Cosine inverse (sin x)
Understand the Problem
The question is asking for the cosine inverse of the sine of x, which involves evaluating the trigonometric function and applying the inverse function. This typically requires an understanding of inverse trigonometric functions and their properties.
Answer
$$ \cos^{-1}(\sin x) = \begin{cases} \frac{\pi}{2} - x & \text{if } x \leq \frac{\pi}{2} \\ \frac{3\pi}{2} - x & \text{if } x > \frac{\pi}{2} \end{cases} $$
Answer for screen readers
The answer is: $$ \cos^{-1}(\sin x) = \begin{cases} \frac{\pi}{2} - x & \text{if } x \leq \frac{\pi}{2} \ \frac{3\pi}{2} - x & \text{if } x > \frac{\pi}{2} \end{cases} $$
Steps to Solve
- Recognize the functions involved
We have to find $\cos^{-1}(\sin x)$. This means we're looking for the angle whose cosine is equal to $\sin x$.
- Apply the identity
We use the fact that $\sin x = \cos\left(\frac{\pi}{2} - x\right)$. Thus, we rewrite our expression: $$ \cos^{-1}(\sin x) = \cos^{-1}\left(\cos\left(\frac{\pi}{2} - x\right)\right) $$
- Determine the value of the inverse cosine
Since $\cos^{-1}(\cos y) = y$ for $y$ in the interval $[0, \pi]$, we find: $$ \cos^{-1}\left(\cos\left(\frac{\pi}{2} - x\right)\right) = \frac{\pi}{2} - x \text{ (if } \frac{\pi}{2} - x \text{ is in } [0, \pi]\text{)} $$
- Adjust for the interval
If $\frac{\pi}{2} - x < 0$, we must adjust by adding $2\pi$ for the general solution: $$ \cos^{-1}(\sin x) = \begin{cases} \frac{\pi}{2} - x & \text{if } x \leq \frac{\pi}{2} \ \frac{3\pi}{2} - x & \text{if } x > \frac{\pi}{2} \end{cases} $$
The answer is: $$ \cos^{-1}(\sin x) = \begin{cases} \frac{\pi}{2} - x & \text{if } x \leq \frac{\pi}{2} \ \frac{3\pi}{2} - x & \text{if } x > \frac{\pi}{2} \end{cases} $$
More Information
The expression for $\cos^{-1}(\sin x)$ varies depending on the value of $x$. This is due to the properties of the inverse cosine and the periodicity of the sine function. Understanding their interplay is crucial in trigonometry.
Tips
- Forgetting to apply the correct range for $\cos^{-1}$. The output of $\cos^{-1}$ must fall within $[0, \pi]$.
- Not considering adjustments for values of $x$ that fall outside of the standard range.
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