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Find cos 2x if cos x = 4/5 - Computers & Internet

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How can I solve Sinx+cosx=7/5?


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

Sep 07, 2014 | Computers & Internet

1 Answer

Cos2x + 3 = 5cosx


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.

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

1 Answer

TI-89 dimension mistmatch error


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


2_23_2012_9_41_18_am.jpg

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

1 Answer

Graphing cos 2x? i have a TI-86 but i think my problem is something that can be done on any TI calculator, i need to know how to graph cos 2x, all i get is a straight line


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

1 Answer

Differentiate each of the following w.r.t.x; 29.sin2xsinx


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
  1. (29)'=0 derivative of a constant is zero
  2. (sin(2X))'=cos(2X)*(2X)'=2*cos(2X)
  3. (sin(X))'=cos(X)
Result is
(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...

1 Answer

Sin x cos x = -1/2 solve for x


sin x cos x = -1/2
=> 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...

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Help


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

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

1 Answer

(sin-cos)(sin+cos)


(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

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

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