Presentation on theme: "Inverse Trig Functions Inverse Trig = Solve for the Angle."— Presentation transcript:
Inverse Trig Functions Inverse Trig = Solve for the Angle
INVERSE vs. RECIPROCAL b Its important not to confuse an INVERSE trig function with a RECIPROCAL trig function. b The reciprocal of SINE is COSECANT: (sin x ) -1 = csc x b The inverse of SINE is ARCSIN: sin -1 x = arcsin x The notation is very important - be careful !!
Evaluating INVERSE Trigs b Evaluating an inverse trig means finding the ANGLE that gives us the stated ratio. For example, to evaluate arcsin 0.5, (read “the arcsine of 0.5”) we must ask ourselves “for what angle does sin = 0.5 ? “ Answer: 30° or π/6 radians
How Many Answers are Right? b There are other angles for which the sine of that angle = 0.5. For example sin 30° = 0.5sin 30° = 0.5 sin 150° = 0.5sin 150° = 0.5 sin 390° = 0.5sin 390° = 0.5 b The way we handle this is to give a general solution, that is written in a way such that all possible solutions are included in it.
General Solutions b Your solution can include all angles that are co-terminal to the solution angle by adding 360k or 2πk to the angle, where k is any integer (positive or negative). In our example, the first obvious angle was 30° or π/6 radians: arcsin 0.5 = 30° + 360k
General Solutions b To be complete, the solution must include angles from other quadrants that produce the same sign and value. Think of this as the reverse of the process we used when we found reference angles --- this time we know the reference angle is 30° and the sign is positive, so we have to find the angle in quadrant II ( the other quadrant where sine is positive ) that shares the same reference angle ( 30° ). The answer ?... 150°
General Solutions b Our complete, general solution is: arcsin 0.5 = 30° + 360k ; 150° + 360k OR, in radiansarcsin 0.5 = 30° + 360k ; 150° + 360k OR, in radians arcsin 0.5 = π/6 + 2πk ; 5π/6 + 2πkarcsin 0.5 = π/6 + 2πk ; 5π/6 + 2πk b Solve these: arccos ( -0.5 )arccos ( -0.5 ) arcsec 1arcsec 1
General Solutions b Solutions: arccos ( -0.5 ) = 120° + 360k; 240° + 360karccos ( -0.5 ) = 120° + 360k; 240° + 360k OR arccos ( -0.5 ) = 2π/3 + 2πk; 4π/3 + 2πk OR arccos ( -0.5 ) = 2π/3 + 2πk; 4π/3 + 2πk arcsec 1 = 0° + 360k OR arcsec 1 = 0 + 2πkarcsec 1 = 0° + 360k OR arcsec 1 = 0 + 2πk b Notice that in our second example there is only one angle between 0° and 360° where secant = 1. The secant of 180° = -1.
The Briefest Answer b Evaluate arctan ( -1 ) First find the Quad I angle where tan = 1: 45° or π/4 radians.First find the Quad I angle where tan = 1: 45° or π/4 radians. Now, in which quadrants is tan negative ? Quadrants II and IVNow, in which quadrants is tan negative ? Quadrants II and IV Angles: 135° + 360k ; 315° + 360k But this answer can be simplified to 135° + 180k. Why does this satisfy ALL the angles ?Angles: 135° + 360k ; 315° + 360k But this answer can be simplified to 135° + 180k. Why does this satisfy ALL the angles ?
b arcsin ( -1 ) b arccos ( √2/2 ) b arctan ( -√3 ) b arccsc √2 b arccot 0 b arcsec ( -2 ) Practice Problems
Summary b When evaluating a trig function for a given angle, there is ONE ANSWER. Remember all the rules for evaluating trig functions: reference & coterminal angles, positive & negative quadrants, etc. b When evaluating an inverse trig function, there can be INFINITE ANSWERS.
Summary b First find the angle in quadrant I that yields the desired ratio ( the number given in the inverse trig function ). b Then determine which quadrant ( s ) the solution is in to achieve the given sign - sometimes there is only 1 angle, that is OK. b State the angle answer ( s ) for those quadrants.
Summary b Add a multiple of k to include all co- terminal solutions ( i.e. + 360k or + 2πk ) for each of your angle solutions. b Finally, it may be possible to make your answer more brief, like we did with arctan. This is usually the case when the solutions differ by exactly 180° or π radians.