sin3sin573. sin 4cos274. sin5cos375. cos5sin376. cos4sin 2(77–84) Convert the following sums or differences to products.77. sin 5sin378. sin 4sin 280. sin 5sin380. sin 4sin 281. cos5cos382. cos4cos283. cos5cos384. cos4cos2
Section 6.1 Trigonometric Identities Section 6.4 Solving Trigonometric EquationsObjectivesUnderstanding 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 zero2.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 EQUATIONSDiscussion 1: Solving a Simple EquationLet’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 θalone.2sin θ− 1 = 02sin θ= 1 (added one)sin θ=12(divided by two)Now, we need to think about what angle input would cause the output of the sine function to be 12.sin θ= yon the unit circle. An angle that has a point on the unit circle with y-value of12is6.We know that6is one angle. Now we need to think about every other angle whose reference angle is6that would cause the output of the sine function to be positive.Sine is positive in the 1stand 2ndquadrants. So, we have6and 56(whose reference angle in the2ndquadrant is6). In addition to these two we also have every 2πrevolutions around the unit circle from these two angles, which will also
Chapter 6 Trigonometric Identities and Equationsmake sine equal12, since sine is periodic with a period of 2π.Final answer.θ=26nand θ=526n(nany 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 = 02sin θ= 1 (added one)sin θ=12(divided by two)But now we could use the calculator andfind the sin-1 12=θ.Notice that 0.52359… is the same as 6which we found earlier as one of the many answers.