Question

Verify the Pythagorean Identity.

1 + cos^2(theta) = csc^2(theta)

Answer 1

@jim_thompson5910

Answer 2

you sure it's cos ? and not cot ?

Answer 3

Yeah its cot, sorry about that

Answer 4

that's ok

Answer 5

use the fact that

cot(x) = cos(x)/sin(x)

Answer 6

Okay

Answer 7

doing so will give you


\[\large 1 + \cot^2(\theta) = \csc^2(\theta)\]


\[\large 1 + \frac{\cos^2(\theta)}{\sin^2(\theta)} = \csc^2(\theta)\]


what's your next step?

Answer 8

Uhmm... would you do something with the 1 ? Im not completly sure

Answer 9

try combine the 1 with \(\large \frac{\cos^2(\theta)}{\sin^2(\theta)} \)

Answer 10

so you need to rewrite 1 as a fraction with the denominator \(\large \sin^2(\theta)\) since that's the LCD

Answer 11

then you can combine the fractions

Answer 12

cot^2(x) + 1 ?

Answer 13

idk how you got cot

Answer 14

Hold on, Im thinking

Answer 15

alright

Answer 16

Okay so I need to have 1 with sin in the denominator?

Answer 17

sin^2 actually

Answer 18

since that's the denominator of \(\large \frac{\cos^2(\theta)}{\sin^2(\theta)} \)

Answer 19

\[\frac{ \cos ^{2}\Theta }{ \sin ^{2}1\Theta } ?\]

Answer 20

or would there be another 1 somewhere

Answer 21

if the denominator was x, then we can write 1 as x/x

Answer 22

Oh, then just put 1 as a fraction with sin^2 ?

Answer 23

yes, what do you get

Answer 24

\[\frac{ \cos ^{2}\theta }{ \frac{ \sin ^{2} }{ 1 } \theta} ? \]

Answer 25

no

Answer 26

\[\large 1 = \frac{\sin^2(\theta)}{\sin^2(\theta)}\]

Answer 27

ie...anything over itself is always 1

Answer 28

Ohh

Answer 29

so what this means is that

\[\large 1 + \frac{\cos^2(\theta)}{\sin^2(\theta)} = \csc^2(\theta)\]

turns into

\[\large \frac{\sin^2(\theta)}{\sin^2(\theta)} + \frac{\cos^2(\theta)}{\sin^2(\theta)} = \csc^2(\theta)\]

Answer 30

Okay I follow you

Answer 31

so what's next

Answer 32

I add? Or do I need to cancel out the sin^2 (theta) ?

Answer 33

you add the fractions

Answer 34

would it be sin^2+cos^2x/sin^2x = csc^2x ?

Answer 35

keep going

Answer 36

Or does the top part need to be sin^2x + cos^2x ?

Answer 37

it should be

\[\large \frac{\sin^2(\theta)+\cos^2(\theta)}{\sin^2(\theta)} = \csc^2(\theta)\]

Answer 38

because remember that

a/c + b/c = (a+b)/c

Answer 39

Okay, now do I cancel out sin^2x ?

Answer 40

no

Answer 41

what does the top simplify to?

Answer 42

\[\large \sin^2(\theta)+\cos^2(\theta) = ??\]

Answer 43

Uh.. it equals 1 ?

Answer 44

Yeah it equals 1 right?

Answer 45

yep

Answer 46

\[\large \sin^2(\theta)+\cos^2(\theta) = 1\]

Answer 47

you now have

\[\large \frac{1}{\sin^2(\theta)} = \csc^2(\theta)\]

Answer 48

Okay

Answer 49

do you see what's next?

Answer 50

No

Answer 51

what is csc

Answer 52

Idk...

Answer 53

csc(x) = 1/sin(x) by definition of cosecant

Answer 54

so we have 1/(sin^2(x)) which will turn into what?

Answer 55

csc^2x ?

Answer 56

good

Answer 57

\[\large \frac{1}{\sin^2(\theta)} = \csc^2(\theta)\]

turns into

\[\large \csc^2(\theta) = \csc^2(\theta)\]

Answer 58

the two sides of the last equation are identical, so we have confirmed the identity

Answer 59

Here are all the steps in one post


\[\large 1 + \cot^2(\theta) = \csc^2(\theta)\]

\[\large 1 + \frac{\cos^2(\theta)}{\sin^2(\theta)} = \csc^2(\theta)\]

\[\large \frac{\sin^2(\theta)}{\sin^2(\theta)} + \frac{\cos^2(\theta)}{\sin^2(\theta)} = \csc^2(\theta)\]

\[\large \frac{\sin^2(\theta)+\cos^2(\theta)}{\sin^2(\theta)} = \csc^2(\theta)\]

\[\large \frac{1}{\sin^2(\theta)} = \csc^2(\theta)\]

\[\large \csc^2(\theta) = \csc^2(\theta)\]

the two sides of the last equation are identical, so we have confirmed the identity

Answer 60

Okay thank you so much!

Answer 61

you're welcome