Question about Nokia 6233 Cellular Phone

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Quardartic equation x2 sin2α cos2θ + 4xy sinα sin θ + y2 [4cosα ? (1+ cosα)2] = 0

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Try change the phone or try reflashing it

Posted on Aug 24, 2008

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Jul 12, 2014 | Super Tutor Trigonometry (ESDTRIG) for PC

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Evaluating a problem using integration with sin x and cos x

4/sqroot2 is the same value as 2sqroot2, about 2.828 . It's just a matter of how the calculator and the textbook chooses to represent the value.

Dec 30, 2012 | Office Equipment & Supplies

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Im trying to graph a parametric equation and whenever the equation has an exponent/power in it the calculator says Syn error so i know how to graph x=2 cos T, y=2 sin T and similar equations without a...

To enter powers of trigonometric functions you must enclose the functions in parentheses and then apply the exponent to the whole. For example X1t =(sin T)^3 , Y1t=(cos T)^3 will give you a shape similar to a rhombus with concave sides. The symbol ^syands for the operation of raising to a power. The key is the one with the caret symbol ^ , and it is wedged between the xsqure and EXIT keys (third row of keys).
As regards the cotangent, use the equivalent definition cot(X)=1/tan(X).

Jul 26, 2011 | Casio FX-9750GPlus Calculator

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Using inverse cosign and inverse sin always bring up domain errrors. For example, I used the equation COS -1(185 / -20) . And it always came up with domain error. This is the same equation my teacher gave...

Make sure your calculator is in degree mode.
It is probably in radian mode so it can't calculate the inverse sin.
To change this, go to Mode -> Deg.
Hope this helps, cheers!

Mar 06, 2011 | Texas Instruments TI-30 XIIS Calculator

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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))'
  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
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...

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Wrong answer produced in sin / cos problems?

5sin(x)+1 = 0 is the equation you want to solve?
5sin(x) = -1
sin(x) = -(1/5)
arcsin( sin(x) ) = arcsin( -(1/5) )
x = -.201 (radians)
x = -11.5369 (degrees)

Jun 04, 2010 | Texas Instruments TI-84 Plus Calculator

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This is a trigonometry problem not a calculator's.
cos(x) +tan(x).sin(x) = ( cos(x) + (sin(x)/cos(x)).sin(x). After reduction to the same denominator (which is cos(x)) you obtain

{cos(x).cos(x) + sin(x).sin(x)} divided by cos(x).
The content of the bracket above is just 1.
Your fraction will have 1 as numerator and cos(x) as denomitor. That is exactly the definition of the secant function i.e. Function sec is the reciprocal (not the inverse) of the cos function, while the arccos is the inverse of cos.

A mild advice: Avoid writing function without specifying the argument (the variable on which a function acts).


Aug 14, 2009 | Casio FX-115ES Scientific Calculator

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sec^4X- sec^2X = 1/cot^4X + 1/cot^2X
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)
=1/Cos^4X - Cos^2X/Cos^4X
=1/Cos^4X - 1/Cos^2X
=Sec^4X - Sec^2X

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

4 Answers

Trig Identities

Change csc to 1/sin. Find a common denominator and add the two left terms.
1/sin - sin = (1 -sin^2)/sin. Rewrite formula
(1 - sin^2)/sin = cos^2/sin Divide out the /sin.
1 - sin^2 = cos^2 Rearange.
1 = cos^2 + sin^2 Yes, that's true. It's like the Pythagorean formula.

May 22, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

1 Answer

Molweid's equation

Mollweide's equations?


Nov 26, 2007 | Televison & Video

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