Math  /  Trigonometry

QuestionAdvanced Functions (MHF4U)
6. Use compound angles to evaluate the exact value for the following. (a) sinπ12\sin \frac{\pi}{12}

6(a) \qquad (b) cos17π12\cos \frac{17 \pi}{12}
6(b) \qquad (c) tan(5π12)\tan \left(-\frac{5 \pi}{12}\right)
6(c) \qquad
7. Simplify and then evaluate each expression (no decimals). (a) cos(7π12)cos(5π12)+sin(7π12)sin(5π12)\cos \left(\frac{7 \pi}{12}\right) \cos \left(\frac{5 \pi}{12}\right)+\sin \left(\frac{7 \pi}{12}\right) \sin \left(\frac{5 \pi}{12}\right)

7(a) simplified: \qquad evaluate: \qquad (b) sin2xcosxcos2xsinx\sin 2 x \cos x-\cos 2 x \sin x, let: x=4π3x=\frac{4 \pi}{3}
7(b) simplified: \qquad evaluate: \qquad (c) 2sinπ8cosπ82 \sin \frac{\pi}{8} \cos \frac{\pi}{8}
7(b) simplified: \qquad evaluate: \qquad
8. Express as a single trigonometric ratio (a) cos2(β4)sin2(β4)\cos ^{2}\left(\frac{\beta}{4}\right)-\sin ^{2}\left(\frac{\beta}{4}\right)

8(a) \qquad (b) 12sin23x1-2 \sin ^{2} 3 x
8(b) \qquad
9. Find the exact value of tanπ10+tanπ151tanπ10tanπ15\frac{\tan \frac{\pi}{10}+\tan \frac{\pi}{15}}{1-\tan \frac{\pi}{10} \tan \frac{\pi}{15}}.
9. \qquad
10. Angles AA and BB are located in the first quadrant. If sinA=513\sin A=\frac{5}{13}
10. \qquad and cosB=35\cos B=\frac{3}{5}, determine the exact value of cos(A+B)\cos (A+B).
11. Given 3cosθ2cosθsinθ=0\sqrt{3} \cos \theta-2 \cos \theta \sin \theta=0, determine the exact angles of rotation
11. \qquad \qquad \qquad \qquad
12. A child swings on a playground swing set. If the length of the
12. \qquad swing is 3 m and the child swings through an angle of π9\frac{\pi}{9}, determine the exact are length through which the child travels.

Studdy Solution
2 Simplify the expression:
Arc Length=3π9=π3 \text{Arc Length} = \frac{3\pi}{9} = \frac{\pi}{3}
The solutions to the problems are as follows: 6(a): sinπ12=624 \sin \frac{\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{4} 6(b): cos17π12=624 \cos \frac{17\pi}{12} = -\frac{\sqrt{6} - \sqrt{2}}{4} 6(c): tan(5π12)=3131 \tan \left(-\frac{5\pi}{12}\right) = \frac{-\sqrt{3} - 1}{\sqrt{3} - 1} 7(a): cos(π6)=32 \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} 7(b): sin(4π3)=32 \sin \left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2} 7(c): sin(π4)=22 \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} 8(a): cos(β2) \cos \left(\frac{\beta}{2}\right) 8(b): cos6x \cos 6x 9: 13 \frac{1}{\sqrt{3}} 10: 1665 \frac{16}{65} 11: θ=π2+kπ \theta = \frac{\pi}{2} + k\pi or θ=π3+2kπ \theta = \frac{\pi}{3} + 2k\pi or θ=2π3+2kπ \theta = \frac{2\pi}{3} + 2k\pi 12: π3 \frac{\pi}{3}

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