macosurement: Go Given that \( \cos x=\frac{12}{13} \) where \( 0^{\circ}

Ask by Flynn Ray. in Nigeria

Mar 11,2025

To solve the problem, we start with the known condition: 1. \( \cos x = \frac{12}{13} \) where \( 0^{\circ} < x < 90^{\circ} \). From this, we can find \( \sin x \) using the Pythagorean identity: \[ \sin^2 x + \cos^2 x = 1 \] Substituting the value of \( \cos x \): \[ \sin^2 x + \left(\frac{12}{13}\right)^2 = 1 \] Calculating \( \left(\frac{12}{13}\right)^2 \): \[ \left(\frac{12}{13}\right)^2 = \frac{144}{169} \] Now substituting this back into the equation: \[ \sin^2 x + \frac{144}{169} = 1 \] To isolate \( \sin^2 x \), we subtract \( \frac{144}{169} \) from both sides: \[ \sin^2 x = 1 - \frac{144}{169} = \frac{169}{169} - \frac{144}{169} = \frac{25}{169} \] Taking the square root to find \( \sin x \): \[ \sin x = \sqrt{\frac{25}{169}} = \frac{5}{13} \] Next, we can find \( \tan x \): \[ \tan x = \frac{\sin x}{\cos x} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12} \] Now we can substitute \( \sin x \), \( \tan x \), and \( \cos x \) into the expression \( 2 \sin x - (\tan x + 3 \cos x) \): \[ 2 \sin x - (\tan x + 3 \cos x) = 2 \left(\frac{5}{13}\right) - \left(\frac{5}{12} + 3 \left(\frac{12}{13}\right)\right) \] Calculating \( 2 \sin x \): \[ 2 \sin x = 2 \cdot \frac{5}{13} = \frac{10}{13} \] Now calculating \( 3 \cos x \): \[ 3 \cos x = 3 \cdot \frac{12}{13} = \frac{36}{13} \] Now substituting these values into the expression: \[ 2 \sin x - \left(\tan x + 3 \cos x\right) = \frac{10}{13} - \left(\frac{5}{12} + \frac{36}{13}\right) \] To combine the terms inside the parentheses, we need a common denominator. The least common multiple of 12 and 13 is 156. Converting \( \frac{5}{12} \) to have a denominator of 156: \[ \frac{5}{12} = \frac{5 \cdot 13}{12 \cdot 13} = \frac{65}{156} \] Converting \( \frac{36}{13} \) to have a denominator of 156: \[ \frac{36}{13} = \frac{36 \cdot 12}{13 \cdot 12} = \frac{432}{156} \] Now we can combine these: \[ \tan x + 3 \cos x = \frac{65}{156} + \frac{432}{156} = \frac{497}{156} \] Now substituting back into the expression: \[ 2 \sin x - \left(\tan x + 3 \cos x\right) = \frac{10}{13} - \frac{497}{156} \] Converting \( \frac{10}{13} \) to have a denominator of 156: \[ \frac{10}{13} = \frac{10 \cdot 12}{13 \cdot 12} = \frac{120}{156} \] Now we can perform the subtraction: \[ \frac{120}{156} - \frac{497}{156} = \frac{120 - 497}{156} = \frac{-377}{156} \] Thus, the final value of \( 2 \sin x - (\tan x + 3 \cos x) \) is: \[ \frac{-377}{156} \]