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Sagot :
To find the value of [tex]\(\sin 60^\circ\)[/tex], let's go through a step-by-step explanation:
1. Understanding the Unit Circle:
- The unit circle is a circle with a radius of 1 centered at the origin (0, 0) of a coordinate system.
- Angles in the unit circle are commonly measured in degrees or radians, and the position of a point on the unit circle corresponding to an angle [tex]\( \theta \)[/tex] can be described using [tex]\((\cos \theta, \sin \theta)\)[/tex].
2. Recollection of Special Angles:
- For certain angles, the values of sine and cosine are well-known and often memorized because of their frequent use in trigonometry.
- [tex]\(60^\circ\)[/tex] is one of these special angles.
3. Special Triangle Relationships:
- One of the classic approaches to finding [tex]\(\sin 60^\circ\)[/tex] involves an equilateral triangle (where all sides and all angles are equal) with a side length of 2.
- Dropping a perpendicular from one vertex to the midpoint of the opposite side splits the equilateral triangle into two 30-60-90 right triangles.
4. Using the 30-60-90 Triangle:
- In a 30-60-90 triangle, the ratios of the sides are known:
- The side opposite the 30° angle is half the hypotenuse.
- The side opposite the 60° angle is [tex]\(\sqrt{3}\)[/tex] times half the hypotenuse.
- Since we started with a side length of 2 for the equilateral triangle, the hypotenuse of the right triangle is also 2.
- Therefore, the side opposite the 60° angle (which is the height of the triangle when split) is [tex]\(\sqrt{3}\)[/tex], and the hypotenuse is 2.
5. Calculating [tex]\(\sin 60^\circ\)[/tex]:
- Recall that sine is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse in a right triangle.
- Hence, [tex]\(\sin 60^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2}\)[/tex].
Therefore, [tex]\(\sin 60^\circ\)[/tex] is [tex]\(\frac{\sqrt{3}}{2}\)[/tex]. Looking at the given options:
A. [tex]\(\frac{\sqrt{3}}{2}\)[/tex]
B. [tex]\(\frac{1}{\sqrt{3}}\)[/tex]
C. [tex]\(\frac{1}{2}\)[/tex]
D. 1
E. [tex]\(\frac{1}{\sqrt{2}}\)[/tex]
F. [tex]\(\sqrt{3}\)[/tex]
We see that the correct answer is:
A. [tex]\(\frac{\sqrt{3}}{2}\)[/tex]
1. Understanding the Unit Circle:
- The unit circle is a circle with a radius of 1 centered at the origin (0, 0) of a coordinate system.
- Angles in the unit circle are commonly measured in degrees or radians, and the position of a point on the unit circle corresponding to an angle [tex]\( \theta \)[/tex] can be described using [tex]\((\cos \theta, \sin \theta)\)[/tex].
2. Recollection of Special Angles:
- For certain angles, the values of sine and cosine are well-known and often memorized because of their frequent use in trigonometry.
- [tex]\(60^\circ\)[/tex] is one of these special angles.
3. Special Triangle Relationships:
- One of the classic approaches to finding [tex]\(\sin 60^\circ\)[/tex] involves an equilateral triangle (where all sides and all angles are equal) with a side length of 2.
- Dropping a perpendicular from one vertex to the midpoint of the opposite side splits the equilateral triangle into two 30-60-90 right triangles.
4. Using the 30-60-90 Triangle:
- In a 30-60-90 triangle, the ratios of the sides are known:
- The side opposite the 30° angle is half the hypotenuse.
- The side opposite the 60° angle is [tex]\(\sqrt{3}\)[/tex] times half the hypotenuse.
- Since we started with a side length of 2 for the equilateral triangle, the hypotenuse of the right triangle is also 2.
- Therefore, the side opposite the 60° angle (which is the height of the triangle when split) is [tex]\(\sqrt{3}\)[/tex], and the hypotenuse is 2.
5. Calculating [tex]\(\sin 60^\circ\)[/tex]:
- Recall that sine is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse in a right triangle.
- Hence, [tex]\(\sin 60^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2}\)[/tex].
Therefore, [tex]\(\sin 60^\circ\)[/tex] is [tex]\(\frac{\sqrt{3}}{2}\)[/tex]. Looking at the given options:
A. [tex]\(\frac{\sqrt{3}}{2}\)[/tex]
B. [tex]\(\frac{1}{\sqrt{3}}\)[/tex]
C. [tex]\(\frac{1}{2}\)[/tex]
D. 1
E. [tex]\(\frac{1}{\sqrt{2}}\)[/tex]
F. [tex]\(\sqrt{3}\)[/tex]
We see that the correct answer is:
A. [tex]\(\frac{\sqrt{3}}{2}\)[/tex]
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