Find values of the sine and cosine functions for each angle measure.
2x, given tan x = 5/3 and sin x < 0
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Identify the given information: \( \tan x = \frac{5}{3} \) and \( \sin x < 0 \). This tells us the tangent ratio and the sign of the sine function, which helps determine the quadrant of angle \( x \).
Recall that \( \tan x = \frac{\sin x}{\cos x} \). Since \( \tan x = \frac{5}{3} \), we can represent \( \sin x = 5k \) and \( \cos x = 3k \) for some positive constant \( k \).
Use the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) to find \( k \). Substitute \( \sin x = 5k \) and \( \cos x = 3k \) into the identity: \( (5k)^2 + (3k)^2 = 1 \).
Solve for \( k \) from the equation \( 25k^2 + 9k^2 = 1 \), which simplifies to \( 34k^2 = 1 \), then find \( k = \pm \frac{1}{\sqrt{34}} \).
Determine the correct sign of \( k \) using the condition \( \sin x < 0 \). Since \( \sin x = 5k \), \( k \) must be negative. Then calculate \( \sin 2x \) and \( \cos 2x \) using the double-angle formulas: \( \sin 2x = 2 \sin x \cos x \) and \( \cos 2x = \cos^2 x - \sin^2 x \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Trigonometric Ratios and Their Relationships
Trigonometric ratios (sine, cosine, tangent) relate the angles of a right triangle to the ratios of its sides. Knowing one ratio, such as tangent, allows calculation of sine and cosine using identities or the Pythagorean theorem, essential for finding values of sine and cosine from tangent.
Double-angle formulas express trigonometric functions of 2x in terms of functions of x. For sine and cosine, these are sin(2x) = 2 sin x cos x and cos(2x) = cos² x - sin² x, enabling calculation of sine and cosine for 2x once sin x and cos x are known.
The signs of sine, cosine, and tangent depend on the angle's quadrant. Given sin x < 0 and tan x = 5/3, the angle x lies in the third quadrant where sine and cosine are negative, but tangent is positive. This information is crucial to assign correct signs to sine and cosine values.