Question

The double-angle identity for tangent in terms of the tangent function is \(\tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x}\) a) Verify numerically that this equation is true for \(x=\frac{\pi}{6}\) b) The expression tan \(2 x\) can also be written using the quotient identity for tangent: \(\tan 2 x=\frac{\sin 2 x}{\cos 2 x} .\) Verify this equation numerically when \(x=\frac{\pi}{6}\). c) The expression \(\frac{\sin 2 x}{\cos 2 x}\) from part b) can be expressed as \(\frac{2 \sin x \cos x}{\cos ^{2} x-\sin ^{2} x}\) using double-angle identities. Show how the expression for tan \(2 x\) used in part a) can also be rewritten in the form \(\frac{2 \sin x \cos x}{\cos ^{2} x-\sin ^{2} x}\).

Short answer

For \(x = \frac{\pi}{6}\), the double-angle identity for tangent holds true numerically, and can also be verified using the quotient identity and double-angle trigonometric identities.

Step-by-step solution

Verify the double-angle identity for tangent

Use the given double-angle identity for tangent to verify the equation. For this, substitute the value of \(x = \frac{\pi}{6}\) into the equation.\[ \tan 2 \left(\frac{\pi}{6}\right) = \frac{2 \tan\left(\frac{\pi}{6}\right)}{1-\tan^2\left(\frac{\pi}{6}\right)} \] \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\)\[ \tan\left(\frac{\pi}{3}\right) = \frac{2 \times \frac{1}{\sqrt{3}}}{1 - \left(\frac{1}{\sqrt{3}}\right)^2} = \frac{2 \times \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{2 \times \frac{1}{\sqrt{3}}}{\frac{2}{3}} = \frac{2 \times \frac{1}{\sqrt{3}}}{\frac{2}{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \] Therefore, the equation is verified as true.

Verify using quotient identity for tangent

Express \(\tan 2x\) using the quotient identity. For this, calculate \( \sin 2x \) and \( \cos 2x \) and substitute \( x = \frac{\pi}{6} \). \[ \sin(2x) = 2 \sin x \cos x \] \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \) and \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \) \[ \sin 2\left(\frac{\pi}{6}\right) = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \] \[ \cos(2x) = \cos^2 x - \sin^2 x \] \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \) and \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \) \[ \cos 2\left(\frac{\pi}{6}\right) = \left( \frac{\sqrt{3}}{2} \right)^2 - \left( \frac{1}{2} \right)^2 = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \] Therefore, \[ \tan 2x = \frac{\sin 2x}{\cos 2x} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \] This verifies the equation for the quotient identity.

Rewrite tan 2x using double-angle identities

Express \(\frac{\sin 2x}{\cos 2x}\) using the identities: \( \sin 2x = 2 \sin x \cos x\) and \( \cos 2x = \cos^2 x - \sin^2 x \) and show they match the form \(\frac{2 \sin x \cos x}{\cos^2 x - \sin^2 x}\). \[ \frac{\sin 2x}{\cos 2x} = \frac{2 \sin x \cos x}{\cos^2 x - \sin^2 x} \] Since \(\tan 2 x = \frac{2 \tan x}{1-\tan ^2 x}\) and \( \tan x = \frac{\sin x}{\cos x}\), substitute \( \sin x = \tan x \cos x \). The original form then becomes: \[ \frac{2 \sin x \cos x}{\cos^2 x - \sin^2 x} \] This shows that all forms can ultimately be expressed equivalently.

Key concepts

tangent function

The tangent function is one of the basic functions in trigonometry. Mathematically, it's defined using the sine and cosine functions. Specifically, the tangent of an angle x, denoted as \(\tan x\), is given by the ratio of the sine and cosine of that angle: \( \tan x = \frac{\text{sin} x}{\text{cos} x} \). This makes it very useful for various trigonometric calculations.
Tangent has some unique properties, such as being periodic with a period of \( \frac{\text{π}}{2} \) and having asymptotes that occur at odd multiples of \( \frac{\text{π}}{2} \). These properties help us understand many trigonometric relationships and are essential when working with trigonometric identities.

quotient identity

The quotient identity in trigonometry uses the definitions of sine and cosine to express the tangent function. Essentially, it states that \( \tan x = \frac{\text{sin} x}{\text{cos} x} \).
This identity is fundamental because it links the tangent function directly to the two primary trigonometric functions: sine and cosine.
Understanding this relationship helps simplify complex trigonometric expressions and solve problems more efficiently.
In part b of the exercise, we verify the identity \( \tan 2x = \frac{\text{sin} 2x}{\text{cos} 2x} \) through numerical substitution, reinforcing the quotient identity.

trigonometric verification

Trigonometric verification involves showing that two different trigonometric expressions are equivalent. It often requires using various identities and substituting values to ensure both sides of an equation are identical.
In this exercise, we verified the double-angle identity for tangent by substituting \( x = \frac{\text{π}}{6} \) and showing both sides equal \( \tan 2 \times \frac{\text{π}}{6} = \frac{2 \tan \frac{\text{π}}{6}}{1 - \tan^2 \frac{\text{π}}{6}} = \text{√3} \).
Similarly, we verified \( \frac{\text{sin} 2x}{\text{cos} 2x} = \tan 2x \) using exact values obtained from sine and cosine functions.

precalculus problems

Precalculus problems often involve the application of trigonometric identities to simplify expressions or solve equations. These problems lay the groundwork for more advanced studies in calculus.
The problem in the exercise involves using multiple trigonometric identities, such as the double-angle and quotient identities.
By verifying these identities numerically and algebraically, students strengthen their understanding of fundamental trigonometric relationships.
Solving these types of problems is crucial for success in calculus because they promote a deeper comprehension of how different trigonometric functions interconnect.