Problem 20 - Algebra 2 and Trigonometry

Question

Use \(\cos A=\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2}\) to show that the exact value of \(\tan \frac{\pi}{8}=\sqrt{2}-1\)

Step-by-Step Solution

Answer

Thus, \( \tan \frac{\pi}{8} = \sqrt{2} - 1 \).

Step 1: Express \\ \tan \\ using half-angle identity

The identity for the tangent of a half-angle is given by \( \tan \frac{A}{2} = \pm \sqrt{\frac{1 - \cos A}{1 + \cos A}} \). We want to apply this identity using \( A = \frac{\pi}{4} \) since \( \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \).

Step 2: Substitute \\ \cos A \\ into the identity

Substitute \( \cos A = \frac{\sqrt{2}}{2} \) into the half-angle identity: \[\tan \frac{\pi}{8} = \pm \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}}\].

Step 3: Simplify the expression inside the square root

Calculate the numerator and denominator separately:- Numerator: \( 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2} \)- Denominator: \( 1 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2} \)This transforms the expression as follows:\[\tan \frac{\pi}{8} = \pm \sqrt{\frac{\frac{2-\sqrt{2}}{2}}{\frac{2+\sqrt{2}}{2}}} = \pm \sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}\].

Step 4: Simplify further by rationalizing the denominator

Multiply the numerator and the denominator by \( 2 - \sqrt{2} \), the conjugate of the denominator:\[\frac{2 - \sqrt{2}}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \frac{(2-\sqrt{2})^2}{4-2} = \frac{4 - 4\sqrt{2} + 2}{2} = \frac{2-2\sqrt{2}}{2}\]. This simplifies to \( 1 - \sqrt{2} \).

Step 5: Determine the correct sign

Since \( \frac{\pi}{8} \) is in the first quadrant, where tangent is positive, we choose the positive sign: \( \tan \frac{\pi}{8} = \sqrt{2} - 1 \).

Key Concepts

Half-Angle Identity

The half-angle identity is a useful tool in trigonometry. It allows us to find the tangent of half an angle if we know the cosine of the original angle. In general, the identity is expressed as: \[ \tan \frac{A}{2} = \pm \sqrt{\frac{1 - \cos A}{1 + \cos A}} \] This equation can seem a bit intimidating at first, but it's quite straightforward once you break it down.

• The \( 1 - \cos A \) creates the numerator of a fraction.
• The \( 1 + \cos A \) stands as the denominator.

After arranging these terms inside the square root, you will either take the positive or negative root, based on the angle's quadrant. For the first quadrant, as in our exercise with \( \frac{\pi}{8} \), where sine and tangent are positive, we choose the positive root.

Tangent Function

The tangent function, \( \tan \), is one of the primary trigonometric functions. It relates the angles of a right triangle to the ratios of its sides. Specifically: \[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \] This is helpful when dealing with angles in standard position, especially with trigonometric identities. For angles not commonly appearing in triangles like \( \frac{\pi}{8} \), identities are crucial. The half-angle identity lets us compute the value of \( \tan \frac{\pi}{8} \) without a calculator, using known values from standard angles. The understanding and application of these identities are key when you dive deeper into trigonometry.

Rationalizing the Denominator

Rationalizing the denominator is a technique used to make expressions easier to work with by eliminating radicals from denominators. This process involves multiplying the numerator and denominator by the conjugate of the denominator. In our exercise:

• We had \( \frac{2-\sqrt{2}}{2+\sqrt{2}} \).
• We wanted to rationalize this by multiplying by the conjugate \( 2-\sqrt{2} \).

The key idea is that multiplying conjugates results in a rational number (because it eliminates the radical). The form of a conjugate difference, \( a-b \) and \( a+b \), when multiplied, becomes \( a^2 - b^2 \), a convenient formula that simplifies calculations, making the expression easier to work with and understand.