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huma Posted on Oct 10, 2013
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64(cos^8x+sin^8x)=cosx+28cos4x+35 - Computers & Internet

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k24674

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  • Computers & ... Master 8,093 Answers
  • Posted on Oct 10, 2013
k24674
Computers & ... Master
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Here is to get you started. For the rest of the algebra, I leave it to you.

64(cos^8x+sin^8x)=cosx+28cos4x+35 - 36cbe960-6a5b-4ab6-a606-fc455dca634e.png

5 Related Answers

A

Anonymous

  • Posted on Oct 22, 2008

SOURCE: trignometery: prove that .....

THIS PROBELM IS TO DIFFICULT SO PLEASE SLOVE THIS PROBELM

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ali_zulfikar

ali_zulfikar

  • 156 Answers
  • Posted on Mar 15, 2009

SOURCE: (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
[email protected]
9899780221

Anonymous

  • 146 Answers
  • Posted on Jan 30, 2010

SOURCE: 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°



Anonymous

  • 146 Answers
  • Posted on Apr 14, 2010

SOURCE: Please help me change:

we can represent 5 sin X --3 cos X as

5.83 sin ( X - 30.96)

if you want steps please leave a comment and don't forget to vote for me thank you

Anonymous

  • 100 Answers
  • Posted on Jun 01, 2010

SOURCE: cos A/1-tan A + sin A/1-cot A = sin A + cos A

you can follow this link and can get the solution http://mathforum.org/dr.math/faq/formulas/faq.trig.html

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Find the derivitive of sin(cosx)

The 9750GII doesn't have symbolic capabilities so it can't give you a symbolic answer for this. It can evaluate it numerically at any point.

The derivative of sin(cos(x) dx is cos(cos(x))-sin(x).
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Derivatives

The derivative of sin(cos(x)) dx is cos(cos(x))-sin(x). Unfortunately the fx-9850GII doesn't do symbolic differentiation so you'll have to get this result some other way. I cheated and used a calculator from another manufacturer, one that does do symbolic differentiation.
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Use the identity tan(x/2)=sinx/1+cosx to solve for the value of tan45 degrees

tan(x/2)=sin(x)/(1+cos(x))
Setting x/2=45, means that x=90 (degrees)
But cos(90)=0 and sin(90)=1. Thus tan(45)=1/(1+0)=1.
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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)]

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Cos2x + 3 = 5cosx

the solution you got is correct it is

cos X =2 and cos X =1/2


but we know maximum value of cosX is 1.so we discard the solution cos X =2


so only solution is cos X=1/2

and X = 60 degrees

hope you r satisfied.please rate the solution high.thank you
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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°



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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
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(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
[email protected]
9899780221
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1 - sin^2 = cos^2 Rearange.
1 = cos^2 + sin^2 Yes, that's true. It's like the Pythagorean formula.

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