If 3 sin θ + 5 cos θ = 5, then what is the value of 5 sin θ

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CDS - 2024

Mathematics

If $3 \sin \theta + 5 \cos \theta = 5$, then what is the value of $5 \sin \theta - 3 \cos \theta$?

This question was previously asked in

CDS 2 2024 Maths Question Paper (01-Sep-2024)

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$ -3$

$ -2$

$5$

$8$

The correct answer is

$ -3$

Solving the Trigonometric Equation

We are given a trigonometric equation and asked to find the value of a related expression. The given equation is:

$3 \sin \theta + 5 \cos \theta = 5 \quad \quad (1)$

We need to find the value of the expression:

$5 \sin \theta - 3 \cos \theta$

Let's denote the value we need to find as $x$.

$5 \sin \theta - 3 \cos \theta = x \quad \quad (2)$

Algebraic Manipulation Strategy

A common technique to solve problems of this type involves squaring both equations and combining them. This utilizes the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$.

Step 1: Square Equation (1)

Squaring both sides of equation (1):

$(3 \sin \theta + 5 \cos \theta)^2 = 5^2$

Expanding 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 \quad (3) $

Step 2: Square Equation (2)

Squaring both sides of equation (2):

$(5 \sin \theta - 3 \cos \theta)^2 = x^2$

Expanding 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 \quad (4) $

Step 3: Combine the Squared Equations

Now, let's add equation (3) and equation (4):

$(9 \sin^2 \theta + 30 \sin \theta \cos \theta + 25 \cos^2 \theta) + (25 \sin^2 \theta - 30 \sin \theta \cos \theta + 9 \cos^2 \theta) = 25 + x^2$

Notice that the cross terms ($30 \sin \theta \cos \theta$ and $-30 \sin \theta \cos \theta$) cancel each other out:

$ 9 \sin^2 \theta + 25 \sin^2 \theta + 25 \cos^2 \theta + 9 \cos^2 \theta = 25 + x^2 $

Group the $\sin^2 \theta$ and $\cos^2 \theta$ terms:

$ (9+25) \sin^2 \theta + (25+9) \cos^2 \theta = 25 + x^2 $

$ 34 \sin^2 \theta + 34 \cos^2 \theta = 25 + x^2 $

Step 4: Apply the Pythagorean Identity

Factor out 34 and use the identity $\sin^2 \theta + \cos^2 \theta = 1$:

$ 34 (\sin^2 \theta + \cos^2 \theta) = 25 + x^2 $

$ 34 (1) = 25 + x^2 $

$ 34 = 25 + x^2 $

Step 5: Solve for $x$

Rearrange the equation to solve for $x^2$:

$ x^2 = 34 - 25 $

$ x^2 = 9 $

Taking the square root of both sides gives:

$ x = \pm \sqrt{9} $

$ x = \pm 3 $

Determining the Specific Value

We have two possible values for $x$: $3$ and $-3$. To find the specific value, let's examine the original equation $3 \sin \theta + 5 \cos \theta = 5$.

Consider the case where $\cos \theta = 1$. For this to be possible, $\sin \theta$ must be $0$ (since $\sin^2 \theta + \cos^2 \theta = 1$).

Let's check if $\sin \theta = 0$ and $\cos \theta = 1$ satisfy the given equation:

$ 3(0) + 5(1) = 0 + 5 = 5 $

This pair of values ($\sin \theta = 0, \cos \theta = 1$) satisfies the given condition.

Now, substitute these values into the expression we want to find ($5 \sin \theta - 3 \cos \theta$):

$ 5(0) - 3(1) = 0 - 3 = -3 $

Therefore, the value of $5 \sin \theta - 3 \cos \theta$ is $-3$. This matches one of the possible values we found for $x$.

Final Answer Summary

The steps involved using algebraic manipulation and the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ to find possible values. By testing a specific valid case ($\sin \theta = 0, \cos \theta = 1$), we determined the unique value of the expression $5 \sin \theta - 3 \cos \theta$ to be $-3$.

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