali

ali_zulfikar@yahoo.com

9899780221

Jan 26, 2009 | Super Tutor Trigonometry (ESDTRIG) for PC

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