6.3 Sum and Difference Identities

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Presentation transcript:

6.3 Sum and Difference Identities Derive the identity for cos(A – B). Let angles A and B be angles in standard position on a unit circle with B < A and S and Q be the points on the terminal sides of angels A and B, respectively. Q has coordinates (cos B, sin B). S has coordinates (cos A, sin A). R has coordinates (cos (A – B), sin (A – B)). Angle SOQ equals A – B. Since SOQ = POR, chords PR and SQ are equal.

6.3 Sum and Difference Identities By the distance formula, chords PR = SQ, Simplifying this equation and using the identity sin² x + cos² x =1, we can rewrite the equation as cos(A – B) = cos A cos B + sin A sin B.

6.3 Sum and Difference Identities To find cos(A + B), rewrite A + B as A – (– B) and use the identity for cos (A – B). Cosine of a Sum Or Difference cos(A – B) = cos A cos B + sin A sin B cos(A + B) = cos A cos B – sin A sin B

6.3 Finding Exact Cosine Values Example Find the exact value of the following. cos 15° cos (or 60° – 45°)

6.3 Sine of a Sum or Difference Using the cofunction relationship and letting  = A + B, Now write sin(A – B) as sin(A + (– B)) and use the identity for sin(A + B).

6.3 Sine of a Sum or Difference sin(A + B) = sin A cos B + cos A sin B sin(A – B) = sin A cos B – cos A sin B

6.3 Tangent of a Sum or Difference Using the identities for sin(A + B), cos(A + B), and tan(–B) = –tan B, we can derive the identities for the tangent of a sum or difference. Tangent of a Sum or Difference

6.3 Example Using Sine and Tangent Sum or Difference Formulas Example Find the exact value of the following. sin 75° tan sin 40° cos 160° – cos 40° sin 160° Solution (a)

6.3 Example Using Sine and Tangent Sum or Difference Formulas (b) (c) sin 40° cos 160° – cos 40° sin 160° = sin(40° – 160°) = sin(–120°) Rationalize the denominator and simplify.

6.3 Finding Function Values and the Quadrant of A + B Example Suppose that A and B are angles in standard position, with Find each of the following. sin(A + B) (b) tan (A + B) (c) the quadrant of A + B Solution (a) Since cos A < 0 in Quadrant II.

6.3 Finding Function Values and the Quadrant of A + B (b) Use the values of sine and cosine from part (a) to get (c) From the results of parts (a) and (b), we find that sin(A + B) is positive and tan(A + B) is also positive. Therefore, A + B must be in quadrant I.

In 1831, Michael Faraday discovers a small electric current when a wire is passed by a magnet. This phenomenon is known as Faraday’s law. This property is used to produce electricity by rotating thousands of wires near large electromagnets. Electric generators supply electricity to most homes at a rate of 60 cycles per second. This rotation causes alternating current in wires and can be modeled by either sine or cosine functions.

6.3 Applying the Cosine Difference Identity to Voltage Example Common household electric current is called alternating current because the current alternates direction within the wire. The voltage V in a typical 115-volt outlet can be expressed by the equation V = 163 sin t, where  is the angular velocity (in radians per second) of the rotating generator at the electrical plant and t is time measured in seconds. It is essential for electric generators to rotate at 60 cycles per second so household appliances and computers will function properly. Determine  for these electric generators. Graph V on the interval 0  t  .05. For what value of  will the graph of V = 163cos(t – ) be the same as the graph of V = 163 sin t?

6.3 Applying the Cosine Difference Identity to Voltage Solution Since each cycle is 2 radians, at 60 cycles per second,  = 60(2) = 120 radians per second. (b) V = 163 sin t = 163 sin 120t. Because amplitude is 163, choose –200  V  200 for the range. (c)