Trigonometric Functions (1 Viewer)

V_L · May 16, 2017

Hi,

Attached is 6 questions from different sections of the Trigonometric Functions chapter. Sorry for the hassle but i haven't been able to figure out the method for these so any help would be appreciated.

Thank you!!

1729 · May 16, 2017

V_L said:
Hi,

Attached is 6 questions from different sections of the Trigonometric Functions chapter. Sorry for the hassle but i haven't been able to figure out the method for these so any help would be appreciated.

Thank you!!

$\noindent \textbf{Q10.} When the sector of a circle is curved around to form a cone, the sector's arc length becomes the cone's base circumference. The circumference, $c$ is $c = r\theta = \frac{5\pi}{3}$. We know that $r = \frac{c}{2\pi}$, and so the radius of the circular base $r = \frac{5}{6}.$ Now the sector's radius becomes the cone's slant height. Using the Pythagorean Theorem, the perpendicular height $h = \sqrt{5^2 - \left(\frac{5}{6}\right)^2} = \sqrt{\frac{875}{36}}.$ Therefore $V = \frac{1}{3}\pi r^2h = \frac{1}{3} \pi \left(\frac{5}{6}\right)^2\sqrt{\frac{875}{36}} = \frac{125\sqrt{35}\pi}{648}$ cm$^3$ and, \\ $A = A_{sector} + \pi r^2 = \frac{1}{2}(5)^2\left(\frac{\pi}{3}\right) + \left(\frac{5}{6}\right)^2\pi= \frac{175\pi}{36}$ cm$^2$

$$\noindent \textbf{Q15.} $1 - \sin^2{\left(\frac{\pi}{2}-\theta\right)} = 1 - \cos^2{\theta} = \sin^2{\theta}$

$$\noindent \textbf{Q16.} If $\frac{\pi}{2} < x < \pi$ then $x$ belongs to Quadrant II. We construct a triangle in this quadrant and use the fact that $\tan{x} = -\frac{3}{4}$ to determine the length of the hypotenuse, which is 5 units. Therefore $\sin{x} = \frac{3}{5}$ and $\cos{x} = -\frac{4}{5}$, noting that all trigonometric ratios except sine (and cosecant) are negative for obtuse angles.$

$$\noindent \textbf{Q17.} $\tan^2{x} = 1 \implies \tan{x} = \pm 1.$ Being positive and negative, $x$ belongs to all Quadrants and therefore from the basic angle $x = \frac{\pi}{4}$ we have, \\ $x = \frac{\pi}{4}, \pi - \frac{\pi}{4}, \pi + \frac{\pi}{4}, 2\pi - \frac{\pi}{4}$. That is, $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$

Trigonometric Functions (1 Viewer)

V_L

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