Question about Casio FX-9860G Graphic Calculator

# Y = -2 cos x - Casio FX-9860G Graphic Calculator

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What is your question? Do you want to plot it? Find its roots? Find its period? Or something else?

Posted on Jul 16, 2013

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Posted on Jan 02, 2017

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## Related Questions:

### 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...

### X=ln(seca+tana) find coshx

Sorry I do not like to work with secant and cosecant.
sec(a)+tan(a)=(1+sin(a))/cos(a)
ln(sec(a)+tan(a))= ln( (1+sin(a))/cos(a))=X
2*cosh(X)= e^(X)+e^(-X)
e^(X)=(1+sin(a))/cos(a)
e^(-X)= cos(a)/(1+sin(a))
2cosh(X)=(1+sin(a))/cos(a) +cos(a)/(1+sin(a))= 2/cos(a)
cosh(X)=1/cos(a)=sec(a)

Now that you see how you can do it, I trust you will discover any mistake I might have made.
If you want to use the classPad function sequence Action>Transformation>simplify(, do it step by step as I have detailed above.
Good Luck.

Dec 07, 2013 | Casio ClassPad 300 Calculator

### I need to rewrite Y=5(sqrt2)sin(x)-5(sqrt2)cos(x) as Y=Asin(Bx-c)

Use the fact that cos(pi/4)=sin(pi/4)= 1/square root(2). Trigonometric identity cos(a+b)=cos(a)cos(b)-sin(a)sin(b).

Nov 07, 2010 | SoftMath Algebrator - Algebra Homework...

### 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...

### How to use the cosine,tangent etc functions of calculator fx-82ES plus

Hello,
1.Set the correct angle unit required by your problem: degrees, radians, or grads. [SHIFT][MODE] [3:deg] or [4:Rad]

2. Press the key for the function COS, SIN, or TAN
[COS] displays Cos(

3.Enter the angle 12 deg Screen shows cos(12
Close the right parenthesis ) Screen shows cos(12)
4.Press [=] Screen displays 0.9781

If you want the inverse trigonometric functions you access them with arccos [SHIFT] [COS] (cos^-1)
arcsin [SHIFT][SIN] (sin^-1)
actan [SHIFT][TAN] (tan^-1)

You have to know the principal domain for the inverse trigonometric functions (see any book on trigonometry) to understand the results.
Hope it helps.

Nov 05, 2009 | Casio Office Equipment & Supplies

### Cos+tan(sin)=sec

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).

Note:

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

### 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

### How to get TI-94 to give me the Cos(20) = anumber not Cos(20)

instead of putting a whole number put in a decamal.... that worked for me
ex.
cos(25.)
not cos(25)

Oct 25, 2008 | Texas Instruments TI-92 Plus Calculator

### 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

### Cosh Squared or Cubed

if you want to square sin(3), press:

(
sin
3
)
)
^
2

Sep 24, 2007 | Texas Instruments TI-89 Calculator

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