Certainly! Let's solve the given trigonometric equations step-by-step.
### Part (a): Solving [tex]\(\sin \theta = \frac{1}{\sqrt{2}}\)[/tex]
1. Identify the standard angle:
The value [tex]\(\frac{1}{\sqrt{2}}\)[/tex] is a common value for the sine function. It can be recognized that:
[tex]\[
\sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}
\][/tex]
2. Consider the general solutions for sine:
The sine function is positive in the first and second quadrants. Therefore, the principal solutions for [tex]\(\sin \theta = \frac{1}{\sqrt{2}}\)[/tex] within one cycle (from [tex]\(0\)[/tex] to [tex]\(2\pi\)[/tex]) are:
[tex]\[
\theta = \frac{\pi}{4} \quad \text{and} \quad \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}
\][/tex]
### Part (b): Solving [tex]\(\tan \theta = 1\)[/tex]
1. Identify the standard angle:
The value [tex]\(1\)[/tex] is a common value for the tangent function. It can be recognized that:
[tex]\[
\tan \left(\frac{\pi}{4}\right) = 1
\][/tex]
2. Consider the general solutions for tangent:
The tangent function is positive in the first and third quadrants. For [tex]\(\tan \theta = 1\)[/tex], the principal solution within one cycle (as [tex]\(\tan\)[/tex] has a period of [tex]\(\pi\)[/tex]) is:
[tex]\[
\theta = \frac{\pi}{4}
\][/tex]
### Summary of Solutions
- For the equation [tex]\(\sin \theta = \frac{1}{\sqrt{2}}\)[/tex], the principal solutions are:
[tex]\[
\theta = \frac{\pi}{4}, \quad \theta = \frac{3\pi}{4}
\][/tex]
- For the equation [tex]\(\tan \theta = 1\)[/tex], the principal solution is:
[tex]\[
\theta = \frac{\pi}{4}
\][/tex]
Thus, the final results are:
1. [tex]\(\sin \theta = \frac{1}{\sqrt{2}}\)[/tex] has principal solutions: [tex]\(\frac{\pi}{4}, \frac{3\pi}{4}\)[/tex].
2. [tex]\(\tan \theta = 1\)[/tex] has the principal solution: [tex]\(\frac{\pi}{4}\)[/tex].
