Exercise 33.1 Prove, Without The Use Of A Calculator, That: ${ \cos 75^{\circ} + \cos 15^{\circ} = \frac{\sqrt{6}}{2} }$3.2 Determine The General Solution Of: ${ 1 + 4 \sin^2 X - 5 \sin X + \cos 2x = 0 }$

Introduction

In this exercise, we will be proving a trigonometric identity and solving a trigonometric equation. The first part involves proving that $\cos 75^{\circ} + \cos 15^{\circ} = \frac{\sqrt{6}}{2}$ without the use of a calculator. The second part involves determining the general solution of the trigonometric equation $1 + 4 \sin^2 x - 5 \sin x + \cos 2x = 0$.

Exercise 33.1: Proving the Trigonometric Identity

To prove the trigonometric identity $\cos 75^{\circ} + \cos 15^{\circ} = \frac{\sqrt{6}}{2}$, we can use the sum-to-product identity for cosine:

$$\cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)$$

In this case, $A = 75^{\circ}$ and $B = 15^{\circ}$.

import math
angle_A = math.radians(75)
angle_B = math.radians(15)

sum_angles = (angle_A + angle_B) / 2
diff_angles = (angle_A - angle_B) / 2

cos_sum_angles = math.cos(sum_angles)
cos_diff_angles = math.cos(diff_angles)

product_cosines = 2 * cos_sum_angles * cos_diff_angles

print("The product of the cosines is:", product_cosines)

Using this identity, we can rewrite the expression as:

$$\cos 75^{\circ} + \cos 15^{\circ} = 2 \cos \left( \frac{75^{\circ} + 15^{\circ}}{2} \right) \cos \left( \frac{75^{\circ} - 15^{\circ}}{2} \right)$$

Simplifying the expression, we get:

$$\cos 75^{\circ} + \cos 15^{\circ} = 2 \cos 45^{\circ} \cos 30^{\circ}$$

Using the values of $\cos 45^{\circ} = \frac{\sqrt{2}}{2}$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$, we can substitute these values into the expression:

$$\cos 75^{\circ} + \cos 15^{\circ} = 2 \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right)$$

Simplifying the expression, we get:

$$\cos 75^{\circ} + \cos 15^{\circ} = \frac{\sqrt{6}}{2}$$

Therefore, we have proved the trigonometric identity $\cos 75^{\circ} + \cos 15^{\circ} = \frac{\sqrt{6}}{2}$.

Exercise 33.2: Solving the Trigonometric Equation

To solve the trigonometric equation $1 + 4 \sin^2 x - 5 \sin x + \cos 2x = 0$, we can start by using the double-angle identity for cosine:

$$\cos 2x = 1 - 2 \sin^2 x$$

Substituting this expression into the equation, we get:

$$1 + 4 \sin^2 x - 5 \sin x + 1 - 2 \sin^2 x = 0$$

Simplifying the equation, we get:

$$2 + 2 \sin^2 x - 5 \sin x = 0$$

Factoring out the common term $2$, we get:

$$2 (1 + \sin^2 x - \frac{5}{2} \sin x) = 0$$

Dividing both sides of the equation by $2$, we get:

$$1 + \sin^2 x - \frac{5}{2} \sin x = 0$$

Rearranging the equation, we get:

$$\sin^2 x - \frac{5}{2} \sin x + 1 = 0$$

This is a quadratic equation in terms of $\sin x$. We can solve this equation using the quadratic formula:

$$\sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $a = 1$, $b = -\frac{5}{2}$, and $c = 1$.

import math

a = 1
b = -5/2
c = 1

discriminant = b**2 - 4ac

solution1 = (-b + math.sqrt(discriminant)) / (2a)
solution2 = (-b - math.sqrt(discriminant)) / (2a)

print("The two solutions are:", solution1, "and", solution2)

Solving the equation, we get:

$$\sin x = \frac{\frac{5}{2} \pm \sqrt{\left( \frac{5}{2} \right)^2 - 4}}{2}$$

Simplifying the expression, we get:

$$\sin x = \frac{\frac{5}{2} \pm \frac{\sqrt{9}}{2}}{2}$$

Simplifying further, we get:

$$\sin x = \frac{\frac{5}{2} \pm \frac{3}{2}}{2}$$

Therefore, we have two possible solutions:

$$\sin x = \frac{\frac{5}{2} + \frac{3}{2}}{2} = 2$$

$$\sin x = \frac{\frac{5}{2} - \frac{3}{2}}{2} = 1$$

However, since the sine function is bounded between $-1$ and $1$, the solution $\sin x = 2$ is extraneous and can be discarded.

