How Do You Simplify Complex Trigonometric Expressions?

[stuck]

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.

 

[rock.freak667]

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.

sint+(cott)(cost)

= sint + \frac{cos^2 t}{sint} = \frac{?+??}{sint}secx*sin^2x*cosecx=\frac{1}{cosx}\times sin^2x \times \frac{1}{sinx}

sin^2x \times \frac{1}{sinx} gives what? and then that times \frac{1}{cosx} gives what?

 

[CompuChip]

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)

 

[BrendanH]

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.

 

[stuck]

sint+(cott)(cost)

= sint + \frac{cos^2 t}{sint} = \frac{?+??}{sint}


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

sin^2x \times \frac{1}{sinx} gives what? and then that times \frac{1}{cosx} gives what?

would it be:

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

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

 

[Tedjn]

You are correct, but rock.freak is showing you that you can simplify sin²x/sinx. Hint: what is y²/y, or 5²/5?