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Inverse Trigonometric Functions

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Inverse Trigonometric Functions
Inverse Sine Function
f(x)=sin⁡x   ↔  f−1(x)=arcsin⁡x   or   sin⁡−1xf\left(x\right)=\sin x\ \ \ \leftrightarrow\ \ f^{-1}\left(x\right)=\arcsin x\ \ \ \text{or}\ \ \ \sin^{-1}xf(x)=sinx   ↔  f−1(x)=arcsinx   or   sin−1x


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Inverse Cosine Function
f(x)=cos⁡x   ↔  f−1(x)=arccos⁡x   or   cos⁡−1xf\left(x\right)=\cos x\ \ \ \leftrightarrow\ \ f^{-1}\left(x\right)=\arccos x\ \ \ \text{or}\ \ \ \cos^{-1}xf(x)=cosx   ↔  f−1(x)=arccosx   or   cos−1x


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Inverse Tangent Function
f(x)tan⁡x   ↔  f−1(x)=arctan⁡x   or   tan⁡−1xf\left(x\right)\tan x\ \ \ \leftrightarrow\ \ f^{-1}\left(x\right)=\arctan x\ \ \ \text{or}\ \ \ \tan^{-1}xf(x)tanx   ↔  f−1(x)=arctanx   or   tan−1x


Inverse Tangent has horizontal asymptotes at y=π2  and  y=−π2\displaystyle y=\frac{\pi}{2} \ \text{ and } \ y=\frac{-\pi}{2}y=2π​  and  y=2−π​


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Dealing with Inverse Trig Functions
If you are given a question with different trig functions like tan⁡(arcsin⁡x)=?\tan\left(\arcsin x\right)=?tan(arcsinx)=?
let θ=trig−1(x)\theta=\text{trig}^{-1}\left(x\right)θ=trig−1(x)
then we have trig(θ)=x\text{trig}  (\theta)=xtrig(θ)=x
Draw a triangle based on this θ\thetaθ

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Example: Trig & Inverse Trig Values
Determine the values of the following:

a.cos⁡(arccos⁡ (12))\cos\left(\arccos\ \left(\frac{1}{2}\right)\right)cos(arccos (21​))

The domain of arccos⁡\arccosarccos is [−1,1]\left[-1,1\right][−1,1].
You can use the cancellation rule directly since the argument is in [−1,1]\left[-1,1\right][−1,1]
Therefore, this evaluates to 12\frac{1}{2}21​.

b. arcsin⁡(sin⁡(π6))\arcsin\left(\sin\left(\frac{\pi}{6}\right)\right)arcsin(sin(6π​))

The range of arcsin⁡\arcsinarcsin is [−π2, π2]\left[-\frac{\pi}{2},\ \frac{\pi}{2}\right][−2π​, 2π​].
You can use the cancellation rule directly since the argument is in [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right][−2π​,2π​]
Therefore, this evaluates to π6\frac{\pi}{6}6π​

c. arcsin⁡(sin⁡3π4)\arcsin\left(\sin\frac{3\pi}{4}\right)arcsin(sin43π​)

The range of arcsin⁡\arcsinarcsin is [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right][−2π​,2π​]
Cancellation rule does not work in this case because the argument is not in [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right][−2π​,2π​]
We need to find an angle xxx in [−π2, π2]\left[-\frac{\pi}{2},\ \frac{\pi}{2}\right][−2π​, 2π​]such that sin⁡x=sin⁡3π4\sin x=\sin\frac{3\pi}{4}sinx=sin43π​.
Notice that sin⁡3π4\sin\frac{3\pi}{4}sin43π​ (quadrant 2) is positive, so the angle we're looking for is in quadrant 1.
Therefore, the angle xxx we need is π4\frac{\pi}{4}4π​

d. sin⁡(cos⁡−10)\sin\left(\cos^{-1}0\right)sin(cos−10).

*Recall that cos⁡−1=arccos⁡\cos^{-1}=\arccoscos−1=arccos
The domain of arccos⁡\arccosarccos is [−1,1]\left[-1,1\right][−1,1]. Notice that 0 is in that domain.
Let θ=arccos⁡0\theta=\arccos0θ=arccos0, 
Then cos⁡θ=0\cos\theta=0cosθ=0.
Since the range of arccos⁡\arccosarccos is [0,π]\left[0,\pi\right][0,π], we get that θ=π2\theta=\frac{\pi}{2}θ=2π​.

Putting this back into the 𝑠𝑖𝑛 function, sin⁡(arccos⁡0)=sin⁡(θ)=sin⁡(π2)=1\sin\left(\arccos0\right)=\sin\left(\theta\right)=\sin\left(\frac{\pi}{2}\right)=1sin(arccos0)=sin(θ)=sin(2π​)=1.

Find the exact value of tan⁡[arcsin⁡(−35)]\tan\left[\arcsin\left(-\frac{3}{5}\right)\right]tan[arcsin(−53​)]

Answer

I don't know

Find the exact value of sin⁡(2cos⁡−1x)\sin\left(2\cos^{-1}x\right)sin(2cos−1x)

A) 21−x22\sqrt{1-x^2}21−x2​

B) 2x1−x22x\sqrt{1-x^2}2x1−x2​

C) 21−x2x\frac{2\sqrt{1-x^2}}{x}x21−x2​​

D) 1−x22x\frac{\sqrt{1-x^2}}{2x}2x1−x2​​

E) None of the above

I don't know