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Here is to get you started. For the rest of the algebra, I leave it to you.

Posted on Oct 10, 2013

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Posted on Jan 02, 2017

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SOURCE: trignometery: prove that .....

THIS PROBELM IS TO DIFFICULT SO PLEASE SLOVE THIS PROBELM

Posted on Oct 22, 2008

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

ali_zulfikar@yahoo.com

9899780221

Posted on Mar 15, 2009

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°

Posted on Jan 30, 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

Posted on Apr 14, 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

Posted on Jun 01, 2010

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

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).

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

Oct 08, 2013 | Casio FX-9750GII Graphing Calculator

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.

Oct 08, 2013 | Casio FX9750GII Graphic Calculator

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.

Setting x/2=45, means that x=90 (degrees)

But cos(90)=0 and sin(90)=1. Thus tan(45)=1/(1+0)=1.

Mar 13, 2013 | SoftMath Algebrator - Algebra Homework...

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...

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

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

Apr 17, 2010 | SoftMath Algebrator - Algebra Homework...

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...

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

if the sides of a triangle are 15cm ,16cm, and 17 cm ,then the area of the triangle is what ?

May 09, 2008 | Texas Instruments TI-30XA Calculator

Oct 19, 2017 | Evga Computers & Internet

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