Sin3 sin5 73 sin 4 cos2 74 sin5 cos3 75 cos5 sin3 76

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  sin3 sin5 73.   sin 4 cos2 74.   sin5 cos3 75.   cos5 sin3 76.   cos4 sin 2 (77–84) Convert the following sums or differences to products. 77. sin 5 sin3 78. sin 4 sin 2 80. sin 5 sin3 80. sin 4 sin 2 81. cos5 cos3 82. cos4 cos2 83. cos5 cos3 84. cos4 cos2
Section 6.1 Trigonometric Identities Section 6.4 Solving Trigonometric Equations Objectives Understanding how to solve trigonometric equations It is now time to solve equations that have trigonometric functions. To solve equations of this type you will need to get the equations into factors equal to zero with each factor containing only one trigonometric function. General Strategy: 1. Get the equation equal to zero 2. Convert all trigonometric functions into the same function by using identities or if that isn’t possible, then factor the equation into factors where each has only one type of trigonometric function. 3. Set each factor equal to zero and solve for the trigonometric function. 4. Lastly identify which angles make each equation true. SOLVING TRIGONOMETRIC EQUATIONS Discussion 1: Solving a Simple Equation Let’s find all the angles that make 2sin θ − 1 = 0. As we think about our strategy for solving this we see that it is equal to zero already and that there is only one function. Therefore, we will begin by getting sin θ alone. 2sin θ − 1 = 0 2sin θ = 1 (added one) sin θ = 1 2 (divided by two) Now, we need to think about what angle input would cause the output of the sine function to be 1 2 . sin θ = y on the unit circle. An angle that has a point on the unit circle with y -value of 1 2 is 6 . We know that 6 is one angle. Now we need to think about every other angle whose reference angle is 6 that would cause the output of the sine function to be positive. Sine is positive in the 1 st and 2 nd quadrants. So, we have 6 and 5 6 (whose reference angle in the 2 nd quadrant is 6 ). In addition to these two we also have every 2 π revolutions around the unit circle from these two angles, which will also
Chapter 6 Trigonometric Identities and Equations make sine equal 1 2 , since sine is periodic with a period of 2 π . Final answer. θ = 2 6 n and θ = 5 2 6 n ( n any integer) We could have used the calculator to help us with this problem as long as we are comfortable with non-exact solutions. We would still need to solve for sine. 2sin θ − 1 = 0 2sin θ = 1 (added one) sin θ = 1 2 (divided by two) But now we could use the calculator and find the sin -1 1 2 = θ . Notice that 0.52359… is the same as 6 which we found earlier as one of the many answers.

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