Answer :
Sure, let's solve this step-by-step.
Given:
[tex]\[ \tan \theta=\frac{1}{7} \][/tex]
[tex]\[ \tan \beta=\frac{1}{3} \][/tex]
To find [tex]\(\cos 2\theta\)[/tex] and [tex]\(\sin 4\beta\)[/tex], we will use trigonometric identities.
1. Finding [tex]\(\cos 2\theta\)[/tex]:
The double-angle identity for cosine is:
[tex]\[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \][/tex]
Substituting [tex]\(\tan \theta = \frac{1}{7}\)[/tex]:
[tex]\[ \cos 2\theta = \frac{1 - \left(\frac{1}{7}\right)^2}{1 + \left(\frac{1}{7}\right)^2} \][/tex]
[tex]\[ \cos 2\theta = \frac{1 - \frac{1}{49}}{1 + \frac{1}{49}} \][/tex]
[tex]\[ \cos 2\theta = \frac{\frac{49}{49} - \frac{1}{49}}{\frac{49}{49} + \frac{1}{49}} \][/tex]
[tex]\[ \cos 2\theta = \frac{\frac{48}{49}}{\frac{50}{49}} \][/tex]
[tex]\[ \cos 2\theta = \frac{48}{50} \][/tex]
[tex]\[ \cos 2\theta = 0.96 \][/tex]
2. Finding [tex]\(\sin 4\beta\)[/tex]:
The multiple-angle identity for sine is:
[tex]\[ \sin 4\beta = 2 \sin 2\beta \cos 2\beta \][/tex]
First, we need [tex]\(\sin 2\beta\)[/tex] and [tex]\(\cos 2\beta\)[/tex]. The double-angle formulas for sine and cosine are:
[tex]\[ \sin 2\beta = \frac{2 \tan \beta}{1 + \tan^2 \beta} \][/tex]
[tex]\[ \cos 2\beta = \frac{1 - \tan^2 \beta}{1 + \tan^2 \beta} \][/tex]
Substituting [tex]\(\tan \beta = \frac{1}{3} \)[/tex]:
[tex]\[ \sin 2\beta = \frac{2 \left(\frac{1}{3}\right)}{1 + \left(\frac{1}{3}\right)^2} \][/tex]
[tex]\[ \sin 2\beta = \frac{\frac{2}{3}}{1 + \frac{1}{9}} \][/tex]
[tex]\[ \sin 2\beta = \frac{\frac{2}{3}}{\frac{10}{9}} \][/tex]
[tex]\[ \sin 2\beta = \frac{2}{3} \cdot \frac{9}{10} \][/tex]
[tex]\[ \sin 2\beta = \frac{18}{30} \][/tex]
[tex]\[ \sin 2\beta = \frac{3}{5} \][/tex]
[tex]\[ \cos 2\beta = \frac{1 - \left(\frac{1}{3}\right)^2}{1 + \left(\frac{1}{3}\right)^2} \][/tex]
[tex]\[ \cos 2\beta = \frac{1 - \frac{1}{9}}{1 + \frac{1}{9}} \][/tex]
[tex]\[ \cos 2\beta = \frac{\frac{9}{9} - \frac{1}{9}}{\frac{9}{9} + \frac{1}{9}} \][/tex]
[tex]\[ \cos 2\beta = \frac{\frac{8}{9}}{\frac{10}{9}} \][/tex]
[tex]\[ \cos 2\beta = \frac{8}{10} \][/tex]
[tex]\[ \cos 2\beta = 0.8 \][/tex]
Now, using these results in the equation for [tex]\(\sin 4\beta\)[/tex]:
[tex]\[ \sin 4\beta = 2 \left(\frac{3}{5}\right) \left(0.8\right) \][/tex]
[tex]\[ \sin 4\beta = 2 \left(\frac{3 \times 8}{5 \times 10}\right) \][/tex]
[tex]\[ \sin 4\beta = 2 \left(\frac{24}{50}\right) \][/tex]
[tex]\[ \sin 4\beta = \frac{48}{50} \][/tex]
[tex]\[ \sin 4\beta = 0.96 \][/tex]
Thus, we have shown that:
[tex]\[ \cos 2\theta = \sin 4\beta \][/tex]
[tex]\[ 0.96 = 0.96 \][/tex]
Hence, [tex]\(\boxed{\cos 2 \theta = \sin 4 \beta}\)[/tex].
