The monthly sales S (in thousands of units) of lawn mowers are approximated by S

Question

The monthly sales S (in thousands of units) of lawn mowers are approximated by S = 745 - 43.75 cos(pi t/6) where t is the time (in months), with t = 1 corresponding to January. Graph the sales function over a 12-month period. (Use radian mode.) Determine the calendar months during which sales exceed 100,000 units. Find the time t when the sales attain a maximum level. Find the time t when the sales are at the lowest level. Use an inverse trig function to find the angle theta in quadrant I such that 5 tan theta = 1024. Give answers in both degrees and radians, accurate to 4 significant digits. Suppose that the angle theta is in quadrant I and that sin theta = x/7. Find formulas for cos theta and tan theta in terms of x. Your answers should be algebraic formulas that are free of trig functions.

Explanation / Answer

8)b)  cos(pi t/6)= (-100+74.50)/43.75 = -0.58286
so pi*t/6 = 2.193 rad and pi*t/6 = 4.09
so t=4.2 approx 4 April and t= 7.8 approx August

9) 5*tan = 1024

  tan = 204.8

= arctan(204.8) = 89.72 degree = 1.566 radian

10) sin = x/7

cos = sqrt[1 - sin^2 ] = sqrt[1 - x^2/49] = [sqrt(49 - x^2)]/7

tan = sin/cos = x/sqrt(49 - x^2)

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