Inverse trig question (1 Viewer)

hs17 · Jan 22, 2021

Solve the following equation

sin^-1 (x) - cos^-1 (x) = sin^-1 (3x+1)

Qeru · Jan 22, 2021

hs17 said:
Solve the following equation

sin^-1 (x) - cos^-1 (x) = sin^-1 (3x+1)

Its always a good idea to state the domain for x before solving so: $-1 \leq x \leq 1$ and $\frac{-2}{3}\leq x \leq 0$ .

Now take the sin of both sides to obtain:
$\sin(\sin^{-1} (x) - \cos^{-1} (x))=\sin{(\sin^{-1} (3x+1))$

$\sin(\sin^{-1} (x))\cos(\cos^{-1} (x)) - \sin(\cos^{-1} (x))\cos(\sin^{-1} (x))=\sin{(\sin^{-1} (3x+1)) \quad \text{By the double angle formula}$

Note that: $\sin{\alpha}=\sqrt{1-\cos^2{\alpha}}$ and $\cos{\alpha}=\sqrt{1-\sin^2{\alpha}}$ by the pythagorean identities, we only have to take the positive case here since $\sin(\cos^{-1} (x))=\cos(\sin^{-1} (x))\geq0$

Therefore:
$\sin(\sin^{-1} (x))\cos(\cos^{-1} (x)) - \sqrt{1-\cos^2\cos^{-1}(x)}\sqrt{1-\sin^2\sin^{-1}(x)}=\sin{(\sin^{-1} (3x+1))$

$x^2 - (\sqrt{1-x^2})^2= 3x+1 \quad \text{Using the fact that} \quad f(f^{-1}(x))=x$

$x^2 - (1-x^2)= 3x+1 \quad \text{Since} \quad 1-x^2\geq0 \quad \text{in the domain}$

$2x^2-3x-2=0$

$(2x+1)(x-2)=0$

$\therefore x=-\frac{1}{2},2$

But $x \leq 0$ so the only solution is: $x=-\frac{1}{2}$

idkkdi · Jan 22, 2021

Qeru said:
Its always a good idea to state the domain for x before solving so: $-1 \leq x \leq 1$ and $\frac{-2}{3}\leq x \leq 0$ .

Now take the sin of both sides to obtain:
$\sin(\sin^{-1} (x) - \cos^{-1} (x))=\sin{(\sin^{-1} (3x+1))$

$\sin(\sin^{-1} (x))\cos(\cos^{-1} (x)) - \sin(\cos^{-1} (x))\cos(\sin^{-1} (x))=\sin{(\sin^{-1} (3x+1)) \quad \text{By the double angle formula}$

Note that: $\sin{\alpha}=\sqrt{1-\cos^2{\alpha}}$ and $\cos{\alpha}=\sqrt{1-\sin^2{\alpha}}$ by the pythagorean identities, we only have to take the positive case here since $\sin(\cos^{-1} (x))=\cos(\sin^{-1} (x))\geq0$

Therefore:
$\sin(\sin^{-1} (x))\cos(\cos^{-1} (x)) - \sqrt{1-\cos^2\cos^{-1}(x)}\sqrt{1-\sin^2\sin^{-1}(x)}=\sin{(\sin^{-1} (3x+1))$

$x^2 - (\sqrt{1-x^2})^2= 3x+1 \quad \text{Using the fact that} \quad f(f^{-1}(x))=x$

$x^2 - (1-x^2)= 3x+1 \quad \text{Since} \quad 1-x^2\geq0 \quad \text{in the domain}$

$2x^2-3x-2=0$

$(2x+1)(x-2)=0$

$\therefore x=-\frac{1}{2},2$

But $x \leq 0$ so the only solution is: $x=-\frac{1}{2}$

In case anyone does not see through why this part of the step is true:

It's because of the domains of inverse trig functions.

Though that solution is a bit long lol.
$\sin^{-1} (x) - \cos^{-1} (x)=\sin^{-1} (3x+1)$
$\frac{\pi}{2} = \sin^{-1} (3x+1) + 2\cos^{-1}(x)$
Obviously:
$\2\cos^{-1} (x) =\cos^{-1} (3x+1)$
Trivially, the equation is equivalent to:
$\2(\pi -\cos^{-1} (u)) = \pi +\cos^{-1} (u)$
Therefore,
$x=\frac{-1}{2}$

idkkdi · Jan 22, 2021

idkkdi said:
In case anyone does not see through why this part of the step is true:

It's because of the domain's of inverse trig functions.

