Solved: Exer. 4756: Find the center and radius of the

Trigonometry exam 3 study guide Pythagorean Identities: 2 2 Sin θ+Cos θ=1 2 2 Tan θ+1=Sec θ 2 2 1+Cot θ=Csc θ X 0 π π π π Conjugates: 6 4 3 2 A+B--------A-B so (a+b)*(a-b)=a -b Sin 0 1 √ 2 √3 1 2 2 2 sin α−sin α=cos α−cos α 4 Co 1 3 2 1 0 √ √ s 2 2 2 sin α+cos α ≠ 1 Ta 0 √ 3 1 √ 3 DN 2 2 Instead: sin α=(1−cos α) and n 3 E sin α=(1−cos α)(1−cos α) or 1−2cos α+cos α4 Sin + All + Csc + Tan + Cos + Cot + Sec + 1 cosθ= 2 Cosine is positive in quadrants 1 and 4 and has two solutions 1 π cosr= 2sr= 3 Use the table to find the value of the reference angles Q1: π Q4: 5π 3 3 Assign the values from the unit circle Q1:θ= +k∗2π 3 Add two rotations to equal θ Q4:θ= 5π +k∗2π 3 Answer for sin and cos problems must have two values Let’s try a tangent problem: tanθ= √3 3 tanrisr= π 6 Q1:θ= +k∗π 6 tangent problems only require one value because tangent problems create a straight diagonal line Now for a more interesting problem: −2sin3θ+1=0 1 sin3θ= Use algebra to create a rational equation (note: sin3θ ≠ 3sinθ) 2 sinr= isr= π 2 6 Q1:3θ= +k∗2π 6 Write answers in terms of 3θ Q2:3θ= 5π +k∗2π 6 Q1:θ= π + k∗2π 18 3 Use division Q2:θ= 5π +∗2π 18 3 Next something even harder: 2 4 csc θ= 3 2 cscθ=+¿− Find the square roots √3 √3 sinθ=+¿− Reciprocate for a sin problem 2 √3 π sinr= isr= 2 3 π Q1:θ= +k∗2π There should be an answer in every quadrant 3 2π Q2:θ= 3 +k∗2π because the reference angle is positive and negative 4π Q3:θ= 3 +k∗2π 5π Q4:θ= +k∗2π 3 Now for Quadratic equations: 2 cos θ−cosθ=0 cosθ(cosθ−1 )=0 Factor cosθ=0∨cosθ=1 cosr=1isr=0 π cosr=0isr= 2 π π θ= +k∗π 2 Quadrants are irrel2van and 0 represent a right θ=k∗2π angle Last one: cosθ=1−sin 2 1−sinθ¿ cosθ¿ =¿ Square both sides ¿ cos θ=1−2sinθ+sin θ 2 1−sin θ=1−2sinθ+si n θ 2 Use Pythagorean identities 2sin θ−2sinθ=0 Move terms to one side 2sinθ(sinθ−1=0 Factor 2sinr=0issinr=0isr=0 π sinr=1isr= 2 θ= +k∗π Remember to check for extraneous solutions 2 θ=k∗2 π Cos(a-b) = cosacosb + sinasinb Cos(a+b) = cosacosb – sinasinb Sin(a-b) = sinacosb - cosasinb Sin(a+b) = sinacosb + cosasinb tana−tanb Tan(a-b) =1+tanatanb tana+tanb Tan(a+b) = 1−tanatanb Here is an example problem: 7π 7π 2π π Find the Exact value ofn12 (hint: 12 = 3 − 4 ) sin 2π− π =sin2π cosπ −cos2π sin (3 4 ) 3 4 3 4 Plug in the values ¿ √3 √2 − −1 √2 Derive reference ( 2 ( 2 (2) (2) angles 6 − 2 ¿ √ − √ Multiply 4 4 6+ 2 √ √ Combine (this is the 4 finished form of the problem Double angle formulas: Half angle formulas(+ and – is either/ or not both): Sin: sin2x=2sinxcosx Sin: x 1−cosx sin2=± √ 2 2 2 Cos: cos2x=cos x−si n x Cos: x 1+cosx cos2=± √ 2 2 cos2x=2cos x−1 Tan: tan = 1−cosx 2 sinx cos2x=1−2si n x tan2 x=2tanx Tan: 1−tan x −29 3π −21 −20 Example: secθ= π