proving trig identities

[visual grove]

Is it $\frac{1}{\csc{x}}-\sin{x}$

[little lark]

Ephesians 2:8-9

[visual grove]

Or $\frac{1}{\csc{x}-\sin{x}}$

[little lark]

Ephesians 2:8-9

[visual grove]

What have you tried so far?

Where are you struggling

$\frac{1}{\frac{1}{\sin{x}}-\sin{x}}$

[little lark]

Ephesians 2:8-9

[visual grove]

So you are here correct?

Lets say you had $\frac12 -2$

[little lark]

Ephesians 2:8-9

[visual grove]

How do we find the common denominator

That's fine yeah. Then what?

$\frac12 - \frac21$

[little lark]

Ephesians 2:8-9

[visual grove]

$\frac12 - \frac42$

[little lark]

Ephesians 2:8-9

[visual grove]

Then we can subtract them and so on

Same thing here, just with sinx

$\frac{1}{\sin{x}}-\sin{x}$

[little lark]

Ephesians 2:8-9

[visual grove]

You can put sinx/1

$\frac{1}{\sin{x}}-\frac{\sin{x}}{1}$

[little lark]

Ephesians 2:8-9

[visual grove]

Then we multiply the second one by what?

Great

$\frac{1}{\sin{x}}-\frac{\sin^2{x}}{\sin{x}}$

[little lark]

Ephesians 2:8-9

[visual grove]

They have the same denominator now

So we can write it as a single fraction

$\frac{1-\sin^2{x}}{\sin{x}}$

[little lark]

Ephesians 2:8-9

[visual grove]

Now all of this was actually our denominator

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Yeah from here

$\frac{1}{\frac{1-\sin^2{x}}{\sin{x}}}$

[little lark]

Ephesians 2:8-9

[visual grove]

What next?

That's true

$\frac{1}{\frac{\cos^2{x}}{\sin{x}}}$

[little lark]

Ephesians 2:8-9

[visual grove]

$\frac{\sin{x}}{\cos^2{x}}$

[little lark]

Ephesians 2:8-9

[visual grove]

Now what

Sure it is, we just gotta manipulate it a little to fit our result

$\frac{\sin{x}}{\cos{x}\cos{x}}$

[little lark]

Ephesians 2:8-9

[visual grove]

This is still true

Now I will split it into two fractions

Almost

We have cosx in the denominator, fix the first one

Nice

And those two are multiplied

This leads to your result and the identity is proven

The hardest part about these kinds of problems are those little tricks at the end

Where I split it into two fractions

I think with everything else you're doing great, you managed to lead me up to that point

And you sort of knew we were almost there but werent sure how to get there

1×sinx

Gets multiplied

And it's still sinx

No need to

If we have $\frac12 × \frac34$

[little lark]

Ephesians 2:8-9

[visual grove]

We dont need to find the common denominator

We multiply the numerators

And the denominators

Or if you had $3×\frac{1}{\sin{x}}$

[little lark]

Ephesians 2:8-9

[visual grove]

Its the same as $\frac31×\frac{1}{\sin{x}}$

[little lark]

Ephesians 2:8-9

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$\frac{3×1}{1×\sin{x}}$

[little lark]

Ephesians 2:8-9

[visual grove]

Which is 3/sinx

So we only need to find the common denominator when adding or subtracting fractions

Multiplication and division do not require it

Great

Sure

Np, ping me when you find it

Ok

$f(x) = \sqrt{2x^2+1}$

Is it that or is the 1 under the square root as well?

[little lark]

Ephesians 2:8-9

[visual grove]

How do you find the inverse? Do you know?

$x = \sqrt{2y^2+1}$

[little lark]

Ephesians 2:8-9

[visual grove]

And now we try and work out y

It's under the root so we cant do that

We can get rid of the root first though

Correct

$x^2 = (\sqrt{2y^2+1})^2$

[little lark]

Ephesians 2:8-9

[visual grove]

On the right, the root and square cancel out

$x^2 = 2y^2+1$

[little lark]

Ephesians 2:8-9

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$\frac{x^2 -1}{2} = y^2$

[little lark]

Ephesians 2:8-9

[visual grove]

And now to just get y, we can do a square root on both sides

$f^{-1}(x)=y= \sqrt{\frac{x^2 -1}{2}}$

Which is our inverse function

[little lark]

Ephesians 2:8-9

[visual grove]

$f\bigg(\sqrt{\frac{x^2 -1}{2}}\bigg)$

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Ephesians 2:8-9

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So in our original function we will put f^-1 instead of x

Almost, it's all still under another square root

[visual grove]

All under this big one

$\sqrt{2 \bigg(\sqrt{\frac{x^2 -1}{2}}\bigg)^2+1}$

There we go

Hard on phone sorry

Also it was +1 mb

[little lark]

Ephesians 2:8-9

[visual grove]

There we go

f(f inverse(x)) should always be x

So this problem can be worked out

[quiet vortex]

except if the range of f-1(x) is NOT the same as domain of f(x)

[visual grove]

Anyways, from here the inner root and square will cancel out

Square is gone

Canceled out with the root

$\sqrt{2\frac{x^2 -1}{2}+1}$

[little lark]

Ephesians 2:8-9

[visual grove]

You have $\sqrt{a^2}$

[little lark]

Ephesians 2:8-9

[quiet vortex]

im not sure if that's multiplying or a mixed fraction

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Doesnt matter what a is

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Multiplication

[quiet vortex]

isnt that |a|

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Yep

[quiet vortex]

also i think there's something missing here

if we squared the square root

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You go ahead and help them then

[quiet vortex]

since the results of a square root is always non-negative

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Start from the beginning of the problem

[quiet vortex]

x should be ≥ 0

and thus the result of |x| we got at the end

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Here

[quiet vortex]

it's just x

[quiet vortex]

no you already did the most of it and if i did it again it just repeating from what you have done

no you need to multiply the 2 into fraction first

[visual grove]

The 2s gotta cancel out first yeah

[quiet vortex]

and you divide 2 to both numerator and denominator you will get?

yeah it should

because remember they both on the same fraction

no youd get x²-1

then there's+1 outside

yeah

yep

not really √(x^2) is |x|

but

the x is always positive or 0

so just x

it's called absolute value

it just turn negative into postive

like |-5| = 5

|3| = 3

|0| = 0

but remember |-x| = |x| because x is unknown

you cant really determine x to be positive or negative if they do not given to be

yeah it is because i had learnt things about absolute value before ive learnt about functions

idk why he skipped it

wow i think you may have skipped a lot of some foundations

do you know anything about set like domain or range?

ok cool

but i think you may need to find a resource to learn about absolute value tho
it's important to know this
because there might be a question that you need to invert a function that have absolute function in it

i understand that
trig is hard for me too

[narrow mortar]

@smoky gale

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