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Need some help: Trig Identities

  • Thread starter stuck
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  • #1
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Theres a few...
Write each expression as a single trigonometric ratio or as the number 1.


1) sint+(cott)(cost)

2) (sec x)(sin^2x)(csc x)


For number one I went like this:
sin t + ((1/cot)(cos/1))
sin t + (cos t/cot t)
sin t + (cos t/1)( sin t/cos x)
(sin t cos t)/1 + (sin t cot t)/1

But then I get stuck.


For number 2 I went like this:
(secx)(sin^2x)(cscx)
(1/cosx)(sin^2x/1)(1/sinx)
sin^2x/(cosx)(sinx)

But then I got stuck again. :confused:
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
Theres a few...
Write each expression as a single trigonometric ratio or as the number 1.


1) sint+(cott)(cost)

2) (sec x)(sin^2x)(csc x)


For number one I went like this:
sin t + ((1/cot)(cos/1))
sin t + (cos t/cot t)
sin t + (cos t/1)( sin t/cos x)
(sin t cos t)/1 + (sin t cot t)/1

But then I get stuck.


For number 2 I went like this:
(secx)(sin^2x)(cscx)
(1/cosx)(sin^2x/1)(1/sinx)
sin^2x/(cosx)(sinx)

But then I got stuck again. :confused:
sint+(cott)(cost)

[tex]= sint + \frac{cos^2 t}{sint} = \frac{?+??}{sint}[/tex]


[tex]secx*sin^2x*cosecx=\frac{1}{cosx}\times sin^2x \times \frac{1}{sinx}[/tex]

[itex]sin^2x \times \frac{1}{sinx}[/itex] gives what? and then that times [itex]\frac{1}{cosx}[/itex] gives what?
 
  • #3
CompuChip
Science Advisor
Homework Helper
4,302
47
Err, what is the question? If I read your notation correctly, first you say it is
sin t+ (cot t)(cos t)
and then you proceed to calculate
sin t + (1/cot t)(cos t)
 
  • #4
63
0
Yes, the above statements are right. Your mistake lies in the fact that you mean to write 1/ tan t , and not 1/cott to represent cott.
 
  • #5
6
0
sint+(cott)(cost)

[tex]= sint + \frac{cos^2 t}{sint} = \frac{?+??}{sint}[/tex]


[tex]secx*sin^2x*cosecx=\frac{1}{cosx}\times sin^2x \times \frac{1}{sinx}[/tex]

[itex]sin^2x \times \frac{1}{sinx}[/itex] gives what? and then that times [itex]\frac{1}{cosx}[/itex] gives what?
would it be:

(sin^2x) x (1/sinx) = sin^2x/sinx

(sin^2x/sinx) x (1/cosx) = (sin^2x)/(sinx cosx)?
 
  • #6
737
0
You are correct, but rock.freak is showing you that you can simplify sin2x/sinx. Hint: what is y2/y, or 52/5?
 

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