1. y=sin3x

2. y=3sin1/2x

3. y=cos(x-Pi/4)

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Posted on Jan 02, 2017

You have several types of graphs

**Function graph**s

Y_1=f(x), Y1=3X^2-4, [X, T, Theta, n] key types**X**

**Polar graphs** r=F(theta), r=r_o*ln(theta). [X, T, Theta, n] types **Theta **

**Parametric graphs** X_1=f(T) and Y_1=g(T). [X.T, Theta, n] types **T**

Examples: X_1= cos(T). Y_1= 2(1--sin(T))

**Sequence graphs **u_n+1= f(u_n), [X,T,Theta,n] types **n**

Y_1=f(x), Y1=3X^2-4, [X, T, Theta, n] key types

Examples: X_1= cos(T). Y_1= 2(1--sin(T))

Nov 21, 2013 | Texas Instruments TI-84 Plus Calculator

You have several types of graphs

**Function graph**s

Y_1=f(x), Y1=3X^2-4, [X, T, Theta, n] key types**X**

**Polar graphs** r=F(theta), r=r_o*ln(theta). [X, T, Theta, n] types **Theta **

**Parametric graphs** X_1=f(T) and Y_1=g(T). [X.T, Theta, n] types **T**

Examples: X_1= cos(T). Y_1= 2(1--sin(T))

**Sequence graphs **u_n+1= f(u_n), [X,T,Theta,n] types **n**

Y_1=f(x), Y1=3X^2-4, [X, T, Theta, n] key types

Examples: X_1= cos(T). Y_1= 2(1--sin(T))

Nov 21, 2013 | Texas Instruments TI-84 Plus Silver...

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

The trig functions depend on the angular mode. Make sure you have the proper one set.

Aug 16, 2012 | Texas Instruments TI-Nspire Graphic...

cos(5PI)=cos(4PI+PI)=cos(PI)=-1

sin(19PI/6)=sin(18PI/6+ PI/6)=sin(3PI +Pi/6)=sin(2PI+PI+PI/6)=sin(PI+PI/6)=sin(-PI/6)=-sin(PI/6)=-1/2

sin(19PI/6)=sin(18PI/6+ PI/6)=sin(3PI +Pi/6)=sin(2PI+PI+PI/6)=sin(PI+PI/6)=sin(-PI/6)=-sin(PI/6)=-1/2

Dec 12, 2011 | Super Tutor Trigonometry (ESDTRIG) for PC

The surface area of the outer surface is the circumference of the outside of the pipe times its length. So, if OR is the radius of the outside of the pipe and SO is the outer surface area,

SO = 2 * pi * OR * 14

Similarly, the surface area of the outer surface is SI = 2 * pi * IR *14 (if SI is the inner surface area and IR is the radius of the inside of the pipe.

The question states that the difference between the outside surface area and the inside surface area is 88 sq. cm:

SO - SI = 88 ; substituting:

(2 * pi * OR * 14) - (2 * pi * IR * 14) = 88 ; factoring

(2 * pi * 14) (OR - IR) = 88 ;dividing both sides by 2*pi*14

(OR - IR) = 88/(2 * pi) = 88/(2 * 3.14159*14) = 1.00

So, the outside radius is 1 cm more than the inside radius.

It's not clear if the volume stated is the volume of the metal in the pipe or the volume of air inside the pipe, so I will solve it both ways:

If volume of 176cc is of the air inside, the formula for this volume is 14 * pi * IR *IR

176 = 14 * 3.14159 * IR * IR ; dividing both sides by 14 * 3.14159

176/(14 * 3.14159) = IR * IR ; doing the arithmetic

4 = IR * IR ; taking the square root of both sides

sqrt(4) = IR

**IR = 2 cm**

substituting back into the first equation, the OR is 1cm more than the IR, so

**OR = 3 cm**

If volume of 176cc is of the iron in the pipe, the formula for that volume is the difference between the volume of the outside of the pipe and the volume of the inside of the pipe, or

(14 * pi * OR * OR) - (14 * pi * IR * IR) = 176 ; factoring

(14 * pi) ((OR * OR) - (IR * IR)) = 176 ; dividing both sides by 14 * pi

((OR * OR) - (IR * IR)) = 176/(14 * 3.14159) = 4

but, since OR is 1 cm more than IR (from above), we can substitute OR = IR + 1

and OR * OR = (IR + 1) * (IR + 1) = (IR*IR) + 2*IR +1

So, ((OR * OR) - (IR * IR)) = 4 becomes

(IR*IR) + 2*IR +1 - (IR*IR) = 4 ; simplifying (IRsquared - IRsquared = 0)

2*IR + 1 = 4 ; subtract 1 from both sides

2*IR = 3 ; divide both sides by 2

IR = 3/2 ;

**IR = 1.5cm**

**OR = 2.5cm**

SO = 2 * pi * OR * 14

Similarly, the surface area of the outer surface is SI = 2 * pi * IR *14 (if SI is the inner surface area and IR is the radius of the inside of the pipe.

