Finding cos(α + 2β) Given Trigonometric Ratios

We are given the values of \(\tan \alpha\) and \(\sin \beta\) for acute angles \(\alpha\) and \(\beta\). Our goal is to find the value of \(\cos(\alpha + 2\beta)\). To do this, we will first find the necessary trigonometric ratios for \(\alpha\) and \(\beta\), then calculate the ratios for \(2\beta\), and finally use the angle addition formula for cosine.

Step 1: Find Trigonometric Ratios for α

We are given \(\tan \alpha = \frac{1}{7}\) and \(0 < \alpha < \frac{\pi}{2}\). Since \(\alpha\) is in the first quadrant, all trigonometric ratios for \(\alpha\) are positive. We can use the identity \(1 + \tan^2 \alpha = \sec^2 \alpha\) to find \(\sec \alpha\), and then \(\cos \alpha\).

\(\sec^2 \alpha = 1 + \left(\frac{1}{7}\right)^2 = 1 + \frac{1}{49} = \frac{49 + 1}{49} = \frac{50}{49}\)

\(\sec \alpha = \sqrt{\frac{50}{49}} = \frac{\sqrt{50}}{\sqrt{49}} = \frac{5\sqrt{2}}{7}\) (since \(\alpha\) is acute, \(\sec \alpha\) is positive)

\(\cos \alpha = \frac{1}{\sec \alpha} = \frac{1}{\frac{5\sqrt{2}}{7}} = \frac{7}{5\sqrt{2}}\)

Now we can find \(\sin \alpha\) using \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\):

\(\sin \alpha = \tan \alpha \cdot \cos \alpha = \frac{1}{7} \cdot \frac{7}{5\sqrt{2}} = \frac{1}{5\sqrt{2}}\)

So, for \(\alpha\):

Step 2: Find Trigonometric Ratios for β

We are given \(\sin \beta = \frac{1}{\sqrt{10}}\) and \(0 < \beta < \frac{\pi}{2}\). Since \(\beta\) is in the first quadrant, \(\cos \beta\) will be positive. We can use the identity \(\sin^2 \beta + \cos^2 \beta = 1\) to find \(\cos \beta\).

\(\cos^2 \beta = 1 - \sin^2 \beta = 1 - \left(\frac{1}{\sqrt{10}}\right)^2 = 1 - \frac{1}{10} = \frac{10 - 1}{10} = \frac{9}{10}\)

\(\cos \beta = \sqrt{\frac{9}{10}} = \frac{\sqrt{9}}{\sqrt{10}} = \frac{3}{\sqrt{10}}\) (since \(\beta\) is acute, \(\cos \beta\) is positive)

So, for \(\beta\):

Step 3: Find Trigonometric Ratios for 2β

We need \(\cos(2\beta)\) and \(\sin(2\beta)\). We can use the double angle formulas:

• \(\cos(2\beta) = \cos^2 \beta - \sin^2 \beta\) or \(1 - 2\sin^2 \beta\) or \(2\cos^2 \beta - 1\)
• \(\sin(2\beta) = 2 \sin \beta \cos \beta\)

Using the values from Step 2:

\(\cos(2\beta) = 1 - 2\sin^2 \beta = 1 - 2 \left(\frac{1}{\sqrt{10}}\right)^2 = 1 - 2 \left(\frac{1}{10}\right) = 1 - \frac{2}{10} = 1 - \frac{1}{5} = \frac{5 - 1}{5} = \frac{4}{5}\)

\(\sin(2\beta) = 2 \sin \beta \cos \beta = 2 \cdot \left(\frac{1}{\sqrt{10}}\right) \cdot \left(\frac{3}{\sqrt{10}}\right) = 2 \cdot \frac{3}{10} = \frac{6}{10} = \frac{3}{5}\)

So, for \(2\beta\):

Step 4: Calculate cos(α + 2β)

Now we use the angle addition formula for cosine:

\(\cos(A + B) = \cos A \cos B - \sin A \sin B\)

Substituting \(A = \alpha\) and \(B = 2\beta\):

\(\cos(\alpha + 2\beta) = \cos \alpha \cos(2\beta) - \sin \alpha \sin(2\beta)\)

Using the values calculated in Step 1 and Step 3:

\(\cos(\alpha + 2\beta) = \left(\frac{7}{5\sqrt{2}}\right) \left(\frac{4}{5}\right) - \left(\frac{1}{5\sqrt{2}}\right) \left(\frac{3}{5}\right)\)

\(\cos(\alpha + 2\beta) = \frac{7 \cdot 4}{5\sqrt{2} \cdot 5} - \frac{1 \cdot 3}{5\sqrt{2} \cdot 5}\)

\(\cos(\alpha + 2\beta) = \frac{28}{25\sqrt{2}} - \frac{3}{25\sqrt{2}}\)

\(\cos(\alpha + 2\beta) = \frac{28 - 3}{25\sqrt{2}}\)

\(\cos(\alpha + 2\beta) = \frac{25}{25\sqrt{2}}\)

\(\cos(\alpha + 2\beta) = \frac{1}{\sqrt{2}}\)

Thus, the value of \(\cos(\alpha + 2\beta)\) is \(\frac{1}{\sqrt{2}}\).

Revision Table: Trigonometric Formulas Used

Additional Information on Acute Angles

When a problem specifies that angles like \(\alpha\) and \(\beta\) are acute (\(0 < \theta < \frac{\pi}{2}\)), it simplifies finding trigonometric ratios.

• All basic trigonometric ratios (\(\sin, \cos, \tan\)) for acute angles are positive.
• This means when taking square roots, we only consider the positive root.
• It also tells us the angle lies in the first quadrant, where \(x\) and \(y\) coordinates are both positive on the unit circle.

Understanding the quadrant of an angle is crucial in trigonometry as it determines the sign of the trigonometric functions. For instance, if \(\beta\) was in the second quadrant, its cosine would be negative even if its sine was positive.