Question

If \(x\) lies in the 2 nd quadrant and \(\tan x=\frac{-4}{3}\), then \(\sin \frac{x}{2} \cos \frac{x}{2}=\) (a) \(\frac{2}{5}\) (b) \(\frac{2}{\sqrt{5}}\) (c) \(\frac{\sqrt{5}}{2}\) (d) \(\frac{1}{2}\)

Short answer

Answer: The correct answer for the product of sine x/2 and cosine x/2 is \(\frac{4}{5}\).

Step-by-step solution

Find the values of sin x and cos x

Since \(\tan x = \frac{-4}{3}\), that means \(\frac{\sin x}{\cos x} = \frac{-4}{3}\). Let's call the opposite side of the angle \(x\) in a right triangle as \(a\) and the adjacent side as \(b\). Then, we can use the Pythagorean theorem to find the hypotenuse \(c\) (the side opposite to the right angle) as follows: \(a = -4\) (The side opposite the angle x is negative in the 2nd quadrant) \(b = 3\) \(a^2 + b^2 = c^2\) \((-4)^2 + 3^2 = c^2\) \(16 + 9 = c^2\) \(25 = c^2\) \(c = 5\) Now we can find the values of \(\sin x\) and \(\cos x\): \(\sin x = \frac{a}{c} = \frac{-4}{5}\) \(\cos x = \frac{b}{c} = \frac{3}{5}\)

Find the values of sin x/2 and cos x/2 using the half-angle formulas

We can use the half-angle formulas for sine and cosine: \(\sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}}\) \(\cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}}\) Since \(x\) is in the 2nd quadrant, \(\frac{x}{2}\) is in the 1st quadrant, where the sine and cosine functions are both positive. Hence, we need the positive square roots: \(\sin \frac{x}{2} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{2}{5}}\) \(\cos \frac{x}{2} = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \sqrt{\frac{8}{5}}\)

Calculate the product of sin x/2 and cos x/2

Finally, we can calculate the product of \(\sin \frac{x}{2}\) and \(\cos \frac{x}{2}\): \(\sin \frac{x}{2} \cos \frac{x}{2} = \sqrt{\frac{2}{5}} \cdot \sqrt{\frac{8}{5}} = \sqrt{\frac{2 \cdot 8}{5 \cdot 5}} = \sqrt{\frac{16}{25}} = \frac{4}{5}\) It seems there is no matching answer among the choices. The correct answer should be \(\frac{4}{5}\), which is not listed among the given options.

Key concepts

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the variable they contain. These identities are fundamental to the study of trigonometry, facilitating the simplification and computation of complex trigonometric expressions and equations.

For instance, one of the most basic identities is the Pythagorean identity, which is derived from the Pythagorean theorem and illustrates the relationship between the sine and cosine of an angle: \[ \text{sin}^2 x + \text{cos}^2 x = 1 \] This identity is key in finding the values of the sine and cosine functions when one of them is known, especially when solving problems set in the context of a right triangle like in our original exercise.

Half-Angle Formulas

The half-angle formulas are useful when we are required to calculate the sine, cosine, or tangent of half an angle, a situation that often emerges in geometric problems. These formulas can be derived from the trigonometric identities and express the trigonometric functions of a half-angle in terms of the square roots of expressions involving trigonometric functions of the original angle.

For example, the half-angle formulas for sine and cosine are given by: \[ \text{sin} \frac{x}{2} = \text{\pm} \text{sqrt}\frac{1 - \text{cos} x}{2} \] \[ \text{cos} \frac{x}{2} = \text{\pm} \text{sqrt}\frac{1 + \text{cos} x}{2} \] Applying the correct sign based on the quadrant where the half-angle is located is crucial. In our exercise, since \(x\) lies in the second quadrant, where sine is positive, we could deduce the sign for the half-angles immediately.

Pythagorean Theorem

The Pythagorean theorem is a cornerstone of geometry, relating the lengths of the sides of a right-angled triangle. Stated simply, the theorem says that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

This theorem can be expressed mathematically as: \[ a^2 + b^2 = c^2 \] Here, \(a\) and \(b\) are the lengths of the legs, and \(c\) is the length of the hypotenuse. In trigonometric terms, when considering a unit circle or a triangle formed with a radius (hypotenuse) of 1, the legs represent the values of sine and cosine for a specific angle, tying back into trigonometric identities. During the exercise, we used the Pythagorean theorem to deduce the hypotenuse of a triangle when given the lengths of the other sides. Doing this allowed us to define the sine and cosine values for the given angle, \(x\), before applying half-angle formulas.