Graphs of trigonomentry

You have several types of graphs
Function graphs
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

You have several types of graphs
Function graphs
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

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.

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

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

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).

Hello,
That habit of TI, Casio, and Sharp to label the inverse trigonometric functions with the -1 superscript can cause confusions.

1. The inverse trigonometric functions arcosine, arcsine, and arctangent (labeled by manufacturers as cos^-1, sin^-1, and tan^-1) should not be confused with the other trigonometric functions known as secant(x) =1/cos(x), cosecant(x)=1/sin(x) and cotangent(x) = 1/tan(x).
2. To avoid errors in the use of the inverse trigonometric functions, one must be careful and set the angle unit to the one required by the problem at hand (degrees, or radians)
3. To make the trigonometric functions really functions, their range is restricted.
4. In this calculator arcosine (x) gives results between 0 and 180 degrees (if angle MODE is Degree) or between 0 and Pi radians (if angle MODE is Radian).
5. The range of results for arcsine(x) and arctangent(x) is between -90 degrees and +90 degrees (if angle MODE Degree) or -Pi/2 and Pi/2 (if angle MODE is Radian)
   

With this information you should be able to set the angle unit correctly ([MODE][Radian] or [MODE][Degree] ) and interpret the results. If you want to extend the angle to other values, use the periodicity of the trigonometric functions.

Hope it helps

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.

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.

-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.