Question about Texas Instruments TI-83 Plus Calculator

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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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Posted on Jan 02, 2017

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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.

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

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...

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

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.

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

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...

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.

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

Hope it helps.

Nov 05, 2009 | Casio Office Equipment & Supplies

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:

{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

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

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

THIS PROBELM IS TO DIFFICULT SO PLEASE SLOVE THIS PROBELM

Oct 07, 2008 | Computers & Internet

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.

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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