Answer on Question #50655 – Math –Trigonometry
sin − 1 ( c o s θ ) = ( π 2 ) − θ or ( π 2 ) + θ I think both are correct but not sure.? \sin^{-1}(cos\theta) = \left(\frac{\pi}{2}\right) - \theta \text{ or } \left(\frac{\pi}{2}\right) + \theta \text{ I think both are correct but not sure.?} sin − 1 ( cos θ ) = ( 2 π ) − θ or ( 2 π ) + θ I think both are correct but not sure.?
Here my thinking
sin ∧ ( − 1 ) ⊗ ^ ( cos ⊗ ^ θ ) = sin ∧ ( − 1 ) ⊗ ^ ( sin ( ( π / 2 ) − θ ) ) = ( π / 2 ) − θ \sin^{\wedge}(-1)\widehat{\otimes}(\cos\widehat{\otimes}\theta) = \sin^{\wedge}(-1)\widehat{\otimes}(\sin((\pi/2) - \theta)) = (\pi/2) - \theta sin ∧ ( − 1 ) ⊗ ( cos ⊗ θ ) = sin ∧ ( − 1 ) ⊗ ( sin (( π /2 ) − θ )) = ( π /2 ) − θ
Or in another way
sin ∧ ( − 1 ) ⊗ ^ ( cos ⊗ ^ θ ) = sin ∧ ( − 1 ) ⊗ ^ ( sin ( ⊗ ^ ( π / 2 ) + θ ) ) = ( π / 2 ) + θ \sin^{\wedge}(-1)\widehat{\otimes}(\cos\widehat{\otimes}\theta) = \sin^{\wedge}(-1)\widehat{\otimes}(\sin(\widehat{\otimes}(\pi/2) + \theta)) = (\pi/2) + \theta sin ∧ ( − 1 ) ⊗ ( cos ⊗ θ ) = sin ∧ ( − 1 ) ⊗ ( sin ( ⊗ ( π /2 ) + θ )) = ( π /2 ) + θ
I want to differentiate sin − 1 ( c o s θ ) \sin^{-1}(cos\theta) sin − 1 ( cos θ ) ( π / 2 ) − θ (\pi/2) - \theta ( π /2 ) − θ ( π / 2 ) + θ (\pi/2) + \theta ( π /2 ) + θ
Solution
You are right.
sin − 1 ( c o s θ ) = ( π 2 ) − cos − 1 ( c o s θ ) = ( π 2 ) − θ . \sin^{-1}(cos\theta) = \left(\frac{\pi}{2}\right) - \cos^{-1}(cos\theta) = \left(\frac{\pi}{2}\right) - \theta. sin − 1 ( cos θ ) = ( 2 π ) − cos − 1 ( cos θ ) = ( 2 π ) − θ .
But
cos θ = cos ( − θ ) . \cos\theta = \cos(-\theta). cos θ = cos ( − θ ) .
So,
sin − 1 ( c o s θ ) = sin − 1 ( c o s ( − θ ) ) = ( π 2 ) − cos − 1 ( c o s ( − θ ) ) = ( π 2 ) + θ . \sin^{-1}(cos\theta) = \sin^{-1}(cos(-\theta)) = \left(\frac{\pi}{2}\right) - \cos^{-1}(cos(-\theta)) = \left(\frac{\pi}{2}\right) + \theta. sin − 1 ( cos θ ) = sin − 1 ( cos ( − θ )) = ( 2 π ) − cos − 1 ( cos ( − θ )) = ( 2 π ) + θ .
You need
d d θ ( sin − 1 ( c o s θ ) ) = d d cos θ ( sin − 1 ( c o s θ ) ) ⋅ d d θ ( ( cos θ ) ) = − 1 1 − ( cos θ ) 2 ⋅ sin θ = − sin θ ∣ sin θ ∣ = − S i g n ( θ ) . \begin{aligned}
\frac{d}{d\theta}\big(\sin^{-1}(cos\theta)\big) &= \frac{d}{d\cos\theta}\big(\sin^{-1}(cos\theta)\big) \cdot \frac{d}{d\theta}\big((\cos\theta)\big) = -\frac{1}{\sqrt{1 - (\cos\theta)^2}} \cdot \sin\theta = -\frac{\sin\theta}{|\sin\theta|} \\
&= -\mathrm{Sign}(\theta).
\end{aligned} d θ d ( sin − 1 ( cos θ ) ) = d cos θ d ( sin − 1 ( cos θ ) ) ⋅ d θ d ( ( cos θ ) ) = − 1 − ( cos θ ) 2 1 ⋅ sin θ = − ∣ sin θ ∣ sin θ = − Sign ( θ ) .
where
S i g n ( θ ) = { 1 , when θ > 0 0 , when θ = 0 − 1 , when θ < 0 \mathrm{Sign}(\theta) = \begin{cases}
1, & \text{when } \theta > 0 \\
0, & \text{when } \theta = 0 \\
-1, & \text{when } \theta < 0
\end{cases} Sign ( θ ) = ⎩ ⎨ ⎧ 1 , 0 , − 1 , when θ > 0 when θ = 0 when θ < 0
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