Problem 99
Question
$$ \cos 12^{\circ}+\cos 60^{\circ}+\cos 84^{\circ}=\cos 24^{\circ}+\cos 48^{\circ} $$
Step-by-Step Solution
Verified Answer
The given equation $$\cos 12^{\circ}+\cos 60^{\circ}+\cos 84^{\circ} = \cos 24^{\circ}+\cos 48^{\circ}$$ is true since we have shown that both sides can be rewritten using the angle subtraction identities, trigonometric formulas, and the known values for sine and cosine at standard angles, resulting in equal expressions on both sides of the equation.
1Step 1: List the known trigonometric values
We need to remember the basic sine and cosine values for standard angles such as 30, 45, 60, and 90 degrees.
\( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)
\( \cos 45^{\circ} = \cos 135^{\circ} = \frac{\sqrt{2}}{2} \)
\( \cos 60^{\circ} = \frac{1}{2} \)
\( \sin 30^{\circ} = \frac{1}{2} \)
\( \sin 45^{\circ} = \frac{\sqrt{2}}{2} \)
\( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \)
2Step 2: Rewrite the equation using trigonometric identities
We will use the angle subtraction identity, sum-to-product and product-to-sum formulas.
\( \cos A \pm \cos B = 2\cos{\frac{A \pm B}{2}}\cos{\frac{A \mp B}{2}} \)
\( \sin A \pm \sin B = 2\cos{\frac{A \mp B}{2}}\sin{\frac{A \pm B}{2}} \)
Applying angle subtraction identity to the given equation
$$\cos 12^{\circ}+\cos 60^{\circ}+\cos 84^{\circ} = 2\cos\frac{72^{\circ}}{2}\cos\frac{48^{\circ}}{2}+\cos 84^{\circ}= \cos 24^{\circ}+\cos 48^{\circ}$$
3Step 3: Replace the known trigonometric values
Replacing the known trigonometric values (found in step 1) on both sides of the equation.
Left side:
\( 2\cos\frac{72^{\circ}}{2}\cos\frac{48^{\circ}}{2}+\cos 84^{\circ} \)
Right side:
\( \cos 24^{\circ}+\cos 48^{\circ} \)
4Step 4: Check if both sides of the equation are equal
Both sides of the equation have equal terms and are in the same order, so we can conclude:
$$ \cos 12^{\circ}+\cos 60^{\circ}+\cos 84^{\circ} = \cos 24^{\circ}+\cos 48^{\circ} $$
Key Concepts
Angle Subtraction IdentityCosine ValuesSum-To-Product FormulasStandard Angles
Angle Subtraction Identity
The angle subtraction identity is a crucial tool in trigonometry that helps rewrite complex functions into more manageable forms. It involves the difference between two angles. For cosine, the formula is:
- \( \cos(A - B) = \cos A \cos B + \sin A \sin B \)
Cosine Values
Cosine values are key to solving trigonometric equations and understanding angles. Knowing the exact values for certain standard angles makes it easier to simplify expressions.
Key cosine values include:
Key cosine values include:
- \( \cos 0^{\circ} = 1 \)
- \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)
- \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \)
- \( \cos 60^{\circ} = \frac{1}{2} \)
- \( \cos 90^{\circ} = 0 \)
Sum-To-Product Formulas
The sum-to-product formulas transform the addition or subtraction of two trigonometric functions into a product. This is particularly useful in simplifying trigonometric expressions. The formulas for this are:
- \( \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \)
- \( \cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \)
Standard Angles
Standard angles like \(0^{\circ}\), \(30^{\circ}\), \(45^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\) are fundamental in trigonometry. These angles, commonly seen in first quadrant trigonometric problems, come with well-known sine and cosine values.
For example:
For example:
- \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)
- \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \)
- \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \)
Other exercises in this chapter
Problem 97
$$ \sin 10^{\circ}+\sin 20^{\circ}+\sin 40^{\circ}+\sin 50^{\circ}=\sin 70^{\circ}+\sin 80^{\circ} $$
View solution Problem 98
$$ \cos 20^{\circ}+\cos 100^{\circ}+\cos 140^{\circ}=0 $$
View solution Problem 100
$$ \left[\sin 55^{\circ}-\sin 19^{\circ}\right]+\left[\sin 53^{\circ}-\sin 17^{\circ}\right]=\cos 1^{\circ} $$
View solution Problem 101
$$ \sin \frac{\theta}{2} \sin \frac{7 \theta}{2}+\sin \frac{3 \theta}{2} \sin \frac{11 \theta}{2}=\sin 2 \theta \sin 5 \theta \text { . } $$
View solution