If \(\sin A = \dfrac{2x}{1+x^2}\), what is \(\cos A\)?
\(\dfrac{1-x^2}{1+x^2}\)
Use the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\).
Step 1 – Compute sin²A:
\(\sin^2 A = \left(\dfrac{2x}{1+x^2}\right)^2 = \dfrac{4x^2}{(1+x^2)^2}\)
Step 2 – Find cos²A:
\(\cos^2 A = 1 - \dfrac{4x^2}{(1+x^2)^2} = \dfrac{(1+x^2)^2 - 4x^2}{(1+x^2)^2}\)
Expand the numerator:
\((1+x^2)^2 - 4x^2 = 1 + 2x^2 + x^4 - 4x^2 = 1 - 2x^2 + x^4 = (1 - x^2)^2\)
So:
\(\cos^2 A = \dfrac{(1-x^2)^2}{(1+x^2)^2}\)
Step 3 – Take the square root:
\(\cos A = \dfrac{1-x^2}{1+x^2}\)
Hence \(\cos A = \dfrac{1-x^2}{1+x^2}\).
Quick verification (Pythagorean triple): this is the standard parameterisation where, for a right triangle with opposite = 2x, hypotenuse = 1+x², the adjacent side is \(\sqrt{(1+x^2)^2 - (2x)^2} = 1-x^2\).
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