Question

If \( 1^\circ = 0.0175 \) radians, then the approximate value of \( \sec 58^\circ \) is

• A. 1.9899
• B. 1.8788
• C. 1.8511
• D. 1.9677

Hint:

Always convert the change in angle (\( \Delta x \)) to radians when using derivative-based approximation formulas, as differentiation rules for trig functions assume radian measure.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

We use the concept of differentials to approximate the value of a function near a known point. We know the exact values for trigonometric functions at \( 60^\circ \). We can approximate \( \sec 58^\circ \) using the tangent line approximation (differentials) at \( x = 60^\circ \).
Step 2: Key Formula or Approach:

Approximation formula: \( f(x + \Delta x) \approx f(x) + f'(x) \Delta x \). Here \( f(x) = \sec x \), \( f'(x) = \sec x \tan x \). Convert degrees to radians for calculations involving derivatives.
Step 3: Detailed Explanation:

Let \( y = \sec x \). Choose \( x = 60^\circ \) because we know its values. Let \( x + \Delta x = 58^\circ \). Then \( \Delta x = 58^\circ - 60^\circ = -2^\circ \). Convert \( \Delta x \) to radians: Given \( 1^\circ = 0.0175 \) radians. \( \Delta x = -2 \times 0.0175 = -0.035 \) radians. Calculate \( f(x) \) and \( f'(x) \) at \( x = 60^\circ \): \( f(60^\circ) = \sec(60^\circ) = 2 \). \( f'(60^\circ) = \sec(60^\circ) \tan(60^\circ) = 2 \cdot \sqrt{3} \). Using \( \sqrt{3} \approx 1.732 \), \( f'(60^\circ) \approx 2 \times 1.732 = 3.464 \). Now apply the approximation formula: \[ \sec(58^\circ) \approx \sec(60^\circ) + (\sec(60^\circ) \tan(60^\circ)) \cdot \Delta x \] \[ \sec(58^\circ) \approx 2 + (3.464) \cdot (-0.035) \] Calculate the correction term: \( 3.464 \times 0.035 = 0.12124 \). \[ \sec(58^\circ) \approx 2 - 0.12124 \] \[ \sec(58^\circ) \approx 1.87876 \] Rounding to 4 decimal places gives 1.8788.
Step 4: Final Answer:

The approximate value is 1.8788.