Given:
[tex]\[ \tan \theta=\frac{1}{7} \][/tex]
[tex]\[ \tan \beta=\frac{1}{3} \][/tex]
To find [tex]\(\cos 2\theta\)[/tex] and [tex]\(\sin 4\beta\)[/tex], we will use trigonometric identities.
1. Finding [tex]\(\cos 2\theta\)[/tex]:
The double-angle identity for cosine is:
[tex]\[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \][/tex]
Substituting [tex]\(\tan \theta = \frac{1}{7}\)[/tex]:
[tex]\[ \cos 2\theta = \frac{1 - \left(\frac{1}{7}\right)^2}{1 + \left(\frac{1}{7}\right)^2} \][/tex]
[tex]\[ \cos 2\theta = \frac{1 - \frac{1}{49}}{1 + \frac{1}{49}} \][/tex]
[tex]\[ \cos 2\theta = \frac{\frac{49}{49} - \frac{1}{49}}{\frac{49}{49} + \frac{1}{49}} \][/tex]
[tex]\[ \cos 2\theta = \frac{\frac{48}{49}}{\frac{50}{49}} \][/tex]
[tex]\[ \cos 2\theta = \frac{48}{50} \][/tex]
[tex]\[ \cos 2\theta = 0.96 \][/tex]
2. Finding [tex]\(\sin 4\beta\)[/tex]:
The multiple-angle identity for sine is:
[tex]\[ \sin 4\beta = 2 \sin 2\beta \cos 2\beta \][/tex]
First, we need [tex]\(\sin 2\beta\)[/tex] and [tex]\(\cos 2\beta\)[/tex]. The double-angle formulas for sine and cosine are:
[tex]\[ \sin 2\beta = \frac{2 \tan \beta}{1 + \tan^2 \beta} \][/tex]
[tex]\[ \cos 2\beta = \frac{1 - \tan^2 \beta}{1 + \tan^2 \beta} \][/tex]
Substituting [tex]\(\tan \beta = \frac{1}{3} \)[/tex]:
[tex]\[ \sin 2\beta = \frac{2 \left(\frac{1}{3}\right)}{1 + \left(\frac{1}{3}\right)^2} \][/tex]
[tex]\[ \sin 2\beta = \frac{\frac{2}{3}}{1 + \frac{1}{9}} \][/tex]
[tex]\[ \sin 2\beta = \frac{\frac{2}{3}}{\frac{10}{9}} \][/tex]
[tex]\[ \sin 2\beta = \frac{2}{3} \cdot \frac{9}{10} \][/tex]
[tex]\[ \sin 2\beta = \frac{18}{30} \][/tex]
[tex]\[ \sin 2\beta = \frac{3}{5} \][/tex]
[tex]\[ \cos 2\beta = \frac{1 - \left(\frac{1}{3}\right)^2}{1 + \left(\frac{1}{3}\right)^2} \][/tex]
[tex]\[ \cos 2\beta = \frac{1 - \frac{1}{9}}{1 + \frac{1}{9}} \][/tex]
[tex]\[ \cos 2\beta = \frac{\frac{9}{9} - \frac{1}{9}}{\frac{9}{9} + \frac{1}{9}} \][/tex]
[tex]\[ \cos 2\beta = \frac{\frac{8}{9}}{\frac{10}{9}} \][/tex]
[tex]\[ \cos 2\beta = \frac{8}{10} \][/tex]
[tex]\[ \cos 2\beta = 0.8 \][/tex]
Now, using these results in the equation for [tex]\(\sin 4\beta\)[/tex]:
[tex]\[ \sin 4\beta = 2 \left(\frac{3}{5}\right) \left(0.8\right) \][/tex]
[tex]\[ \sin 4\beta = 2 \left(\frac{3 \times 8}{5 \times 10}\right) \][/tex]
[tex]\[ \sin 4\beta = 2 \left(\frac{24}{50}\right) \][/tex]
[tex]\[ \sin 4\beta = \frac{48}{50} \][/tex]
[tex]\[ \sin 4\beta = 0.96 \][/tex]
Thus, we have shown that:
[tex]\[ \cos 2\theta = \sin 4\beta \][/tex]
[tex]\[ 0.96 = 0.96 \][/tex]
Hence, [tex]\(\boxed{\cos 2 \theta = \sin 4 \beta}\)[/tex].
