# Need some help: Trig Identities

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.

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rock.freak667
Homework Helper
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
Homework Helper
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)

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.

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)?

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