If 3 sin θ + 5 cos θ = 5, then what is the value of 5 sin θ - 3 cos θ equal to ?
-3
Solving Trigonometric Equations to Find Value
We are given a trigonometric equation involving sine and cosine of an angle \( \theta \), and we need to find the value of another expression involving sine and cosine of the same angle.
The given equation is:
\( 3 \sin \theta + 5 \cos \theta = 5 \quad \text{(Equation 1)} \)
We need to find the value of the expression:
\( 5 \sin \theta - 3 \cos \theta \)
Let's assume the value of this expression is \( x \):
\( 5 \sin \theta - 3 \cos \theta = x \quad \text{(Equation 2)} \)
A common technique when dealing with expressions of the form \( a \sin \theta + b \cos \theta \) and \( b \sin \theta - a \cos \theta \) is to square both expressions and add them. This utilizes the identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
Squaring Equation 1
Square both sides of Equation 1:
\( (3 \sin \theta + 5 \cos \theta)^2 = 5^2 \)
Expand the left side using \( (a+b)^2 = a^2 + 2ab + b^2 \):
\( (3 \sin \theta)^2 + 2(3 \sin \theta)(5 \cos \theta) + (5 \cos \theta)^2 = 25 \)
\( 9 \sin^2 \theta + 30 \sin \theta \cos \theta + 25 \cos^2 \theta = 25 \quad \text{(Equation 3)} \)
Squaring Equation 2
Square both sides of Equation 2:
\( (5 \sin \theta - 3 \cos \theta)^2 = x^2 \)
Expand the left side using \( (a-b)^2 = a^2 - 2ab + b^2 \):
\( (5 \sin \theta)^2 - 2(5 \sin \theta)(3 \cos \theta) + (3 \cos \theta)^2 = x^2 \)
\( 25 \sin^2 \theta - 30 \sin \theta \cos \theta + 9 \cos^2 \theta = x^2 \quad \text{(Equation 4)} \)
Adding Equation 3 and Equation 4
Now, add Equation 3 and Equation 4 together:
\( (9 \sin^2 \theta + 30 \sin \theta \cos \theta + 25 \cos^2 \theta) + (25 \sin^2 \theta - 30 \sin \theta \cos \theta + 9 \cos^2 \theta) = 25 + x^2 \)
Combine the terms:
- Terms with \( \sin^2 \theta \): \( 9 \sin^2 \theta + 25 \sin^2 \theta = (9+25) \sin^2 \theta = 34 \sin^2 \theta \)
- Terms with \( \cos^2 \theta \): \( 25 \cos^2 \theta + 9 \cos^2 \theta = (25+9) \cos^2 \theta = 34 \cos^2 \theta \)
- Terms with \( \sin \theta \cos \theta \): \( +30 \sin \theta \cos \theta - 30 \sin \theta \cos \theta = 0 \)
So, the sum becomes:
\( 34 \sin^2 \theta + 34 \cos^2 \theta = 25 + x^2 \)
Using the Pythagorean Identity
Factor out 34 from the left side:
\( 34 (\sin^2 \theta + \cos^2 \theta) = 25 + x^2 \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( 34 (1) = 25 + x^2 \)
\( 34 = 25 + x^2 \)
Solving for x
Rearrange the equation to solve for \( x^2 \):
\( x^2 = 34 - 25 \)
\( x^2 = 9 \)
Take the square root of both sides:
\( x = \pm \sqrt{9} \)
\( x = \pm 3 \)
Thus, the value of \( 5 \sin \theta - 3 \cos \theta \) can be either \( 3 \) or \( -3 \).
Based on the options provided, the possible value is \( -3 \).
Revision Table: Key Trigonometric Identities
| Identity Type | Identity |
|---|---|
| Pythagorean Identity | \( \sin^2 \theta + \cos^2 \theta = 1 \) |
| Pythagorean Identity | \( 1 + \tan^2 \theta = \sec^2 \theta \) |
| Pythagorean Identity | \( 1 + \cot^2 \theta = \csc^2 \theta \) |
| Expansion Formula | \( (a+b)^2 = a^2 + 2ab + b^2 \) |
| Expansion Formula | \( (a-b)^2 = a^2 - 2ab + b^2 \) |
Additional Information: Alternative Approaches
While squaring and adding is an effective method here, other approaches exist for solving trigonometric equations or finding values of related expressions:
- Converting to a Single Trigonometric Function: Expressions like \( a \sin \theta + b \cos \theta \) can be written in the form \( R \sin (\theta + \alpha) \) or \( R \cos (\theta - \beta) \), where \( R = \sqrt{a^2 + b^2} \). This can sometimes simplify solving equations. In this case, \( 3 \sin \theta + 5 \cos \theta = \sqrt{3^2+5^2} \sin (\theta + \alpha) = \sqrt{34} \sin (\theta + \alpha) \). So, \( \sqrt{34} \sin (\theta + \alpha) = 5 \). Similarly, \( 5 \sin \theta - 3 \cos \theta \) relates to \( \sqrt{5^2+(-3)^2} \cos (\theta + \gamma) = \sqrt{34} \cos (\theta + \gamma) \). This approach might involve finding the values of \( \alpha \) and \( \gamma \) and relating them.
- Solving for \( \sin \theta \) and \( \cos \theta \): In some cases, you might be able to set up a system of equations if you had two independent linear equations in terms of \( \sin \theta \) and \( \cos \theta \). However, here we have one linear equation and need to find a value of another linear expression. The squaring method effectively bypasses finding the individual values of \( \sin \theta \) and \( \cos \theta \).
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