Question about Texas Instruments TI-89 Calculator

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Hi danagasta,

If my last instructions did not work then you need to reset your RAM. You are missing the negative sign and, I have no idea where you got the absolute value inside the brackets.

Yes, the absolute value are used in books and, on integral tables but, I can asure you that the Ti 89 should not have produce them in this answer.

I'll stand behind that In( -cos(x) / (sin(x)-1)) is the correct answer. You can go ahead and simplify after the calculation to get the answer you want but, that is the good answer that Ti 89 is going to give you for the integral of sec(x).

Errors that do not match the suspected, I am pretty sure that Texas Instruments would have you clear the RAM. I do not like resetting the RAM but, I could not find the cause to why you got the answer you did. And so far no other expert has produced a solution to convince me otherwise.

I was not able to come up with any of the 3 answers MiB8888 did on my Ti 89. I am not saying that they are not good answers, they are book and integral tables answers. You are not going to get them on the Ti 89.

Posted on Jan 09, 2011

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See cap image below

Posted on Feb 05, 2016

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Actually, it should be any of:

- ln ' ( sin(x) + 1 ) / cos(x) '
- ln ' sec(x) + tan(x) '
- ln ' cos(x) / ( sin(x) - 1 ) '.

Posted on Jan 07, 2011

Hi there,

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Posted on Jan 02, 2017

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.

sec(a)+tan(a)=(1+sin(a))/cos(a)

ln(sec(a)+tan(a))=

2*cosh(X)= e^(X)+e^(-X)

2cosh(X)=(1+sin(a))/cos(a) +cos(a)/(1+sin(a))= 2/cos(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

Good Luck.

Dec 07, 2013 | Casio ClassPad 300 Calculator

Does it refuse to do so or does it give an error message?

Three common errors:

Three common errors:

- Not having the correct angle unit.
**Wrong result, No error message** - Confusing reciprocal of sine (1/sin(x) with arc sine (x) ,sin^-1(x). Confusing the reciprocal of cosine, 1/cos(x) with arc cosine (cos^-1(x)).
**Wrong result, No error message** - Taking the argument of the inverse sine and/or inverse cosine functions outside the interval [-1,1].
**This gives a domain error.**

Oct 28, 2013 | Texas Instruments TI-81 Calculator

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

I believe you are using the wrong syntax for the commands.

diff ( function, variable) Example: diff (sin(x^2), x) sgould give you 2.x.cos(x^2)

expand( will expand an expression that has parentheses: expand( (a+b)^2) will give a^2+2a.b+b^2

Integrate $

$ ( tan^-1(x), x) willl give x.tan^-1(x) -(1/2) ln ('x^2+1')

Check the syntax of a command before using it. You will find the list of all comands and their syntax in the reference manual.

diff ( function, variable) Example: diff (sin(x^2), x) sgould give you 2.x.cos(x^2)

expand( will expand an expression that has parentheses: expand( (a+b)^2) will give a^2+2a.b+b^2

Integrate $

$ ( tan^-1(x), x) willl give x.tan^-1(x) -(1/2) ln ('x^2+1')

Check the syntax of a command before using it. You will find the list of all comands and their syntax in the reference manual.

Aug 22, 2011 | Casio ClassPad 300 Calculator

What are you trying to do? The syntax you use depend on what you want to achieve.

Indefinite integral : Integral( ln(x),x) gives the correct answer xln(x)-x

Definite integral : Integral(ln(x),x,1,2) gives the integral of ln(x) between x=1 and x=2

Definite integral between 1 and y : Integral (ln(x), x,1,y) gives the correct expression yln(y)-y +1

By inserting the =0 you are in fact assigning the result of the integration to value 0, and the calculator calculates the integral and assigns 0 to it. This is what my TIVoyage 200 PLT gives me:

{Integral((lnx),x,1) = 0.} returns {x.ln(x)-x+1=0}

which is an equation.

Indefinite integral : Integral( ln(x),x) gives the correct answer xln(x)-x

Definite integral : Integral(ln(x),x,1,2) gives the integral of ln(x) between x=1 and x=2

Definite integral between 1 and y : Integral (ln(x), x,1,y) gives the correct expression yln(y)-y +1

By inserting the =0 you are in fact assigning the result of the integration to value 0, and the calculator calculates the integral and assigns 0 to it. This is what my TIVoyage 200 PLT gives me:

{Integral((lnx),x,1) = 0.} returns {x.ln(x)-x+1=0}

which is an equation.

Apr 17, 2011 | Texas Instruments TI-89 Calculator

5sin(x)+1 = 0 is the equation you want to solve?

so

5sin(x) = -1

sin(x) = -(1/5)

arcsin( sin(x) ) = arcsin( -(1/5) )

x = -.201 (radians)

x = -11.5369 (degrees)

so

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

You didn't specify what calculator you have.

On my TI30XA, 0.6339869 '2nd' 'COS-1' gives 50.655114

On my Windows XP calculator, 0.6339869 'Inv' 'cos' gives 50.655114

On a Casio FX-300ES, 0.6339869 'Shift' 'cos-1'

I hope one of these helps. If not, let me know what calculator you have and I'll figure it out.

On my TI30XA, 0.6339869 '2nd' 'COS-1' gives 50.655114

On my Windows XP calculator, 0.6339869 'Inv' 'cos' gives 50.655114

On a Casio FX-300ES, 0.6339869 'Shift' 'cos-1'

I hope one of these helps. If not, let me know what calculator you have and I'll figure it out.

Sep 08, 2009 | Office Equipment & Supplies

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)

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

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

Hello,

Clear all variables [F6] [ENTER]

Here is a screen capture of the display for your integral.

Hope it helps

Clear all variables [F6] [ENTER]

Here is a screen capture of the display for your integral.

Hope it helps

Dec 27, 2008 | Texas Instruments TI-89 Calculator

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Are you sure there's not a - sign in front of cos(x) such as -cos(x). like in ln[ |-cos(x) / (sin(x)-1)| ] which is the correct answer.

I've tried everything to figure out what could be wrong to give you that answer. I'll keep tring and, let you know.

Where you getting the absolute from? That may be the problem. I'll see if I can assimulate.

I know you probably can't wait around for me to hit the right combination to give you a exact solution. It may have something to do with folder your working in or, not updated os. Make sure you start your problems with the NewProb command entered at the Home screen. Try working out of the main folder. Somethings corrupt.. I'll keep trying too.

The Ti89 microprocessor calculations are inversed g functions unless certain questions are answered first

Once you have simplified to ln[ |-cos(x) / (sin(x)-1)| ] , don't forget get the + C as in,ln[ |-cos(x) / (sin(x)-1)| ] + C

Or simplified to ln[ |(sin(x)-1)/ cos(x)| ] , don't forget get the + C as in, ln[ | (sin(x)-1)/ (cos(x)| ] + C . This may be the answer the instructor is looking for

Charles your work is correct, I just added a few things to it. g doesn't not mean guassian or something simple you might think

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