Therefore, the only valid solution is:

$$\sin x = 1$$

This implies that:

$$x = \sin^{-1} (1)$$

Using the inverse sine function, we get:

$$x = \frac{\pi}{2}$$

Therefore, the general solution of the trigonometric equation $1 + 4 \sin^2 x - 5 \sin x + \cos 2x = 0$ is:

$$x = \frac{\pi}{2} + 2 \pi n$$

where $n$ is an integer.

Conclusion

$$$$$$$$

Q: What is the main goal of Exercise 33.1?

A: The main goal of Exercise 33.1 is to prove the trigonometric identity $\cos 75^{\circ} + \cos 15^{\circ} = \frac{\sqrt{6}}{2}$ without the use of a calculator.

Q: What is the sum-to-product identity for cosine?

A: The sum-to-product identity for cosine is:

$$\cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)$$

Q: How do you simplify the expression $\cos 75^{\circ} + \cos 15^{\circ}$ using the sum-to-product identity?

A: To simplify the expression, we can use the sum-to-product identity for cosine:

$$\cos 75^{\circ} + \cos 15^{\circ} = 2 \cos \left( \frac{75^{\circ} + 15^{\circ}}{2} \right) \cos \left( \frac{75^{\circ} - 15^{\circ}}{2} \right)$$

Simplifying the expression, we get:

$$\cos 75^{\circ} + \cos 15^{\circ} = 2 \cos 45^{\circ} \cos 30^{\circ}$$

Q: What is the value of $\cos 45^{\circ}$ and $\cos 30^{\circ}$?

A: The value of $\cos 45^{\circ}$ is $\frac{\sqrt{2}}{2}$ and the value of $\cos 30^{\circ}$ is $\frac{\sqrt{3}}{2}$.

Q: How do you simplify the expression $2 \cos 45^{\circ} \cos 30^{\circ}$?

A: To simplify the expression, we can substitute the values of $\cos 45^{\circ}$ and $\cos 30^{\circ}$:

$$2 \cos 45^{\circ} \cos 30^{\circ} = 2 \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right)$$

Simplifying the expression, we get:

$$2 \cos 45^{\circ} \cos 30^{\circ} = \frac{\sqrt{6}}{2}$$

Q: What is the main goal of Exercise 33.2?

A: The main goal of Exercise 33.2 is to solve the trigonometric equation $1 + 4 \sin^2 x - 5 \sin x + \cos 2x = 0$.

Q: How do you simplify the equation $1 + 4 \sin^2 x - 5 \sin x + \cos 2x = 0$?

A: To simplify the equation, we can use the double-angle identity for cosine:

$$\cos 2x = 1 - 2 \sin^2 x$$

Substituting this expression into the equation, we get:

$$1 + 4 \sin^2 x - 5 \sin x + 1 - 2 \sin^2 x = 0$$

Simplifying the equation, we get:

$$2 + 2 \sin^2 x - 5 \sin x = 0$$

Q: How do you solve the quadratic equation $2 + 2 \sin^2 x - 5 \sin x = 0$?

A: To solve the quadratic equation, we can use the quadratic formula:

$$\sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $a = 1$, $b = -\frac{5}{2}$, and $c = 1$.

Q: What is the general solution of the trigonometric equation $1 + 4 \sin^2 x - 5 \sin x + \cos 2x = 0$?

A: The general solution of the trigonometric equation is $x = \frac{\pi}{2} + 2 \pi n$, where $n$ is an integer.

Q: What is the final answer to Exercise 33.1 and 33.2?

A: The final answer to Exercise 33.1 is $\cos 75^{\circ} + \cos 15^{\circ} = \frac{\sqrt{6}}{2}$ and the final answer to Exercise 33.2 is $x = \frac{\pi}{2} + 2 \pi n$, where $n$ is an integer.