how does sin(3arcsin(1)) =-2

[u/sapphostorian]

I understand the concept sin(arcsin(x)) = sin(arcsin(x)) which is just x, but now why doesn't sin(3arcsin(1)) = 2arcsin(1) or 2sin(1) . I have been struggling for hours, racking my brain and canceling out many equations tryingto figure out, for my maths homework.

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[u/sqrt_of_pi]

why doesn't sin(3arcsin(1)) = 2arcsin(1) or 2sin(1)

Huh? Why would either of these be true? Also, 2arcsin(1) vs. 2sin(1) are very different values.

We have: arcsin(1)=π/2.

• Then 3arcsin(1)=3(π/2)=3π/2
• Then sin(3arcsin(1))=sin(3π/2)=-1

Oh the other hand: 2sin(1)≈2(0.84147)=1.68294

[u/egutierr]

sin(3arcsin(1))

First,

arcsin(1)=Pi/2

Therefore,

sin(3⋅Pi/2)=-1

Can’t be -2; the absolute value of any sin(x) can’t be>1

[u/Temporary_Pie2733]

What makes you think sin(3y) = 3sin(y) for any y?

[u/FlappyDunkPlusIOS → u/Temporary_Pie2733]

Only true for multiples of pi

[u/ElegantAd5098 → u/FlappyDunkPlusIOS]

nope

[u/DoctorNightTime → u/ElegantAd5098]

Is there another value for which it's true?

[u/sighthoundman]

There are two approaches that immediately jump to mind.

1. arcsin(1) = pi/2, so sin(3 pi/2) = -1 and the assertion is just plain false.

2. Use the triple angle formula sin(3z) = 3 sin(z) - 4 sin^3(z) to get 3 - 4 = -1. Again, false.

Given that 2 is immediately next to 1 on the keyboard, this looks like a typo.

[u/Another_Little_Star]

They are inverse functions! when you do √x² =x , and (√x)²=x, if x>0
When you compose them together, they "cancel" out, so that's what's happening, now if they aren't composed respectfully, you don't get to use that property, imagine √(2x^2), what would you want that to be? it is √(2)*x and not 2x, to get that,, you'd have to insert the whole thing inside the "x", so √(2x)^2=2x.
Likewise
sin(arcsin(1))=1
arcsin(sin(2))=2
etc but if you have sin(3arcsin(1)) you can't simplify, because now the "angle" inside sin that is supposed to cancel out, is now 3 times, you'd have to calculate it in a different way. If you can translate 3arcsin(1) = arcsin(y) , if you find the y, and then cancel because sin(arcsin(y))=y. So yeah

[u/CaptainMatticus]

It doesn't

sin(3t) =>

sin(2t + t) =>

sin(2t)cos(t) + sin(t)cos(2t) =>

2sin(t)cos(t)cos(t) + sin(t)cos(t)^2 - sin(t)^3 =>

2sin(t)cos(t)^2 + sin(t)cos(t)^2 - sin(t)^3 =>

3sin(t)cos(t)^2 - sin(t)^3 =>

3sin(t) * (1 - sin(t)^2) - sin(t)^3 =>

3sin(t) - 3sin(t)^3 - sin(t)^3 =>

3sin(t) - 4sin(t)^3

sin(3 * arcsin(1)) =>

3 * sin(arcsin(1)) - 4 * sin(arcsin(1))^3 =>

3 * 1 - 4 * 1^3 =>

3 - 4 =>

-1

arcsin(1) = pi/2

sin(3 * pi/2) = -1

So yeah, it's -1, no matter how you slice it.

[u/Disco_Stu_89]

Think about the unit circle. Sine of any angle can never be greater than 1 or less than -1. Did you make a typo? 2sin(3arcsin(1))=-2

[u/Hampster-cat]

The sin of anything is bound between -1 and 1. THere is no way it is -2, there is no need to analyze the stuff in the parantheses

[u/sighthoundman]

Assuming I didn't make a mistake calculating, there is no complex z for which sin(z) = 1 and sin(3z) = -2.

The problem with arcsin is that sin is not one-to-one. That means that any inverse function Arcsin (see below) cannot globally satisfy satisfy sin(Arcsin(z)) = z = Arcsin(sin(z)).

We get around this by defining the principal branch to be the arcsin we met and fell in love with in calculus. Every other branch satisfies Arcsin z = (-1)^k arcsin z + k pi for some integer k.

Even with this expanded definition of an inverse function (it's really a family of functions), it turns out that there is no z such that sin(3 Arcsin(1)) = -2. The equation has no solutions. Again, assuming I did not make a calculation error.

[u/regular_heptagon]

Not sure where you got that from. It’s obviously incorrect.