What is the value of \(\frac{3 \sin 58^{\circ}}{\cos 32^{\circ}}+\frac{3 \sin 42^{\circ}}{\cos 48^{\circ}}\) ?
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Understanding the Trigonometric Expression
The question asks us to find the value of the expression: \( \frac{3 \sin 58^{\circ}}{\cos 32^{\circ}}+\frac{3 \sin 42^{\circ}}{\cos 48^{\circ}} \). This expression involves trigonometric ratios of specific angles. To solve this, we need to look for relationships between the angles given in the expression.
Identifying Complementary Angles
Let's examine the angles in each term:
- In the first term, we have \( 58^{\circ} \) and \( 32^{\circ} \). Notice that \( 58^{\circ} + 32^{\circ} = 90^{\circ} \). Angles that add up to \( 90^{\circ} \) are called complementary angles.
- In the second term, we have \( 42^{\circ} \) and \( 48^{\circ} \). Notice that \( 42^{\circ} + 48^{\circ} = 90^{\circ} \). These are also complementary angles.
Using Trigonometric Identities for Complementary Angles
There are standard trigonometric identities for complementary angles. For any angle \( \theta \), we have:
- \( \sin (90^{\circ} - \theta) = \cos \theta \)
- \( \cos (90^{\circ} - \theta) = \sin \theta \)
- \( \tan (90^{\circ} - \theta) = \cot \theta \)
- \( \cot (90^{\circ} - \theta) = \tan \theta \)
- \( \sec (90^{\circ} - \theta) = \csc \theta \)
- \( \csc (90^{\circ} - \theta) = \sec \theta \)
We can use the identities involving sine and cosine to simplify the given expression.
Simplifying the First Term
The first term is \( \frac{3 \sin 58^{\circ}}{\cos 32^{\circ}} \).
We know that \( 58^{\circ} = 90^{\circ} - 32^{\circ} \). Using the identity \( \sin (90^{\circ} - \theta) = \cos \theta \), we can write:
\[ \sin 58^{\circ} = \sin (90^{\circ} - 32^{\circ}) = \cos 32^{\circ} \]Now substitute this back into the first term:
\[ \frac{3 \sin 58^{\circ}}{\cos 32^{\circ}} = \frac{3 \cos 32^{\circ}}{\cos 32^{\circ}} \]Assuming \( \cos 32^{\circ} \neq 0 \) (which is true since \( 32^{\circ} \) is not a multiple of \( 90^{\circ} + 90^{\circ}k \)), we can cancel out \( \cos 32^{\circ} \):
\[ \frac{3 \cos 32^{\circ}}{\cos 32^{\circ}} = 3 \times 1 = 3 \]So, the value of the first term is 3.
Simplifying the Second Term
The second term is \( \frac{3 \sin 42^{\circ}}{\cos 48^{\circ}} \).
We know that \( 42^{\circ} = 90^{\circ} - 48^{\circ} \). Using the identity \( \sin (90^{\circ} - \theta) = \cos \theta \), we can write:
\[ \sin 42^{\circ} = \sin (90^{\circ} - 48^{\circ}) = \cos 48^{\circ} \]Now substitute this back into the second term:
\[ \frac{3 \sin 42^{\circ}}{\cos 48^{\circ}} = \frac{3 \cos 48^{\circ}}{\cos 48^{\circ}} \]Assuming \( \cos 48^{\circ} \neq 0 \) (which is true since \( 48^{\circ} \) is not a multiple of \( 90^{\circ} + 90^{\circ}k \)), we can cancel out \( \cos 48^{\circ} \):
\[ \frac{3 \cos 48^{\circ}}{\cos 48^{\circ}} = 3 \times 1 = 3 \]So, the value of the second term is 3.
Calculating the Final Value
The original expression is the sum of these two terms:
\[ \frac{3 \sin 58^{\circ}}{\cos 32^{\circ}}+\frac{3 \sin 42^{\circ}}{\cos 48^{\circ}} = 3 + 3 = 6 \]Therefore, the value of the given trigonometric expression is 6.
Summary of Steps
- Recognize that the angles in each fraction are complementary.
- Use the identity \( \sin (90^{\circ} - \theta) = \cos \theta \) to rewrite the numerator of each fraction in terms of the denominator's angle.
- Simplify each fraction.
- Add the simplified values to get the final result.
| Term | Angles | Relationship | Identity Used | Simplification | Value |
|---|---|---|---|---|---|
| \( \frac{3 \sin 58^{\circ}}{\cos 32^{\circ}} \) | \( 58^{\circ}, 32^{\circ} \) | \( 58^{\circ} = 90^{\circ} - 32^{\circ} \) | \( \sin 58^{\circ} = \cos 32^{\circ} \) | \( \frac{3 \cos 32^{\circ}}{\cos 32^{\circ}} \) | 3 |
| \( \frac{3 \sin 42^{\circ}}{\cos 48^{\circ}} \) | \( 42^{\circ}, 48^{\circ} \) | \( 42^{\circ} = 90^{\circ} - 48^{\circ} \) | \( \sin 42^{\circ} = \cos 48^{\circ} \) | \( \frac{3 \cos 48^{\circ}}{\cos 48^{\circ}} \) | 3 |
Revision Table: Trigonometric Identities
| Identity | Description |
|---|---|
| \( \sin(90^{\circ} - \theta) = \cos \theta \) | Sine of an angle is equal to the cosine of its complement. |
| \( \cos(90^{\circ} - \theta) = \sin \theta \) | Cosine of an angle is equal to the sine of its complement. |
| \( \tan(90^{\circ} - \theta) = \cot \theta \) | Tangent of an angle is equal to the cotangent of its complement. |
| \( \cot(90^{\circ} - \theta) = \tan \theta \) | Cotangent of an angle is equal to the tangent of its complement. |
Additional Information: Applications of Complementary Angles
The concept of complementary angles and their trigonometric identities is fundamental in trigonometry. It is often used to simplify expressions, solve equations, and prove other identities. These relationships are derived directly from the definitions of trigonometric ratios in a right-angled triangle. If one acute angle in a right triangle is \( \theta \), the other acute angle is \( 90^{\circ} - \theta \). The side opposite \( \theta \) is adjacent to \( 90^{\circ} - \theta \), and the side adjacent to \( \theta \) is opposite \( 90^{\circ} - \theta \). This geometric relationship leads directly to the identities.
Important Questions from Trigonometric Ratios and Identities
If cosec θ = 13/12, then sin θ + cos θ - tan θ is equal to:
If \(\frac{{\tan \theta + \sin \theta }}{{\tan \theta - \sin \theta }} = \frac{{k + 1}}{{k - 1}},\) then k = ?
If α + β = 90° and α = 2β, then the value of 3 cos 2 α - 2 sin 2 β is equal to:
If \(\sqrt{3}\) tan θ = 3 sin θ, then what is the value of sin 2θ − cos 2θ ?
- \(\frac{(1 + tan\theta + sec\theta)(1 + cot\theta - cosec\theta)}{(sec\theta + tan\theta)(1 - sin\theta)}\) is equal to:
