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Hello all,
I am struggling with this question:
7. What is 1/cot(x) in terms of sine?
Here is what I have so far, please tell me if I'm starting off incorrectly!
1/cot(x) = tan
tan = sin/cos
1/cot = sin/cos
And then I'm not sure where to go from here. I am supposed to be using the basic identities listed below to help me express 1/cot in terms of sine:
cscΘ = 1/sinΘ
secΘ = 1/cosΘ
cotΘ = 1/tanΘ
sinΘ = 1/cscΘ
cosΘ = 1/secΘ
tanΘ = 1/cotΘ
sinΘ * cscΘ = 1
cosΘ * secΘ = 1
tanΘ * cotΘ = 1
Ratio Identities
sine = opposite / hypotenuse
cosine = adjacent / hypotenuse
tangent = opposite / adjacent
Here is an example of one problem solved correctly in the way my teacher wants me to do it:
11. Express tangent in terms of cosine.
secΘ = 1/cosΘ
->Tan^2Θ + 1 = sec^2Θ
->Tan^2Θ = sec^2Θ – 1
->Tan^2Θ = (secΘ +1) (secΘ -1)
-> Tan^2Θ = (1/cosΘ +1) (1/cosΘ -1)
-> Tan^2Θ = ([1/cosΘ] +[cos Θ /cos Θ]) ([1/cosΘ] –[cos Θ/cos Θ])
-> Tan^2Θ = (1+cosΘ /cosΘ) (1 -cosΘ /cosΘ)
-> Tan^2Θ = (1-cos^2Θ /cosΘ)
-> √Tan^2Θ = √(1-cos^2Θ /cosΘ)
-> tanΘ = √(1-cos^2Θ) /cosΘ
Tangent in terms of cosine is tanΘ = √(1-cos^2Θ) /cosΘ
I hope that someone could give me some direction on where to go from here. We did this earlier in the year and now I am having to solve this problem as part of a review and I just can't remember how to do it. Thanks everyone
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hi camicat
In Q11 you have made use of the trig. identity:
There are several versions of this. They come directly from Pythagoras Theorem. If you scroll down this page:
https://www.mathsisfun.com/algebra/trig … ities.html
you'll find them just after the place where the theorem is first mentioned.
So from 1/cot = sin/cos you can express cos in terms of sin and you're done.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi Bob,
I think I understand! Here is what I did:
1/cot = sin/cos
sin²θ + cos²θ = 1 (a trig identity I learned in class)
-> cos²θ = 1 - sin²θ
-> cosθ = √(1 - sin²θ)
So therefore
sin/cos = √(1 - sin²θ)
So,
1/cot = √(1 - sin²θ)
Did I do this right?
I also have another question that I am confused about.
14. Find the principal value of Arctan (-1.44) to the nearest minute.
I have NO clue where to start on this, I don't even know what a principal value is. Could you maybe show me an example of how to solve a question similar to this so that I see the steps of how to find a principal value?
Thank you!
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hi camicat
You had a fraction
.....(1)and then it's gone!
......(2)is correct so substitute (2) into (1) and you will still have a fraction.
Trig functions are periodic so that, for example sin(370) = sin (360 + 10) = sin(10)
So when working with inverse trig, such as arctan, there are many angles that would all give rise to tan(angle) = -1.44
So mathematicians have introduced the idea of principal values to make the inverse into proper functions ie. a single input has just one output.
The domain of the principal values is given here:
https://www.mathsisfun.com/algebra/trig … s-tan.html
You'll have to scroll almost to the bottom of the page but you'll find graphs that show the inverse trig functions (If a 'mapping' has multiple values it isn't called a function).
When you use a calculator, as it has to give a single value, it will have been set up to give the principal value. For arctangent the domain is -90 to + 90.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Did you use a calculator to calculate?
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Yes.
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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