715 7.4 Double-Angle and Half-Angle Identities An Application EXAMPLE 5 Determining Wattage Consumption When a toaster is plugged into a common household outlet, the wattage consumed is not constant. Instead, it varies at a high frequency according to the model W= V2 R , where V is the voltage and R is a constant that measures the resistance of the toaster in ohms. (Data from Bell, D., Fundamentals of Electric Circuits, Fourth Edition, Prentice-Hall.) Graph the wattage W consumed by a toaster with the constant R = 15 and V = 163 sin 120pt in the window 30, 0.054 by 3-500, 20004. How many oscillations are there? SOLUTION W= V2 R = 1163 sin 120pt22 15 Substituting the given values into the wattage equation. Note that sin 120pt has maximum value 1, so the expression for W has maximum value 1632 15 ≈1771. The minimum value is 0. The graph in Figure 7 shows that there are six oscillations. S Now Try Exercise 107. −500 2000 0 0.05 For x = t, W(t) = (163 sin 120pt)2 15 Figure 7 Product-to-Sum and Sum-to-Product Identities We can add the corresponding sides of the identities for cos1A + B2 and cos1A - B2 to derive a product-to-sum identity that is useful in calculus. cos1A + B2 = cos A cos B - sin A sin B cos1A - B2 = cos A cos B + sin A sin B cos1A + B2 + cos1A - B2 = 2 cos A cos B Add. cos A cos B = 1 2 3 cos1A +B2 +cos1A −B2 4 Similarly, subtracting cos1A + B2 from cos1A - B2 gives sin A sin B = 1 2 3 cos1A −B2 −cos1A +B2 4 . Using the identities for sin1A + B2 and sin1A - B2 in the same way, we obtain two more identities. Those and the previous ones are now summarized. LOOKING AHEAD TO CALCULUS The product-to-sum identities are used in calculus to find integrals of functions that are products of trigonometric functions. The classic calculus text by Earl Swokowski includes the following example: Evaluate Lcos 5x cos 3x dx. The first solution line reads: “We may write cos 5x cos 3x = 1 2 3cos 8x +cos 2x4.” Product-to-Sum Identities cos A cos B = 1 2 3 cos1A +B2 +cos1A −B2 4 sin A sin B = 1 2 3 cos1A −B2 −cos1A +B2 4 sin A cos B = 1 2 3 sin1A +B2 +sin1A −B2 4 cos A sin B = 1 2 3 sin1A +B2 −sin1A −B2 4
RkJQdWJsaXNoZXIy NjM5ODQ=