Trig Functions Review (Including Trig Quiz Solutions) MHF4UI Friday November 16th, 2012 Convert the following angles to Radians. Provide Exact Answers π 1° = 180 270π 270° = 180 3π = 2 π 1° = 180 900π 900° = 180 = 5π Convert the following angles to Degrees. Round to the nearest degree. 180 1πππ = ( )° π 8 8 180 =( • )° π π π ≈ 146° 180 1πππ = ( )° π 5.75 • 180 5.75 = ( )° π ≈ 329° Question 3: Finding the Arc Length π π= π π = ππ 11π π = 15 6 π ≈ 86.39 Therefore he travelled a distance of about 86.39 metres. 11π ππππ’πππ π£ππππππ‘π¦ = πππ βππ’π 6 11π ππππ’πππ π£ππππππ‘π¦ = πππ πππ 6 • 60 Therefore his angular velocity is 11π 360 per minute. Question 6: Solve the Equation 5 sin π₯ − 6 sin π₯ + 5 = 6 − sin π₯ = 1 sin π₯ = −1 Why must we have sec π ≥ 1 ππ sec π ≤ −1 sec π = βπ¦π πππ In a right angled triangle, the absolute value of the length of the hyp will always be greater than or equal to absolute value of the length of the adj βπ¦π ≥1 πππ Therefore sec π ≥ 1 ππ sec π ≤ −1 What is a Radian? Much like a degree, a Radian is a measurement of an angle. The radian measure of an angle ,Ο΄, is defined as the length, a, of the arc that “subtends” the angle divided by the radius of the arc ,r π π= π Radian Relationship to Degrees 180 1 ππππππ = ( )° π π 1° = πππππππ 180 Word Problems ππππ£πππ πππ π΄ππ πΏππππ‘β, πππππ’π ππ π πΉππππππ π‘βπ ππππ’πππ π£ππππππ‘π¦ ππ π πππ‘ππ‘πππ ππππππ‘ Trig Ratios Within a right angled triangle we will have that: πππ π πππ = βπ¦π πππ πππ π = βπ¦π πππ π‘πππ = πππ βπ¦π ππ ππ = πππ βπ¦π π πππ = πππ πππ πππ‘π = πππ Finding Trig Ratios We found Trig Ratios using our Related Acute angles of: π π π , , 4 3 6 We found trig ratios by drawing our angles in standard position. We found trig ratios when our terminal arm lies on the x or y axis. We also used trig ratios to solve word problems (Kite Example) Solving Trig Equations We solved for simple trig equations by using our special acute angles. We solved for trig equations by using our trig inverse function and finding ππ ππ π₯π We also encountered cases where we had to rearrange our equation and isolate our trig function. We used factoring or the quadratic formula to solve trig equations. When we solved Trig Equations we must note the restriction on our solution: - 0 ≤ π ≤ 2π - No restrictions (infinite solutions) - - other restrictions (we added or subtracted 2π) Graphing Trig Functions We sketched the graphs of all 6 trig functions We found the characteristics of each of these trig functions: - Max/Min - Amplitude - Period - Intercepts - Asymptotes Transformations of Sinusoidal Functions The general form of a Sinusoidal Function can be written as: π¦ = a • sin π π₯ − π + π OR π¦ = a • cos π π₯ − π + π We discussed the effects of variables a, k, c and d on the function We used mapping notation to graph our transformed functions - (x, π¦) 1 π ( π₯ + π, ππ¦ + π) We found the 5 key points for both sine and cosine functions We also used formulas of amplitude, period, vertical shift as well as our knowledge of the behaviour of sine and cosine functions to find the equation in general form Applications of Sinusoidal Functions We applied sine and cosine functions to real life scenarios where we: • Sketched the function • Found the characteristics of the function • Found the equation of the function that would model the scenario • Solved for specific values of the function