Trigonometry Exercise 2.12 for 11th Grade Students

Math: Exploring Trigonometry and Exercise 2.12 for 11th Grade Students

Trigonometry, a branch of mathematics that deals with the relationships between the angles and sides of triangles, is a vital concept to understand as you progress through high school math. This article will focus on 11th grade trigonometry, specifically delving into the Exercise 2.12 from common textbooks. We'll begin by covering the basics of trigonometry and then dive into the exercise itself, providing explanations and examples.

Trigonometric Functions

In trigonometry, you'll encounter three primary functions: sine (sin), cosine (cos), and tangent (tan). These functions are defined in terms of the ratios of the sides of a right triangle.

• Sine (sin): The sine of an angle θ in a right triangle is the ratio of the length of the side opposite the angle to the length of the hypotenuse.

  [ \sin \theta = \frac{opposite}{hypotenuse} ]

• Cosine (cos): The cosine of an angle θ in a right triangle is the ratio of the length of the side adjacent to the angle to the length of the hypotenuse.

  [ \cos \theta = \frac{adjacent}{hypotenuse} ]

• Tangent (tan): The tangent of an angle θ in a right triangle is the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle.

  [ \tan \theta = \frac{opposite}{adjacent} ]

For example, in the right triangle shown above, θ is 30°, and the lengths are 3 (adjacent), 4 (opposite), and 5 (hypotenuse). In this case, (\sin 30^\circ = \frac{3}{5}), (\cos 30^\circ = \frac{4}{5}), and (\tan 30^\circ = \frac{3}{4}).

Exercise 2.12

Exercise 2.12 for 11th grade students typically involves manipulating trigonometric functions of complementary angles and solving trigonometric equations. Below we'll go through a few examples to help you understand the concepts and techniques involved.

Example 1:

Given (\sin 45^\circ = \frac{1}{2}), find the values of (\cos 45^\circ) and (\tan 45^\circ).

(\sin 45^\circ = \frac{1}{2} = \frac{opposite}{hypotenuse}). Since right triangles with 45° angles have two legs of equal length, the opposite and adjacent sides are equal. Therefore, the adjacent side is also 1, and:

[\cos 45^\circ = \frac{adjacent}{hypotenuse} = \frac{1}{2} \ \ \text{and} \ \ \tan 45^\circ = \frac{opposite}{adjacent} = \frac{1}{1} = 1]

Example 2:

Solve the equation (\cos x = 0.6) for (0^\circ \leq x \leq 360^\circ).

To solve this equation, we'll first find the angles for which (\cos x = 0.6). We know that the cosine function reaches its maximum and minimum values at specific angles.

[\cos 60^\circ = \cos 300^\circ = 0.5 > 0.6]

[\cos 300^\circ = \cos 360^\circ - 300^\circ = -0.5 < 0.6]

Now we look for angles between (0^\circ) and (360^\circ) where (\cos x = 0.6). We'll use the fact that (\cos (180^\circ - x) = -\cos x).

[\cos (180^\circ - x) = -\cos x = -0.6]

[\cos x = 0.6 = \cos (180^\circ - x)]

Now we can find the angles (x) by solving for (180^\circ - x):

[180^\circ - x = 120^\circ \Rightarrow x = 60^\circ]

Thus, the solution to the equation (\cos x = 0.6) is (x = 60^\circ).

Example 3:

Solve the equation (\tan x = \frac{1}{2}) for (0^\circ \leq x \leq 360^\circ).

To solve this equation, we'll first find the angles for which (\tan x = \frac{1}{2}). We know that the tangent function reaches its maximum and minimum values at specific angles:

[\tan 45^\circ = \tan 315^\circ = 1 > \frac{1}{2}]

[\tan 315^\circ = \tan (360^\circ - 45^\circ) = -1 < \frac{1}{2}]

Now we look for angles between (0^\circ) and (360^\circ) where (\tan x = \frac{1}{2}). We'll use the fact that (\tan (180^\circ - x) = -\tan x).

[\tan (180^\circ - x) = -\tan x = -\frac{1}{2}]

[\tan x = \frac{1}{2} = \tan (180^\circ - x)]

Now we can find the angles (x) by solving for (180^\circ - x):

[180^\circ - x = 135^\circ \Rightarrow x = 45^\circ]

Thus, the solution to the equation (\tan x = \frac{1}{2}) is (x = 45^\circ).

This exercise is just a small sample of what you’ll cover in 11th grade trigon