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Using the approximations sin x ≈ x and cos x ≈ 1 - x²/2, where x is small and measured in radians, show that (1 - cos 2x)/sin x ≈ 2x.

Mathematics

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Answer

b
0.1396

Step-by-step solution

9 steps

Part a· 2 steps

To show that 1−cos⁡2xsin⁡x≈2x\frac{1 - \cos 2x}{\sin x} \approx 2x using the approximations sin⁡x≈x\sin x \approx x and cos⁡x≈1−x22\cos x \approx 1 - \frac{x^2}{2}

  1. Apply the given approximations to the terms in the expression.

    For cos⁡2x\cos 2x, substitute 2x2x for xx in the approximation for cos⁡x\cos x:

    cos⁡2x≈1−(2x)22=1−4x22=1−2x2\cos 2x \approx 1 - \frac{(2x)^2}{2} = 1 - \frac{4x^2}{2} = 1 - 2x^2

    For sin⁡x\sin x, use the given approximation:

    sin⁡x≈x\sin x \approx x

  2. Substitute these approximations into the given expression and simplify.

    1−cos⁡2xsin⁡x≈1−(1−2x2)x=1−1+2x2x=2x2x=2x\frac{1 - \cos 2x}{\sin x} \approx \frac{1 - (1 - 2x^2)}{x} = \frac{1 - 1 + 2x^2}{x} = \frac{2x^2}{x} = 2x

    Thus, we have shown that 1−cos⁡2xsin⁡x≈2x\frac{1 - \cos 2x}{\sin x} \approx 2x.

Part b0.1396

To obtain an approximate value of 1−cos⁡8∘sin⁡4∘\frac{1 - \cos 8^\circ}{\sin 4^\circ}

  1. Identify the value of xx from the expression.

    Comparing 1−cos⁡8∘sin⁡4∘\frac{1 - \cos 8^\circ}{\sin 4^\circ} with 1−cos⁡2xsin⁡x\frac{1 - \cos 2x}{\sin x}, we see that x=4∘x = 4^\circ.

  2. Convert the angle xx from degrees to radians, as the approximations are valid for radians.

    x=4∘×π180∘=4π180=π45 radiansx = 4^\circ \times \frac{\pi}{180^\circ} = \frac{4\pi}{180} = \frac{\pi}{45} \text{ radians}

  3. Use the approximation derived in part (a).

    1−cos⁡8∘sin⁡4∘≈2x=2×π45\frac{1 - \cos 8^\circ}{\sin 4^\circ} \approx 2x = 2 \times \frac{\pi}{45} 1−cos⁡8∘sin⁡4∘≈2π45\frac{1 - \cos 8^\circ}{\sin 4^\circ} \approx \frac{2\pi}{45}

    Calculating the numerical value:

    2π45≈2×3.1415945≈6.2831845≈0.139626\frac{2\pi}{45} \approx \frac{2 \times 3.14159}{45} \approx \frac{6.28318}{45} \approx 0.139626

    The approximate value is 0.1396\boxed{0.1396}.

Part c· 4 steps

To show that ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x using first principles and the given limits

  1. Write the definition of the derivative from first principles.

    ddx(sin⁡x)=lim⁡h→0sin⁡(x+h)−sin⁡xh\frac{d}{dx}(\sin x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}

  2. Expand sin⁡(x+h)\sin(x+h) using the sum identity sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A \cos B + \cos A \sin B.

    ddx(sin⁡x)=lim⁡h→0(sin⁡xcos⁡h+cos⁡xsin⁡h)−sin⁡xh\frac{d}{dx}(\sin x) = \lim_{h \to 0} \frac{(\sin x \cos h + \cos x \sin h) - \sin x}{h}

  3. Rearrange the terms and split the limit.

    ddx(sin⁡x)=lim⁡h→0sin⁡x(cos⁡h−1)+cos⁡xsin⁡hh\frac{d}{dx}(\sin x) = \lim_{h \to 0} \frac{\sin x (\cos h - 1) + \cos x \sin h}{h} ddx(sin⁡x)=lim⁡h→0(sin⁡xcos⁡h−1h+cos⁡xsin⁡hh)\frac{d}{dx}(\sin x) = \lim_{h \to 0} \left( \sin x \frac{\cos h - 1}{h} + \cos x \frac{\sin h}{h} \right)

  4. Apply the given limits lim⁡h→0sin⁡hh=1\lim_{h \to 0} \frac{\sin h}{h} = 1 and lim⁡h→0cos⁡h−1h=0\lim_{h \to 0} \frac{\cos h - 1}{h} = 0.

    ddx(sin⁡x)=sin⁡x⋅(lim⁡h→0cos⁡h−1h)+cos⁡x⋅(lim⁡h→0sin⁡hh)\frac{d}{dx}(\sin x) = \sin x \cdot \left( \lim_{h \to 0} \frac{\cos h - 1}{h} \right) + \cos x \cdot \left( \lim_{h \to 0} \frac{\sin h}{h} \right) ddx(sin⁡x)=sin⁡x⋅(0)+cos⁡x⋅(1)\frac{d}{dx}(\sin x) = \sin x \cdot (0) + \cos x \cdot (1) ddx(sin⁡x)=0+cos⁡x\frac{d}{dx}(\sin x) = 0 + \cos x ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x

    This shows the derivative of sin⁡x\sin x is cos⁡x\cos x.

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