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Determine the following limit: limit x->5 (3x4 - 375x) / (3x2 - 15x)

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2 parts
2.2

2+22\frac{\sqrt{2 + \sqrt{2}}}{2}

2.3

21.08∘21.08^\circ

158.92∘158.92^\circ

Step-by-step solution

13 steps

Here's the solution to Question 2:

Part 2

TRIGONOMETRY

Part 2.1· 3 steps

Prove the following trigonometric identity using the double-angle identity

cos⁡2β2=(cos⁡β+1)2\cos^2 \frac{\beta}{2} = \frac{(\cos \beta + 1)}{2}

  1. Start with the double-angle identity for cosine.

    The double-angle identity for cosine is given by:

    cos⁡(2θ)=2cos⁡2θ−1\cos(2\theta) = 2\cos^2\theta - 1

  2. Substitute θ=β2\theta = \frac{\beta}{2} into the identity.

    If θ=β2\theta = \frac{\beta}{2}, then 2θ=β2\theta = \beta.
    Substituting these into the identity:

    cos⁡β=2cos⁡2β2−1\cos \beta = 2\cos^2 \frac{\beta}{2} - 1

  3. Rearrange the equation to solve for cos⁡2β2\cos^2 \frac{\beta}{2}.

    Add 1 to both sides:

    cos⁡β+1=2cos⁡2β2\cos \beta + 1 = 2\cos^2 \frac{\beta}{2}

    Divide by 2:

    cos⁡2β2=cos⁡β+12\cos^2 \frac{\beta}{2} = \frac{\cos \beta + 1}{2}

    The identity is proven.

Part 2.2

2+22\frac{\sqrt{2 + \sqrt{2}}}{2}

Calculate the following without the use of a calculator: cos⁡22.5∘\cos 22.5^\circ.

  1. Use the identity derived in Question 2.1.

    We have cos⁡2β2=cos⁡β+12\cos^2 \frac{\beta}{2} = \frac{\cos \beta + 1}{2}.
    Let β2=22.5∘\frac{\beta}{2} = 22.5^\circ.
    Then β=2×22.5∘=45∘\beta = 2 \times 22.5^\circ = 45^\circ.

  2. Substitute β=45∘\beta = 45^\circ into the identity.

    cos⁡222.5∘=cos⁡45∘+12\cos^2 22.5^\circ = \frac{\cos 45^\circ + 1}{2}

    We know that cos⁡45∘=22\cos 45^\circ = \frac{\sqrt{2}}{2}.

    cos⁡222.5∘=22+12\cos^2 22.5^\circ = \frac{\frac{\sqrt{2}}{2} + 1}{2}

  3. Simplify the expression.

    cos⁡222.5∘=2+222\cos^2 22.5^\circ = \frac{\frac{\sqrt{2} + 2}{2}}{2} cos⁡222.5∘=2+24\cos^2 22.5^\circ = \frac{\sqrt{2} + 2}{4}

  4. Take the square root of both sides.

    Since 22.5∘22.5^\circ is in the first quadrant, cos⁡22.5∘\cos 22.5^\circ is positive.

    cos⁡22.5∘=2+24\cos 22.5^\circ = \sqrt{\frac{2 + \sqrt{2}}{4}} cos⁡22.5∘=2+24\cos 22.5^\circ = \frac{\sqrt{2 + \sqrt{2}}}{\sqrt{4}} cos⁡22.5∘=2+22\cos 22.5^\circ = \boxed{\frac{\sqrt{2 + \sqrt{2}}}{2}}

Part 2.3

21.08∘21.08^\circ

Solve for xx if 4−7sin⁡x=2cos⁡2x4 - 7 \sin x = 2 \cos 2x; 0∘≤x≤360∘0^\circ \le x \le 360^\circ. NB use the derived identity in QUESTION 2.1.

  1. Use the derived identity from Question 2.1 to express cos⁡2x\cos 2x in terms of sin⁡x\sin x.

    From cos⁡2β2=cos⁡β+12\cos^2 \frac{\beta}{2} = \frac{\cos \beta + 1}{2}, we can rearrange to get cos⁡β=2cos⁡2β2−1\cos \beta = 2\cos^2 \frac{\beta}{2} - 1.
    Let β=2x\beta = 2x. Then β2=x\frac{\beta}{2} = x.
    So, cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1.
    Substitute this into the given equation:

    4−7sin⁡x=2(2cos⁡2x−1)4 - 7 \sin x = 2(2\cos^2 x - 1) 4−7sin⁡x=4cos⁡2x−24 - 7 \sin x = 4\cos^2 x - 2

  2. Convert cos⁡2x\cos^2 x to sin⁡2x\sin^2 x using the identity cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x.

    4−7sin⁡x=4(1−sin⁡2x)−24 - 7 \sin x = 4(1 - \sin^2 x) - 2 4−7sin⁡x=4−4sin⁡2x−24 - 7 \sin x = 4 - 4\sin^2 x - 2 4−7sin⁡x=2−4sin⁡2x4 - 7 \sin x = 2 - 4\sin^2 x

  3. Rearrange the equation into a quadratic form in terms of sin⁡x\sin x.

    4sin⁡2x−7sin⁡x+4−2=04\sin^2 x - 7 \sin x + 4 - 2 = 0 4sin⁡2x−7sin⁡x+2=04\sin^2 x - 7 \sin x + 2 = 0

  4. Solve the quadratic equation for sin⁡x\sin x using the quadratic formula u=−b±b2−4ac2au = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where u=sin⁡xu = \sin x.

    Here, a=4a=4, b=−7b=-7, c=2c=2.

    sin⁡x=−(−7)±(−7)2−4(4)(2)2(4)\sin x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(4)(2)}}{2(4)} sin⁡x=7±49−328\sin x = \frac{7 \pm \sqrt{49 - 32}}{8} sin⁡x=7±178\sin x = \frac{7 \pm \sqrt{17}}{8}

  5. Evaluate the possible values for sin⁡x\sin x.

    sin⁡x=7+178orsin⁡x=7−178\sin x = \frac{7 + \sqrt{17}}{8} \quad \text{or} \quad \sin x = \frac{7 - \sqrt{17}}{8}

    Approximate values: 17≈4.123\sqrt{17} \approx 4.123.

    sin⁡x≈7+4.1238=11.1238≈1.390\sin x \approx \frac{7 + 4.123}{8} = \frac{11.123}{8} \approx 1.390

    This value is not possible since −1≤sin⁡x≤1-1 \le \sin x \le 1.

    sin⁡x≈7−4.1238=2.8778≈0.3596\sin x \approx \frac{7 - 4.123}{8} = \frac{2.877}{8} \approx 0.3596

    This value is valid.

  6. Find the values of xx in the range 0∘≤x≤360∘0^\circ \le x \le 360^\circ.

    Let sin⁡x=7−178\sin x = \frac{7 - \sqrt{17}}{8}.
    First, find the reference angle α\alpha:

    α=arcsin⁡(7−178)≈arcsin⁡(0.3596)≈21.08∘\alpha = \arcsin\left(\frac{7 - \sqrt{17}}{8}\right) \approx \arcsin(0.3596) \approx 21.08^\circ

    Since sin⁡x\sin x is positive, xx lies in Quadrant I or Quadrant II.
    In Quadrant I:

    x1=α≈21.08∘x_1 = \alpha \approx \boxed{21.08^\circ}

    In Quadrant II:

    x2=180∘−α≈180∘−21.08∘=158.92∘x_2 = 180^\circ - \alpha \approx 180^\circ - 21.08^\circ = \boxed{158.92^\circ}
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