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Cos -1 - Texas Instruments TI-83 Plus Calculator

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Is there a question? Why so secretive, spill it out.
To access the arc cosine (cos^-1) press 2nd COS and enter the argument. Make sure you are within range, because the domain of arc cosine is from -1 to +1. Any value outside of this closed interval will warrant a DOMAIN ERROR.

Posted on May 29, 2011

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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

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

2 Answers

How to solve for the trigonometric functions in the casio scientific calculator?


Press the relevant function key [SIN],[COS],[TAN] followed by the angle value.
For inverse trigonometric functions [SIN^-1], COS^-1], or [TAN^-1], Press [SHIFT][SIN], [SHIFT][COS], or [SHIFT][TAN].
For other functions
sec(x)=1/cos(x)
csc(x)=1/sin(x)
cot(x)=1/tan(x)

When calculating trigonometric functions one must make sure that the angle unit the calculator is using is the correct one.

Aug 26, 2010 | Casio FX-115ES Scientific 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

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

1 Answer

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

1 Answer

(1+cotx-cosecx)(1+tanx+secx)=2


I shall attempt :D
1) cosec A + cot A = 3
we know that (cot A)^2 + 1 = (cosec A)^2
Hence, (cosec A)^2 - (cot A)^2 = 1
thus, (cosec A + cot A) (cosec A - cot A) = 1
3 (cosec A - cot A) = 1
(cosec A - cot A) = 1/3

(cosec A - cot A) = 1/3
(cosec A + cot A) = 3
Summing them, 2 cosec A = 3 1/3
cosec A = 6 2/3 = 5/3
sin A = 0.15
Thus, cos A = sqrt (1 - (sin A)^2) = 0.989


2) Prove that (1+tan x - sec x)(1 + cot x + cosec x) =2
expand
LHS= 1 + cot x + cosec x + tan x + 1 + tan x cosec x - sec x - sec x cot x - sec x cosec x
We can calculate that
tan x cosec x = sec x (since tan x = sin x / cos x)
sec x cot x = cosec x
so the above is
LHS = 1 + cot x + cosec x + tan x + 1 + sec x - sec x - cosec x - sec x cosec x
LHS = 2 + cot x + tan x - sec x cosec x
LHS = 2 + cos x / sin x + sin x / cos x - 1 / (sin x cos x)
LHS = 2 + [{cos x}^2 + {sin x}^2 - 1] / (sin x cos x)
LHS = 2 (proved)

May 12, 2009 | ValuSoft Bible Collection (10281) for PC

1 Answer

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

Trignometery: prove that .....


THIS PROBELM IS TO DIFFICULT SO PLEASE SLOVE THIS PROBELM

Oct 07, 2008 | Computers & Internet

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

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