help with inverse trig problem

[lucid bramble]

how do I solve this?

I know that if the 2 wasnt in front of the arcsin, I would just solve it by looking at the cos in a 5-12-13 triangle, but I don't know how that 2 changes things

[muted wren]

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[magic flint]

well the 2 just multiples whatever arcsin(5/13) is by 2

although the way to solve this is probably some random trig identity

[magic mirage]

There is a trig identity for double angles that might help you here.
$$\cos(2x)=1-(\sin x)^2$$

[halcyon quest]

PyroManiac2653

[magic flint]

case in point :)

[magic mirage]

I'm not to great with trig to be honest. But I remembered there were formulas for double angles so I googled them 😋

[lucid bramble]

I'll try this thanks

[marble viper]

$\cos(2x) = 1 - 2(\sin x)^2$

[halcyon quest]

Yojda

[magic mirage]

Oh, that's right, thank you!
@lucid bramble , see this correction.

[open osprey]

@magic mirage has given 1 rep to @lucid bramble @marble viper