# Inverse trig question (1 Viewer)

#### hs17

##### Member
Solve the following equation

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

#### Qeru

##### Well-Known Member
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: $\bg_white -1 \leq x \leq 1$ and $\bg_white \frac{-2}{3}\leq x \leq 0$.

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

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

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

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

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

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

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

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

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

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

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#### idkkdi

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

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

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

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

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

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

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

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

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

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

But $\bg_white x \leq 0$ so the only solution is:$\bg_white x=-\frac{1}{2}$
In case anyone does not see through why this part of the step is true:
$image=https://latex.codecogs.com/png.latex?\bg_white%20\sin(\cos^{-1}%20(x))=\cos(\sin^{-1}%20(x))\geq0&hash=01793b0b07ce561d931410e234043286$

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

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

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#### idkkdi

##### Well-Known Member
In case anyone does not see through why this part of the step is true:
$image=https://latex.codecogs.com/png.latex?\bg_white%20\sin(\cos^{-1}%20(x))=\cos(\sin^{-1}%20(x))\geq0&hash=01793b0b07ce561d931410e234043286$

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

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

$\bg_white \therefore cos^{-1} (x) = \frac{2\pi}{3}, \frac{4\pi}{3} [\TEX] and "/>"/>
fk latex

#### Qeru

##### Well-Known Member
In case anyone does not see through why this part of the step is true:
$image=https://latex.codecogs.com/png.latex?\bg_white%20\sin(\cos^{-1}%20(x))=\cos(\sin^{-1}%20(x))\geq0&hash=01793b0b07ce561d931410e234043286$

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

Though that solution is a bit long lol.
$\bg_white \sin(\sin^{-1} (x) - \cos^{-1} (x))=\sin{(\sin^{-1} (3x+1))$
$\bg_white \frac{\pi}{2} = \sin^{-1} (3x+1) + 2cos^{-1}(x)$
$\bg_white \text{Obviously} \2cos^{-1} (x) = cos^{-1} (3x+1)$
$\bg_white \pi + cos^{-1} (u) = 2(\pi - cos^{-1} (u)$
$\bg_white \therefore cos^{-1} (x) = \frac{2\pi}{3}, \frac{4\pi}{3} \text{,and} x=\frac{-1}{2}$[/TEX]
You cant assume $\bg_white \sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}$ without proof. Nor can you assume $\bg_white 2\cos^{-1} (x) = \cos^{-1} (3x+1)$

#### idkkdi

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

#### idkkdi

##### Well-Known Member
You cant assume $\bg_white \sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}$ without proof. Nor can you assume $\bg_white 2\cos^{-1} (x) = \cos^{-1} (3x+1)$
is this actually not assumable. it's so trivial.

#### vernburn

##### Active Member
is this actually not assumable. it's so trivial.
If it’s not on the formula sheet you can’t assume it!

#### idkkdi

##### Well-Known Member
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

##### Well-Known Member
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

##### Well-Known Member
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 $\bg_white \sin^2x+\cos^2x=1$)

#### idkkdi

##### Well-Known Member
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

##### Well-Known Member
Only things you can assume are basic definitions and axioms and results that are on the data sheet (like $\bg_white \sin^2x+\cos^2x=1$)
this is why real men take BOS Trials. Screw HSC.

#### Qeru

##### Well-Known Member
no just draw a triangle.
Yeah u need double angle then 'draw a triangle,?' Prove $\bg_white 2\cos^{-1}(x)=\cos^{-1}(3x+1)$ without double angle?

#### idkkdi

##### Well-Known Member
Yeah u need double angle then 'draw a triangle,?' Prove $\bg_white 2\cos^{-1}(x)=\cos^{-1}(3x+1)$ 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

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

##### Well-Known Member
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

##### Active Member
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

##### Well-Known Member
Daily struggles of a 4U kid....
trying to write working out is harder than solving the question *sigh*