Trig Problem: Simplifying cot(θ)sin(-θ) - Homework Help

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Homework Help Overview

The discussion revolves around simplifying the expression cot(θ)sin(-θ) into a single trigonometric function of positive θ. Participants are exploring the properties of trigonometric functions, particularly focusing on the behavior of sine with negative angles.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to rewrite cot(θ) in terms of sine and cosine, questioning the relationship between sin(θ) and sin(-θ). There is a discussion about the implications of sine being an odd function and how that affects the simplification process.

Discussion Status

There is an ongoing exploration of the properties of sine and cosine functions, with some participants recognizing the need to account for the negative sign in the simplification. Guidance has been provided regarding the nature of odd and even functions, and the requirement to express the final answer in terms of positive θ is acknowledged.

Contextual Notes

Participants are grappling with the distinction between sin(θ) and sin(-θ), and how this affects their simplification efforts. The requirement to express the result as a single trigonometric function of positive θ is emphasized.

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Homework Statement



Simplify cot(θ)sin(-θ) so that it is a single trig function of positive θ

Homework Equations




The Attempt at a Solution


I changed cot(θ) to
cos(θ)/sin(θ) X Sin(-θ)/1

I stated that sin(-θ) and sin(θ) were the same and then cancled out the sin(θ) and sin (-θ) to get cos(θ)/1 or Cos(θ)


I feel like this is right but I only got half the points I was supposed to get, not quite sure what I did wrong.
 
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TheKracken said:

Homework Statement



Simplify cot(θ)sin(-θ) so that it is a single trig function of positive θ

Homework Equations




The Attempt at a Solution


I changed cot(θ) to
cos(θ)/sin(θ) X Sin(-θ)/1

I stated that sin(-θ) and sin(θ) were the same and then cancled out the sin(θ) and sin (-θ) to get cos(θ)/1 or Cos(θ)


I feel like this is right but I only got half the points I was supposed to get, not quite sure what I did wrong.

sin(-θ) and sin(θ) are not quite the same. What is different about them?
 
The degree angle is negative in one of them, which would give you a value between 0 and negative one? I just can't seem to grasp how to simplify this.
 
TheKracken said:
The degree angle is negative in one of them, which would give you a value between 0 and negative one? I just can't seem to grasp how to simplify this.

Is the sin() function even or odd? Look at a plot of sin(x) over -360 degrees to +360 degrees. What does it do around x=0?
 
berkeman said:
sin(-θ) and sin(θ) are not quite the same. What is different about them?
Wait this would then give me
cosθ/sinθ X -sinθ/1
yeah?
Or...maybe I am on the wrong track...
 
TheKracken said:
Wait this would then give me
cosθ/sinθ X -sinθ/1
yeah?

:smile: So do you see why you only got partial points?
 
Sin is a odd function, though I do not understand why. Or what that quite means, I just found it in my notes. except maybe it is opposite of it such as it is the same value except with a negative sign on it.
 
berkeman said:
:smile: So do you see why you only got partial points?

Ok well if this is right then the sinθ would cancel out and wouldn't the negative still be there? As in it would then simplify to -cosθ which is not what it asked for. It asked that "so that it is a single trig function of positive θ "
 
TheKracken said:
Ok well if this is right then the sinθ would cancel out and wouldn't the negative still be there? As in it would then simplify to -cosθ which is not what it asked for. It asked that "so that it is a single trig function of positive θ "

For odd functions, f(-x) = -f(x). For even functions, f(-x) = f(x). Can you see how sin() is an odd function, and cos() is an even function?

The question is asking for the answer to be expressed in terms of positive θ. -cos(θ) is a function of positive θ.
 
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OH! Thank you so very much, yes that makes lots of sense to me. Alright, thank you.
 

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