Inverse Trigonometric Functions PDF

HANDOUT 3.4 INVERSE OF SINE AND COSINE FUNCTIONS Inverse Sine Function The domain of the sine function is the set ℝ of real numbers, and its range is the closed interval [−1, 1]. As observed in the previous lessons, the sine function is not one-to- one, and the first...

HANDOUT 3.4 INVERSE OF SINE AND COSINE FUNCTIONS Inverse Sine Function The domain of the sine function is the set ℝ of real numbers, and its range is the closed interval [−1, 1]. As observed in the previous lessons, the sine function is not one-to- one, and the first step is to restrict its domain (by agreeing what the convention is with the following conditions: (1) the sine function is one-to-one in that restricted domain, and (2) the range remains the same. The inverse of the (restricted) sine function 𝑦 = sin 𝑥, where the domain is restricted to 𝜋 𝜋 the closed interval [− 2 , 2 ], is called the inverse sine function or arcsine function, denoted by 𝜋 𝜋 𝑦 = 𝑠𝑖𝑛−1 𝑥 or 𝑦 = arcsin 𝑥. Here, the domain of 𝑦 = arcsin 𝑥 is [−1, 1], and its range is [− 2 , 2 ]. Thus, 𝒚 = 𝒔𝒊𝒏−𝟏 𝒙 (read as “y is the inverse sine of x”) or 𝒚 = 𝐚𝐫𝐜𝐬𝐢𝐧 𝒙 (read as “y is an angle whose sine is x.”) if and only if sin 𝑦 = 𝑥; 𝜋 𝜋 where −1 ≤ 𝑥 ≤ 1 and − 2 ≤ 𝑦 ≤ 2 Graph of Inverse Sine Function 1 Example 1. Find the exact value of each expression. 1 (a) 𝑠𝑖𝑛−1 (2) Solution: Let 𝜃 = 𝑠𝑖𝑛−1 𝑥, thus, 1 𝜃 = 𝑠𝑖𝑛−1 (2). 1 This is equivalent to sin 𝜃 = 2. This means that we are looking for the 𝜋 𝜋 1 𝜋 number 𝜃 in the closed interval [− 2 , 2 ] whose sine is 2. We get 𝜃 = 6. 1 𝜋 Thus, we have 𝑠𝑖𝑛−1 (2) = 6. √3 (b) 𝑠𝑖𝑛−1 (− ) 2 Solution: Let 𝜃 = 𝑠𝑖𝑛−1 𝑥, thus, √3 𝜃 = 𝑠𝑖𝑛−1 (− ). 2 √3 This is equivalent to sin 𝜃 = −. This means that we are looking for the 2 𝜋 𝜋 √3 number 𝜃 in the closed interval [− , ] whose sine is −. We get 2 2 2 𝜋 √3 𝜋 𝜃 = − 3. Thus, we have 𝑠𝑖𝑛−1 (− ) = − 3. 2 2 Example 2. Sketch the graph of each equation. 𝜋 (a) 𝑦 = 𝑠𝑖𝑛−1 𝑥 + 2 (b) 𝑦 = 𝑠𝑖𝑛−1 (𝑥 − 1) Inverse Cosine Function The development of the inverse cosine function follows similarly from that of the inverse sine function. 𝑦 = 𝑐𝑜𝑠 −1 𝑥 or 𝑦 = arccos 𝑥 means cos 𝑦 = 𝑥 where −1 ≤ 𝑥 ≤ 1 and 0 ≤ 𝑦 ≤ 𝜋. Example 3. Find the exact value of each expression. √2 (a) 𝑐𝑜𝑠 −1 ( ) 2 Solution: √2 Find the angle 𝜃, 0 ≤ 𝑦 ≤ 𝜋, whose cosine value equals. 2 3  √2 𝜋 The angle between 0 and 𝜋, inclusive, whose cosine value is is 4. 2 √2 𝜋 Thus, 𝑐𝑜𝑠 −1 ( 2 ) = 4. 1 (b) 𝑐𝑜𝑠 −1 (− 2) Solution: 1 2𝜋 The angle between 0 and 𝜋, inclusive, whose cosine value is − 2 is 3. 1 2𝜋 Thus, 𝑐𝑜𝑠 −1 (− 2) =. 3 Example 4. Sketch the graph of each equation. 𝑥 (a) 𝑦 = 2𝑐𝑜𝑠 −1 (3) (b) 𝑦 = 𝑐𝑜𝑠 −1 (𝑥 − 2) 𝜋 (c) 𝑦 = 𝑐𝑜𝑠 −1 𝑥 + 2 Example 5. Find the exact value, if possible, of each expression. 