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Use the conversion factor pi radians180^ to convert degrees to radians.

Mathematics

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3 parts
i

−5π6 radians-\frac{5\pi}{6} \text{ radians}

ii

47 radians\frac{4}{7} \text{ radians}

iii
Shown:

: sin⁡2θcos⁡θ+cos⁡θ=sec⁡θ\frac{\sin^2 \theta}{\cos \theta} + \cos \theta = \sec \theta

Step-by-step solution

10 steps

Here are the solutions to the questions:

Part i· 2 steps

Express −150∘-150^\circ into radians.

  1. Use the conversion factor π radians180∘\frac{\pi \text{ radians}}{180^\circ} to convert degrees to radians.

    −150∘×π180∘-150^\circ \times \frac{\pi}{180^\circ}

  2. Simplify the fraction.

    −150π180=−15π18=−5π6-\frac{150\pi}{180} = -\frac{15\pi}{18} = -\frac{5\pi}{6}

    The angle in radians is −5π6 radians\boxed{-\frac{5\pi}{6} \text{ radians}}.

Part ii

47 radians\frac{4}{7} \text{ radians}

Find θ\theta, when l=2cml=2\text{cm} and r=3.5cmr=3.5\text{cm}.

  1. Use the formula for arc length, l=rθl = r\theta, where θ\theta is in radians.

    Rearrange the formula to solve for θ\theta:

    θ=lr\theta = \frac{l}{r}

  2. Substitute the given values for ll and rr.

    θ=2 cm3.5 cm\theta = \frac{2\text{ cm}}{3.5\text{ cm}}

  3. Calculate the value of θ\theta.

    θ=23.5=272=47\theta = \frac{2}{3.5} = \frac{2}{\frac{7}{2}} = \frac{4}{7}

    The angle θ\theta is 47 radians\boxed{\frac{4}{7} \text{ radians}}.

Part iii

: sin⁡2θcos⁡θ+cos⁡θ=sec⁡θ\frac{\sin^2 \theta}{\cos \theta} + \cos \theta = \sec \theta

Verify that: sin⁡2θcos⁡θ+cos⁡θ=sec⁡θ\frac{\sin^2 \theta}{\cos \theta} + \cos \theta = \sec \theta.

  1. Start with the left-hand side (LHS) of the equation.

    LHS=sin⁡2θcos⁡θ+cos⁡θ\text{LHS} = \frac{\sin^2 \theta}{\cos \theta} + \cos \theta

  2. Find a common denominator for the terms, which is cos⁡θ\cos \theta.

    LHS=sin⁡2θcos⁡θ+cos⁡θ⋅cos⁡θcos⁡θ\text{LHS} = \frac{\sin^2 \theta}{\cos \theta} + \frac{\cos \theta \cdot \cos \theta}{\cos \theta} LHS=sin⁡2θcos⁡θ+cos⁡2θcos⁡θ\text{LHS} = \frac{\sin^2 \theta}{\cos \theta} + \frac{\cos^2 \theta}{\cos \theta}

  3. Combine the terms over the common denominator.

    LHS=sin⁡2θ+cos⁡2θcos⁡θ\text{LHS} = \frac{\sin^2 \theta + \cos^2 \theta}{\cos \theta}

  4. Apply the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1.

    LHS=1cos⁡θ\text{LHS} = \frac{1}{\cos \theta}

  5. Recognize that 1cos⁡θ\frac{1}{\cos \theta} is equal to sec⁡θ\sec \theta.

    LHS=sec⁡θ\text{LHS} = \sec \theta

    This matches the right-hand side (RHS) of the equation.
    Therefore, the identity is verified. sin⁡2θcos⁡θ+cos⁡θ=sec⁡θ\boxed{\frac{\sin^2 \theta}{\cos \theta} + \cos \theta = \sec \theta}

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