Trig Problem, Help please.

  • Thread starter TheKracken
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  • #1
TheKracken
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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.
 

Answers and Replies

  • #2
berkeman
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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.

sin(-θ) and sin(θ) are not quite the same. What is different about them?
 
  • #3
TheKracken
356
6
The degree angle is negative in one of them, which would give you a value between 0 and negative one? I just cant seem to grasp how to simplify this.
 
  • #4
berkeman
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62,875
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The degree angle is negative in one of them, which would give you a value between 0 and negative one? I just cant 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?
 
  • #5
TheKracken
356
6
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...
 
  • #6
berkeman
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Wait this would then give me
cosθ/sinθ X -sinθ/1
yeah?

:smile: So do you see why you only got partial points?
 
  • #7
TheKracken
356
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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.
 
  • #8
TheKracken
356
6
: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 "
 
  • #9
berkeman
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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.
 
  • #10
TheKracken
356
6
OH!!!! Thank you so very much, yes that makes lots of sense to me. Alright, thank you.
 

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