To solve for [tex]\(\sin 45^{\circ}\)[/tex], we can use the properties of a 45-degree angle in a right triangle.
1. Basic Properties of 45-degree Angles:
- In a 45-degree right triangle, both the angles apart from the right angle are 45 degrees. This means that the triangle is isosceles and the two legs are of equal length.
2. Constructing the Triangle:
- Let’s consider an isosceles right triangle where each of the two legs (let’s name them both [tex]\( a \)[/tex]) is equal. For simplicity, assume that [tex]\( a = 1 \)[/tex].
3. Finding the Hypotenuse:
- By the Pythagorean theorem, the hypotenuse [tex]\( c \)[/tex] of a right triangle with legs [tex]\( a \)[/tex] and [tex]\( b \)[/tex] (both equal to 1) is:
[tex]\[
c = \sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{2}
\][/tex]
4. Expressing [tex]\(\sin 45^{\circ}\)[/tex]:
- The sine of an angle in a right triangle is given by the ratio of the length of the opposite side to the length of the hypotenuse.
- For a 45-degree angle in our triangle:
[tex]\[
\sin 45^{\circ} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{2}}
\][/tex]
5. Simplifying the Expression:
- Rationalizing the denominator is a common step, but in this context, we can directly use the simplified values.
- We already observed that:
[tex]\[
\sin 45^{\circ} = \frac{1}{\sqrt{2}}
\][/tex]
Given the options:
- [tex]\( A. 1 \)[/tex]
- [tex]\( B. \frac{1}{2} \)[/tex]
- [tex]\( C. \frac{1}{\sqrt{2}} \)[/tex]
- [tex]\( D. \sqrt{2} \)[/tex]
We find that the correct answer is [tex]\( \frac{1}{\sqrt{2}} \)[/tex]. When calculated numerically, this evaluates to approximately 0.7071067811865475.
Thus, the answer is:
[tex]\(\boxed{\frac{1}{\sqrt{2}}}\)[/tex]
