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[real frost]

$sin^2 (x) \cdot cos^2 (x) = \frac {1-cos (x)}{8}$

[cold belfry]

Francois

[light plover]

not always true

[real frost]

What do you mean?

@light plover

Just simplify and explain step by step

I plugged this into the graph

And it's not the same graph

[light plover]

yeah

You ask to prove it. I think you mean that those two terms are equal.

but it is not.

Do you mean solving for x?

[real frost]

Here i have the equation true

Look

https://media.discordapp.net/attachments/998271336013111396/1054416207811457204/20221219_181253.jpg

@light plover

[light plover]

you wrote cos x on the right term

It is $$\sin^2 x = \frac{1-\cos 2x}{2}$$

[cold belfry]

k12byda5h

[light plover]

and similar to the cos one

not $\cos^2 x$

[cold belfry]

k12byda5h

[real frost]

I know

I was fast

[light plover]

ah ok

[real frost]

But I did it as $cos 2(x)$

[cold belfry]

Francois

[light plover]

so what is the problem? YOu did it right.

[real frost]

I don't understand it

I just copied it from the board

And I want someone to explain it step by step

[light plover]

ok

[real frost]

If you don't mind

[light plover]

so, I guess you knwo that $\sin^2 x = \frac{1-\cos 2x}{2}$ and $\cos^2 x = \frac{1+\cos 2x}{2}$

[cold belfry]

k12byda5h

[light plover]

Thus, $\sin^2 x \cos^2 x = \frac{1-\cos 2x}{2}\cdot \frac{1+\cos 2x}{2} = \frac{1-\cos^2 2x}{4}$

[cold belfry]

k12byda5h

[light plover]

but $\cos^2 2x = \frac{1+\cos 4x}{2}$

[cold belfry]

k12byda5h

[light plover]

we substitute it back, $$\sin^2 x \cos^2 x = \frac{1-\cos^2 2x}{4} = \frac{1-\frac{1+\cos 4x}{2}}{4}=\frac{1-\cos 4x}{8}$$

[cold belfry]

k12byda5h

[real frost]

By the way how can we write theta on this bot?

[real frost]

Thank you so much

[light plover]

$\theta$

[cold belfry]

k12byda5h

[light plover]

$\Theta$

[cold belfry]

k12byda5h

[dreamy lily]

@quasi jolt has given 1 rep to @light plover