Though that solution is a bit long lol.
$\sin(\sin^{-1} (x) - \cos^{-1} (x))=\sin{(\sin^{-1} (3x+1))$
$\frac{\pi}{2} = \sin^{-1} (3x+1) + 2cos^{-1}(x)$
Obviously $\2cos^{-1} (x) = cos^{-1} (3x+1) {/TEX] Trivially, the equation is equivalent to [TEX] \pi + cos^{-1} (u) = 2(\pi - cos^{-1} (u)$
$\therefore cos^{-1} (x) = \frac{2\pi}{3}, \frac{4\pi}{3} [\TEX] and [TEX]\x=\frac{-1}{2}$ [/TEX][/TEX]

fk latex

Qeru · Jan 22, 2021

idkkdi said:
In case anyone does not see through why this part of the step is true:

It's because of the domain's of inverse trig functions.

Though that solution is a bit long lol.
$\sin(\sin^{-1} (x) - \cos^{-1} (x))=\sin{(\sin^{-1} (3x+1))$
$\frac{\pi}{2} = \sin^{-1} (3x+1) + 2cos^{-1}(x)$
$\text{Obviously} \2cos^{-1} (x) = cos^{-1} (3x+1)$
$\pi + cos^{-1} (u) = 2(\pi - cos^{-1} (u)$
$\therefore cos^{-1} (x) = \frac{2\pi}{3}, \frac{4\pi}{3} \text{,and} x=\frac{-1}{2}$ [/TEX]

You cant assume $\sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}$ without proof. Nor can you assume $2\cos^{-1} (x) = \cos^{-1} (3x+1)$

idkkdi · Jan 22, 2021

Qeru said:
You cant assume $\sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}$ without proof. Nor can you assume $2\cos^{-1} (x) = \cos^{-1} (3x+1)$

trivial

idkkdi · Jan 22, 2021

idkkdi said:
fk latex

ahhh stupid latex good now i think

idkkdi · Jan 22, 2021

Qeru said:
You cant assume $\sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}$ without proof. Nor can you assume $2\cos^{-1} (x) = \cos^{-1} (3x+1)$

is this actually not assumable. it's so trivial.

vernburn · Jan 22, 2021

idkkdi said:
is this actually not assumable. it's so trivial.

If it’s not on the formula sheet you can’t assume it!

idkkdi · Jan 22, 2021

vernburn said:
If it’s not on the formula sheet you can’t assume it!

that's like all of 4u that you cant assume bruh

Qeru · Jan 22, 2021

idkkdi said:
is this actually not assumable. it's so trivial.

Both results need the double angle formula which is similar to what I did so will result in the same length of working.

Qeru · Jan 22, 2021

idkkdi said:
that's like all of 4u that you cant assume bruh

Only things you can assume are basic definitions and axioms and results that are on the data sheet (like $\sin^2x+\cos^2x=1$ )

idkkdi · Jan 22, 2021

Qeru said:
Both results need the double angle formula which is similar to what I did so will result in the same length of working.

no just draw a triangle.

idkkdi · Jan 22, 2021

Qeru said:
Only things you can assume are basic definitions and axioms and results that are on the data sheet (like $\sin^2x+\cos^2x=1$ )

this is why real men take BOS Trials. Screw HSC.

Qeru · Jan 22, 2021

idkkdi said:
no just draw a triangle.

Yeah u need double angle then 'draw a triangle,?' Prove $2\cos^{-1}(x)=\cos^{-1}(3x+1)$ without double angle?

idkkdi · Jan 22, 2021

Qeru said:
Yeah u need double angle then 'draw a triangle,?' Prove $2\cos^{-1}(x)=\cos^{-1}(3x+1)$ without double angle?

bruh draw a triangle, its so obvious that sin inverse + cos inverse x = 90 degrees it's not even funny.
as for that:
pi/2 = inverse sin + inverse cos same thing.

Trebla · Jan 22, 2021

idkkdi said:
this is why real men take BOS Trials. Screw HSC.

FYI we mark the BoS trials in similar way to HSC lol. You can’t just assume results like that without proof, otherwise you can argue the result you’re trying to prove in the first place is also “trivial”.

idkkdi · Jan 22, 2021

Trebla said:
FYI we mark the BoS trials in similar way to HSC lol. You can’t just assume results like that without proof, otherwise you can argue the result you’re trying to prove in the first place is also “trivial”.

ur trolling.
inverse sin + inverse cos = pi/2 is like by definition basically.
is this assuming?

vernburn · Jan 22, 2021

Trebla said:
FYI we mark the BoS trials in similar way to HSC lol. You can’t just assume results like that without proof, otherwise you can argue the result you’re trying to prove in the first place is also “trivial”.

Daily struggles of a 4U kid....

idkkdi · Jan 22, 2021

vernburn said:
Daily struggles of a 4U kid....

trying to write working out is harder than solving the question *sigh*

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