The question states that the difference between the outside surface area and the inside surface area is 88 sq. cm:

SO - SI = 88 ; substituting:

(2 * pi * OR * 14) - (2 * pi * IR * 14) = 88 ; factoring

(2 * pi * 14) (OR - IR) = 88 ;dividing both sides by 2*pi*14

(OR - IR) = 88/(2 * pi) = 88/(2 * 3.14159*14) = 1.00

So, the outside radius is 1 cm more than the inside radius.

It's not clear if the volume stated is the volume of the metal in the pipe or the volume of air inside the pipe, so I will solve it both ways:

If volume of 176cc is of the air inside, the formula for this volume is 14 * pi * IR *IR

176 = 14 * 3.14159 * IR * IR ; dividing both sides by 14 * 3.14159

176/(14 * 3.14159) = IR * IR ; doing the arithmetic

4 = IR * IR ; taking the square root of both sides

sqrt(4) = IR

substituting back into the first equation, the OR is 1cm more than the IR, so

If volume of 176cc is of the iron in the pipe, the formula for that volume is the difference between the volume of the outside of the pipe and the volume of the inside of the pipe, or

(14 * pi * OR * OR) - (14 * pi * IR * IR) = 176 ; factoring

(14 * pi) ((OR * OR) - (IR * IR)) = 176 ; dividing both sides by 14 * pi

((OR * OR) - (IR * IR)) = 176/(14 * 3.14159) = 4

but, since OR is 1 cm more than IR (from above), we can substitute OR = IR + 1

and OR * OR = (IR + 1) * (IR + 1) = (IR*IR) + 2*IR +1

So, ((OR * OR) - (IR * IR)) = 4 becomes

(IR*IR) + 2*IR +1 - (IR*IR) = 4 ; simplifying (IRsquared - IRsquared = 0)

2*IR + 1 = 4 ; subtract 1 from both sides

2*IR = 3 ; divide both sides by 2

IR = 3/2 ;

Jan 11, 2011 | Mathsoft StudyWorks! Mathematics Deluxe...

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...

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

Hello,

There are no dedicated keys for these trigonometric functions, for the simple reason that they can be obtained from the tan, sin, and cos by a simple division.

**cotangent (x) =1/tan(x) . **Do not confuse with the arc tangent tan^(-1)

**cosecant (x)** = 1/sin(x) . Do not confuse with the arcsine sin^(-1)

**secant(x) **=1/cos(x) Do not confuse with the arccosine cos^(-10)

If you know how to use the tan, cos, and sin, with angle unit in degrees or radians, then there will not be any problem

If angle unit is degree, any number you give a trigonometric function is interpreted as degree. For instance if mode is in degree , and you calculate cos(PI) do not expect the value -1. You will have the value corresponding to the cosine of of 3.14159 degrees, namely 0.99849715

Now for you if you are interested.

If [MODE] is in degrees you can still enter angles in radians

You use the [2nd][ANGLE] [3: raised r] [ENTeR].

Here is a screen capture to show you more clearly.

The raised r is obtained by [2nd][ANGLE][3: raised r] [ENTER]

Hope it helps.

There are no dedicated keys for these trigonometric functions, for the simple reason that they can be obtained from the tan, sin, and cos by a simple division.

If you know how to use the tan, cos, and sin, with angle unit in degrees or radians, then there will not be any problem

If angle unit is degree, any number you give a trigonometric function is interpreted as degree. For instance if mode is in degree , and you calculate cos(PI) do not expect the value -1. You will have the value corresponding to the cosine of of 3.14159 degrees, namely 0.99849715

Now for you if you are interested.

If [MODE] is in degrees you can still enter angles in radians

You use the [2nd][ANGLE] [3: raised r] [ENTeR].

Here is a screen capture to show you more clearly.

The raised r is obtained by [2nd][ANGLE][3: raised r] [ENTER]

Hope it helps.

Oct 13, 2009 | Texas Instruments TI-83 Plus Calculator

-1E-13 is a very small number. When doing this kind of a problem you can regard 1E-13 as 0.

Remember that pi is an irrational number. It is only estimated on your calculator. I just played around with a TI-83 and found the following answers:

cos(pi/2) = 0

cos(2*pi + pi/2) = 0

cos(4*pi + pi/2) = 1E-13

cos(20*pi + pi/2) = -1E-13

As you know, that correct answer to each of these is 0. The calculator gives non-zero answers because some very small errors are accumulating. There is nothing wrong with your calculator.

Remember that pi is an irrational number. It is only estimated on your calculator. I just played around with a TI-83 and found the following answers:

cos(pi/2) = 0

cos(2*pi + pi/2) = 0

cos(4*pi + pi/2) = 1E-13

cos(20*pi + pi/2) = -1E-13

As you know, that correct answer to each of these is 0. The calculator gives non-zero answers because some very small errors are accumulating. There is nothing wrong with your calculator.

Nov 20, 2007 | Texas Instruments TI-83 Plus Calculator

Jan 17, 2017 | Facebook Websites

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