Surdic Form: Solving for Sin, Cos, and Tan of 60', 30', and 45' Degrees

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Homework Help Overview

The discussion revolves around finding the sine, cosine, and tangent values for the angles of 30°, 45°, and 60° in surdic form. The original poster expresses confusion regarding the question's requirements and the concept of surds, indicating a lack of recent practice in mathematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to understand the definitions of sine, cosine, and tangent, and how to represent these values in surdic form. There are attempts to draw triangles corresponding to the angles to visualize the relationships between the sides and the trigonometric functions. Questions arise about the correct representation of values and the simplification of expressions, particularly regarding radicals in denominators.

Discussion Status

Participants are actively engaging with the problem, sharing their attempts at drawing triangles and calculating values. Some guidance has been provided regarding the use of unit triangles and the importance of avoiding square roots in the denominator. There is an ongoing exploration of how to properly format the answers in a table without decimal points.

Contextual Notes

There is mention of homework constraints and the need to present answers in a specific format, which includes avoiding decimal representations and ensuring that radicals are not present in denominators.

Venito
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Homework Statement


Hey,

Right my problem is this. I have not been doing maths for over a year and a bit. And I am very rusty.

So I am stuck on what this question is meaning, and what to do.

Find the sin, cos, and tan of 60', 30' and 45'. [' is degree.]
Leave your answer in surdic form. [Root 2 is called a surd.]

Homework Equations


The words in bold are the question from my paper I have been sent.
Now this is what I am stuck on. What in heck are they on about? And what are they meaning.

The Attempt at a Solution


I tried it and thought they were meaning that the values for the sin cos tan, were root3/2, root3/2, 1. But I am far from sure if it the answer or not.


Cheers,

venito
 
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Venito said:

Homework Statement


Hey,

Right my problem is this. I have not been doing maths for over a year and a bit. And I am very rusty.

So I am stuck on what this question is meaning, and what to do.

Find the sin, cos, and tan of 60', 30' and 45'. [' is degree.]
Leave your answer in surdic form. [Root 2 is called a surd.]

Homework Equations


The words in bold are the question from my paper I have been sent.
Now this is what I am stuck on. What in heck are they on about? And what are they meaning.

The Attempt at a Solution


I tried it and thought they were meaning that the values for the sin cos tan, were root3/2, root3/2, 1. But I am far from sure if it the answer or not.


Cheers,

venito

Thread moved from calc to pre-calc math forum.

Welcome to the PF. To put the answers in a form that has the square roots in it, remember the Pythagorean theorem and draw triangles with those degree angles in them. For example, the right triangle with 30' and 60' angles in it has side lengths of what?
 
Usually, trig students memorize these values. Remember to not have a square root in the denominator!
 
berkeman said:
Thread moved from calc to pre-calc math forum.

Thanks did not know were to place it.

Welcome to the PF. To put the answers in a form that has the square roots in it, remember the Pythagorean theorem and draw triangles with those degree angles in them. For example, the right triangle with 30' and 60' angles in it has side lengths of what?

Thanks mate,

Right so I am trying it now. So it is a triangle with A C B. A is on the left hand side, and is the tip of the triangle. C and B are the bottom of it, which is too the right.

Length of A-C is 4.8. C-B is 3.9.

So now I guess I do this. I have drawn a triangle, and added the values of 30' and 60 to it. Which has lengths of 4.8 and 2.9. And will do another triangle for the 45' angle. And I guess with the 45' and angel has the same lengths?

Right so far I just have a triangle, with numbers. Is there a formula or something?
 
Venito said:
Thanks mate,

Right so I am trying it now. So it is a triangle with A C B. A is on the left hand side, and is the tip of the triangle. C and B are the bottom of it, which is too the right.

Length of A-C is 4.8. C-B is 3.9.

So now I guess I do this. I have drawn a triangle, and added the values of 30' and 60 to it. Which has lengths of 4.8 and 2.9. And will do another triangle for the 45' angle. And I guess with the 45' and angel has the same lengths?

Right so far I just have a triangle, with numbers. Is there a formula or something?

