If sinθ = (m 2 – n 2 )/(m 2 + n 2 ) and 0 < θ < π/2, then what is the value of cosθ?

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CDS - 2017

Mathematics

If sinθ = (m ²– n ²)/(m ²+ n ²) and 0 < θ < π/2, then what is the value of cosθ?

This question was previously asked in

CDS I 2017 General Knowledge Previous Year Paper (05-Feb-2017)

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2mn/(m ²+ n ²)

2mn/(m ²- n ²)

(m ²+ n ²)/2mn

(m ²- n ²)/2mn

The correct answer is

2mn/(m ²+ n ²)

Finding cos(θ) Given sin(θ)

This problem requires us to find the value of cosine (&cosθ) when the value of sine (&sinθ) is given, along with the range for the angle θ. We are given:

sinθ = $\frac{m^{2} - n^{2}}{m^{2} + n^{2}} $$
The angle range is 0 < θ < π/2.

The condition 0 < θ < π/2 tells us that the angle θ lies in the first quadrant. In the first quadrant, both the sine and cosine values are positive.

Using the Fundamental Trigonometric Identity

We can use the fundamental Pythagorean identity which relates sine and cosine:

$\sin^{2} θ + \cos^{2} θ = 1 $$

We need to solve for cosθ. Rearranging the identity, we get:

$\cos^{2} θ = 1 - \sin^{2} θ $$

Substituting the Value of sin(θ)

Now, substitute the given value of sinθ into the equation:

$\cos^{2} θ = 1 - {(\frac{m^{2} - n^{2}}{m^{2} + n^{2}})}^{2} $$

Algebraic Simplification

To simplify, find a common denominator:

$\cos^{2} θ = \frac{{(m^{2} + n^{2})}^{2}}{{(m^{2} + n^{2})}^{2}} - \frac{{(m^{2} - n^{2})}^{2}}{{(m^{2} + n^{2})}^{2}} $$

Combine the fractions:

$\cos^{2} θ = \frac{{(m^{2} + n^{2})}^{2} - {(m^{2} - n^{2})}^{2}}{{(m^{2} + n^{2})}^{2}} $$

Use the algebraic identity a² - b² = (a - b)(a + b) for the numerator, where a = (m² + n²) and b = (m² - n²):

Numerator = [(m² + n²) - (m² - n²)] * [(m² + n²) + (m² - n²)]

Simplify the terms inside the brackets:

Numerator = [m² + n² - m² + n²] * [m² + n² + m² - n²]

Numerator = [2n²] * [2m²]

Numerator = 4m²n²

Substitute this back into the equation for cos²θ:

$\cos^{2} θ = \frac{4 m^{2} n^{2}}{{(m^{2} + n^{2})}^{2}} $$

Finding the Value of cos(θ)

Now, take the square root of both sides to find cosθ:

$\cos θ = \sqrt{\frac{4 m^{2} n^{2}}{{(m^{2} + n^{2})}^{2}}} $$

$\cos θ = \frac{2 mn}{m^{2} + n^{2}} $$

Since we established that θ is in the first quadrant (0 < θ < π/2), the value of cosθ must be positive. The result $\frac{2 mn}{m^{2} + n^{2}} $ is positive (assuming m and n are such that the original sin value is valid, typically m > n > 0 for standard triangle definitions, but the identity holds generally).$

Conclusion

Therefore, the value of cosθ is $\frac{2 mn}{m^{2} + n^{2}} $. This matches the first option.$

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