SECTION 7.6 Double-angle and Half-angle Formulas 533 45° R u (a) Show that θ θ θ ( ) ( ) ( ) [ ] = − − R v 2 32 sin 2 cos 2 1 0 2 (b) In calculus, you will be asked to find the angle θ that maximizes R by solving the equation θ θ ( ) ( ) + = sin 2 cos 2 0 Solve this equation for θ. (c) What is the maximum distance R if = v 32 0 feet per second? (d) Graph θ θ ( ) = ° ≤ ≤ ° R R , 45 90 , and find the angle θ that maximizes the distance R. Also find the maximum distance. Use = v 32 0 feet per second. Compare the results with the answers found in parts (b) and (c). 106. Sawtooth Curve An oscilloscope often displays a sawtooth curve. This curve can be approximated by sinusoidal curves of varying periods and amplitudes.A first approximation to the sawtooth curve is given by π π ( ) ( ) = + y x x 1 2 sin 2 1 4 sin 4 Show that π π ( ) ( ) = y x x sin2 cos . 2 50mv V1 2B. Gm.V Trig TVline OH1 Obase1 107. Area of an Isosceles Triangle Show that the area A of an isosceles triangle whose equal sides are of length s, and where θ is the angle between them, is θ = A s 1 2 sin 2 [Hint: See the figure. The height h bisects the angle θ and is the perpendicular bisector of the base.] h s s u (b) The area A of a regular octagon is also given by the formula π = A a2 cot 8 , 2 where a is the length of a side. Find the exact area of a regular octagon whose side is 9 centimeters. 102. Constructing a Rain Gutter A rain gutter is to be constructed of aluminum sheets 12 inches wide. After marking off a length of 4 inches from each edge, the builder bends this length up at an angle θ. See the figure. The area A of the opening as a function of θ is given by θ θ θ θ ( ) ( ) = + ° < < ° A 16 sin cos 1 0 90 4 in. 4 in. 4 in. 12 in. u u (a) In calculus, you will be asked to find the angle θ that maximizes A by solving the equation θ θ θ ( ) + = ° < < ° cos 2 cos 0 0 90 Solve this equation for θ. (b) What is the maximum area A of the opening? (c) Graph θ θ ( ) = ° ≤ ≤ ° A A , 0 90 , and find the angle θ that maximizes the area A.Also find the maximum area. 103. Laser Projection In a laser projection system, the optical angle or scanning angle θ is related to the throw distance D from the scanner to the screen and the projected image width W by the equation θ θ = − D W 1 2 csc cot (a) Show that the projected image width is given by θ = W D2 tan 2 (b) Find the optical angle if the throw distance is 15 feet and the projected image width is 6.5 feet. Source: Pangolin Laser Systems, Inc. 104. Product of Inertia The product of inertia for an area about inclined axes is given by the formula θ θ θ θ θ θ ( ) = − + − I I I I sin cos sin cos cos sin uv x y xy 2 2 Show that this is equivalent to θ θ ( ) ( ) = − + I I I I 2 sin 2 cos 2 uv x y xy Source: Adapted from Hibbeler, Engineering Mechanics: Statics, 13th ed., Pearson © 2013. 105. Projectile Motion An object is propelled upward at an angle θ θ ° < < ° , 45 90 , to the horizontal with an initial velocity of v0 feet per second from the base of a plane that makes an angle of ° 45 with the horizontal. See the figure atop the right column. If air resistance is ignored, the distance R that it travels up the inclined plane is given by the function θ θ θ θ ( ) ( ) = − R v 2 16 cos sin cos 0 2
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