Question

question 6 points 4 find the values of sine, cosine, and tangent of special angles in 45° - 90° triangle.

Explanation:

Step1: Define the 45 - 45 - 90 triangle

A 45° - 45° - 90° triangle is an isosceles right triangle, so the two legs are of equal length. Let the length of each leg be \( a \), and the hypotenuse be \( c \). By the Pythagorean theorem, \( c^{2}=a^{2}+a^{2}=2a^{2} \), so \( c = a\sqrt{2} \).

Step2: Calculate sine of 45°

The sine of an angle in a right triangle is defined as \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \). For a 45° angle, the opposite side and the adjacent side (since it's isosceles) are both \( a \), and the hypotenuse is \( a\sqrt{2} \). So \( \sin45^{\circ}=\frac{a}{a\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \) (rationalizing the denominator).

Step3: Calculate cosine of 45°

The cosine of an angle in a right triangle is defined as \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \). For a 45° angle, the adjacent side is \( a \), and the hypotenuse is \( a\sqrt{2} \). So \( \cos45^{\circ}=\frac{a}{a\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \).

Step4: Calculate tangent of 45°

The tangent of an angle in a right triangle is defined as \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \). For a 45° angle, the opposite side and the adjacent side are both \( a \). So \( \tan45^{\circ}=\frac{a}{a} = 1 \).

Answer:

For a \( 45^{\circ}-45^{\circ}-90^{\circ} \) triangle:

• \( \sin45^{\circ}=\frac{\sqrt{2}}{2} \)
• \( \cos45^{\circ}=\frac{\sqrt{2}}{2} \)
• \( \tan45^{\circ}=1 \)