# Trig Identities

Simplify in terms of cosine and sine only.

$$tan^2x - {csc^2x\over cot^2x}$$

From here, I assume you can flip the fraction and make it

$${tan^2x\over sin^2x}$$

next, i reduce it to:

$$(sec^2x-1) - ({sec^2x -1\over 1 - cos^2x})$$

and anyway..im lost; i dont know where to stop and how simplified this can go..

thanks

that page didn't seem to load for me.

Also, haha, i thought this was beginner stuff..

youre right on the grade level, but we jumped right into this, so i figured it was the first level of trig

first of all

Dont EXPAND using the 1+ crap if you can solve by just making it into a fraction

first of all what is tan in fraction form, a ratio of what two trig functions

secondly what are csc and cot in fraction form

simplify from there

dekoi
jai6638 said:
dude back in India, i did this in 9th grade..

In Canada, we have only done the very basics of trig identities.

KEep in mind I'm not talking about trigonemtric ratios and such.

I am in discrete math and physics in gr12 right now, and i still have found little use for trig identities. :S

I guess it really depends on where you are. I'm in Canada and I did this in grade 10.

dekoi said:
In Canada, we have only done the very basics of trig identities.

KEep in mind I'm not talking about trigonemtric ratios and such.

I am in discrete math and physics in gr12 right now, and i still have found little use for trig identities. :S

trig identities will come in use when you reach university and learn integration (the opposite of differentiation) and then you'll need to use trig substitution which needs you to know how to solve trig identities.

Remember eventually something you learnt will come into use

stunner5000pt said:
first of all

Dont EXPAND using the 1+ crap if you can solve by just making it into a fraction

first of all what is tan in fraction form, a ratio of what two trig functions

secondly what are csc and cot in fraction form

simplify from there

ok $$tanx={sinx\over cosx}$$

and $$cscx={1\over sinx}, cot={1\over tanx}$$

so then im seeing..

$${sin^2x\over cos^2x} - {tan^2x\over sin^2x}$$

since it has to be in terms of cos and sin only, will

$${sin^2x\over cos^2x} - {sin^2x\over cos^2xsin^2x}$$ work?

you didnt do the second part properly

$$\frac{csc^2 x}{cot^2 x} = \frac{\frac{1}{sin^2 x}}{\frac{cos^2 x}{sin^2 x}}$$

now solve using the fact that

$$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}$$

further there is one more identity youll have to use after you simplify these expressions!

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robphy
Homework Helper
Gold Member
You can still simplify it... a lot!

Try plugging in various values for x into your starting expression.

(Try expressing it in terms of sine alone or cosine alone. You may be amused to see the result.)

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stunner5000pt said:
you didnt do the second part properly

$$\frac{csc^2 x}{cot^2 x} = \frac{\frac{1}{sin^2 x}}{\frac{cos^2 x}{sin^2 x}}$$

now solve using the fact that

$$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}$$

further there is one more identity youll have to use after you simplify these expressions!

isnt that what i did??

soccerjayl said:
isnt that what i did??

this is what you did

$${sin^2x\over cos^2x} - {sin^2x\over cos^2xsin^2x}$$

now what can be cancelled out?

what other identity canbe used after the cancellation?

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stunner5000pt said:
this is what you did

$${sin^2x\over cos^2x} - {sin^2x\over cos^2xsin^2x}$$

this is what im telling you t od

$$\frac{csc^2 x}{cot^2 x} = \frac{\frac{1}{sin^2 x}}{\frac{cos^2 x}{sin^2 x}}$$
using
$$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}$$

here a = 1
b = sin^2 x
c = cos^2 x
d = sin^x x

$${sin^2x\over cos^2x} - {sin^2x\over cos^2xsin^2x}$$
is what i got for the entire solution

the part you are telling me to do is the segment i have in the equation above, or $${sin^2x\over cos^2xsin^2x}$$

soccerjayl said:
$${sin^2x\over cos^2x} - {sin^2x\over cos^2xsin^2x}$$
is what i got for the entire solution

the part you are telling me to do is the segment i have in the equation above, or $${sin^2x\over cos^2xsin^2x}$$

stunner5000pt said:
this is what you did

$${sin^2x\over cos^2x} - {sin^2x\over cos^2xsin^2x}$$

now what can be cancelled out?

what other identity canbe used after the cancellation?

ok i was just checkin...

yes $$sin^2x$$ cancels out..

so u have

$${sin^2x -1\over cos^2x}$$

$$sin^2x -1 = -cos^2x$$

so $${-cos^2x\over cos^2x}=-1$$

good?

awesome now you got it
remember dont use the 1+ identities first, try and make the ratios first and then see what happens

but in some cases the opposite may work better

stick with one approach if one takes longer then abandon it for then and return (if on test)

ok thanks a lot!

i have another question if u could just lead me in the right way..

The problem is to use a grapher to see if it is an identity; if not, make a counter example.

$$sin^4xcos^2x = cos^2x + sin^4x - cos^6x$$

when i graphed them, i found there was not an identity, or the same function..

im not exactly sure how i make a counter example; can u lead me in the right way or start me off?

soccerjayl said:
ok thanks a lot!

i have another question if u could just lead me in the right way..

The problem is to use a grapher to see if it is an identity; if not, make a counter example.

$$sin^4xcos^2x = cos^2x + sin^4x - cos^6x$$

when i graphed them, i found there was not an identity, or the same function..

im not exactly sure how i make a counter example; can u lead me in the right way or start me off?

i'm not sure what the counter example is supposed to mean? are you supposed to make an identity out of the function sin^4 x cos^2 x? if so you could do that several ways. just substitute the cos^2 or sin^2 as appropriate to express the product as a sum.

if not i've got no clue on what a 'counter example' is. maybe doing the same for the right side, but i think that would really suck eggs.

oh someone on S.O.S. Mathematics Forum told me that a counter example is when you just plug in a value for x and prove that the sides are not equal..just in case u wanted to know

thanks for the help

ahh the horor of grade 11 math X_X