Try drawing your triangles with unit values for the sides. Like, the simplest triangle with 90', 45' and 45' angles, has sides of 1, 1, ____. And the simplest right triangle with angles of 30' and 60' has sides of length 1, ____, ____.
 
berkeman said:
Try drawing your triangles with unit values for the sides. Like, the simplest triangle with 90', 45' and 45' angles, has sides of 1, 1, ____. And the simplest right triangle with angles of 30' and 60' has sides of length 1, ____, ____.

Right so to answer this.

Like, the simplest triangle with 90', 45' and 45' angles, has sides of 1, 1, __root 2__. And the simplest right triangle with angles of 30' and 60' has sides of length 1, __root 3__, _ _2__.

Okay now I have the triangles like that. So do I now plug in the other values I have or do they get some sort of multiplication?

mac
 
Just use the definitions of the sin, cos and tan functions, and those triangles to fill out the table of answers. And remember Pinu7's comment about getting rid of any roots in denominators (those are usually not considered simplified).
 
berkeman said:
Just use the definitions of the sin, cos and tan functions, and those triangles to fill out the table of answers. And remember Pinu7's comment about getting rid of any roots in denominators (those are usually not considered simplified).

Right. I will let you know how I go.

Thanks.
 
Venito said:
Right. I will let you know how I go.

Thanks.

Right so I played with it.

And have come up with this. 1/2, 0.707106781, √3/2. √3/2, 0.707106781, 1/2. 0.577350269, 1, √3.

Which is the memorized functions or answers to sin cos tan 30 45 60. But with no square roots on the denominator.

So in other words they just want all 9 forms of the above numbers?
 
  • #10
Venito said:
Right so I played with it.

And have come up with this. 1/2, 0.707106781, √3/2. √3/2, 0.707106781, 1/2. 0.577350269, 1, √3.

Which is the memorized functions or answers to sin cos tan 30 45 60. But with no square roots on the denominator.

So in other words they just want all 9 forms of the above numbers?

They want a 3x3 table with the answers... put 30' 45' 60' across the top as colunm headings, and sin, cos, tan as row labels. Fill in each of the 9 boxes with an answer that does not have a decimal point in it.
 
  • #11
berkeman said:
They want a 3x3 table with the answers... put 30' 45' 60' across the top as colunm headings, and sin, cos, tan as row labels. Fill in each of the 9 boxes with an answer that does not have a decimal point in it.

Right so that means I am correct. So what is Pinu7 meaning then about the roots in the denominator?

Should they be taken out or not, from the box of nine I have, or what I mean the results from sin 30 45 60 cos 30 45 60 tan 30 45 60.
 
  • #12
Venito said:
Right so that means I am correct. So what is Pinu7 meaning then about the roots in the denominator?

Should they be taken out or not, from the box of nine I have, or what I mean the results from sin 30 45 60 cos 30 45 60 tan 30 45 60.

I didn't work out the table, so I don't know offhand if you will run into the situation or not. But generally in algebra, when they ask you to simplify an expression, and you end up something like:

\frac{1}{\sqrt{2}}

then you need to get rid of the radical in the denominator to get full credit for the simplification. What could you multiply both the numerator and denominator by, in order to get rid of the square root in the denominator?
 
  • #13
berkeman said:
I didn't work out the table, so I don't know offhand if you will run into the situation or not. But generally in algebra, when they ask you to simplify an expression, and you end up something like:

\frac{1}{\sqrt{2}}

then you need to get rid of the radical in the denominator to get full credit for the simplification. What could you multiply both the numerator and denominator by, in order to get rid of the square root in the denominator?

So I would then take the problem.

\frac{1}{\sqrt{2}}

Then multiply top and bottom by √2/√2. Which would then give me a √2.
Is that right?
 
  • #14
Venito said:
So I would then take the problem.

\frac{1}{\sqrt{2}}

Then multiply top and bottom by √2/√2. Which would then give me a √2.
Is that right?

Not quite. Be sure to multiply both top and bottom by √2...
 
  • #15
berkeman said:
Not quite. Be sure to multiply both top and bottom by √2...

What a idiot. It is √2/2.

Which is now correct.
 
  • #16
Venito said:
What a idiot. It is √2/2.

Which is now correct.

Correct-amundo!
 

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