1 (a) sin(𝑠𝑖𝑛−1 2) Solution: 1 𝜋 1 sin(𝑠𝑖𝑛−1 ) = sin = 2 6 2 4  √3 (b) cos (𝑐𝑜𝑠 −1 (− )) 2 Solution: √3 5𝜋 √3 √3 cos (𝑐𝑜𝑠 −1 (− )) = cos ( 6 ) = − because − ∈ [−1,1] 2 2 2 √2 (c) sin (𝑐𝑜𝑠 −1 ) 2 Solution: √2 𝜋 √2 sin (𝑐𝑜𝑠 −1 ) = sin = 2 4 2 Example 6. Evaluate each of the following. 3 (a) cos (arcsin 5) Solution: 3 3 Let 𝜃 = arcsin 5, then sin θ = 5, 𝜃 being a first-quadrant angle. See figure below. 3 4 cos (arcsin ) = cos 𝜃 = 5 5 2 (b) sin (arccos(− 3)) Solution: 2 2 Let 𝜃 = arccos − 3, then cos θ = − 3, 𝜃 being a second-quadrant angle. See figure. 2 4 sin (arccos(− 3)) = sin 𝜃 = 5 2 1 Example 7. Simplify: sin (arcsin 3 + arccos 2) Solution: 1 𝜋 We know that 𝑎𝑟𝑐𝑐𝑜𝑠 2 = 3. Using the Sine Sum Identity, we have 2 1 = sin (arcsin 3 + arccos 2) 2 𝜋 = sin(arcsin 3 + 3 ) 2 𝜋 2 𝜋 = sin(arcsin ) cos + cos(arcsin ) sin 3 3 3 3 2 1 2 √3 = 3 ∙ 2 + cos (𝑎𝑟𝑐𝑠𝑖𝑛 3) ∙ 2 1 √3 2 =3+ cos (arcsin ) 2 3 5  2 2 We compute cos (𝑎𝑟𝑐𝑠𝑖𝑛 3). Let 𝜃 = arcsin 3, where 𝜃 lies in QI. Using the Pythagorean identity, we have 2 √5 cos (𝑎𝑟𝑐𝑠𝑖𝑛 3) = cos 𝜃 = √1 − 𝑠𝑖𝑛2 𝜃 = 3 Going back to the original computations above, we have 2 1 1 √3 2 sin (arcsin 3 + arccos 2) = 3 + cos (arcsin 3) 2 1 √3 √4 = + 2 ∙ 3 3 2+√15 = 6 General Values of Inverse Sine and Cosine Functions 1 Example 8. Write expressions for the general value of (𝑎) arcsin 2 and (𝑏) arccos(−1). 1 (a) arcsin 2 1 𝜋 The principal value of arcsin is , and a second value (not coterminal with the principal 2 6 5𝜋 1 value) is. The general value of arcsin 2 is given by 6 𝜋 5𝜋 + 2𝑛𝜋 + 2𝑛𝜋 6 6 where n is any positive and negative integer or zero. (b) arccos(−1) The principal value of arccos(−1) is 𝜋, and there is no other value not coterminal with it. Thus, the general value of arccos(−1) is given by 𝜋 + 2𝑛𝜋 where n is any positive and negative integer or zero. Solving Equations Involving Inverse Sine and Cosine Functions 3 𝑦 𝜋 (a) Solve: 4 𝑐𝑜𝑠 −1 (3 ) = 2 Solution: 4 3 −1 𝑦 𝜋 42 ∙ 𝑐𝑜𝑠 ( ) = ∙ 3 4 3 2 3 𝑦 2𝜋 𝑐𝑜𝑠 −1 (3 ) = 3 𝑦 2𝜋 cos (𝑐𝑜𝑠 −1 (3 )) = cos ( 3 ) 𝑦 1 3 = −2 3 𝑦 = −2 𝜋 (b) Solve: 𝑠𝑖𝑛−1 (𝑥 2 − 2𝑥) = − 2 Solution: 𝜋 sin(𝑠𝑖𝑛−1 (𝑥 2 − 2𝑥)) = sin (− ) 2 𝑥 2 − 2𝑥 = −1 𝑥 2 − 2𝑥 + 1 = 0 6  (𝑥 − 1)2 = 0 𝑥=1 Inverse Properties 1. sin(𝑠𝑖𝑛−1 𝑥) = 𝑥 for every x in the interval [−1,1] 𝜋 𝜋 2. 𝑠𝑖𝑛−1 (sin 𝑥) = 𝑥 for every x in the interval [− 2 , 2 ] 3. cos(𝑐𝑜𝑠 −1 𝑥) = 𝑥 for every x in the interval [−1,1] 2. 𝑐𝑜𝑠 −1 (cos 𝑥) = 𝑥 for every x in the interval [0, 𝜋] Practice Exercises: A. Find the exact radian value. 1. 𝑠𝑖𝑛−1 1 6. arccos (−1) 1 2. 𝑐𝑜𝑠 (− 2) −1 7. arcsin 0 √3 3. 𝑠𝑖𝑛−1 (− ) 8. arccos 0 2 √3 √2 4. 𝑐𝑜𝑠 −1 ( 2 ) 9. arccos (− ) 2 √2 5. 𝑠𝑖𝑛−1 ( 2 ) 10. arcsin(−1) B. Evaluate each of the following. 3𝜋 3 1. 𝑐𝑜𝑠 −1 (cos 4 ) 6. cos (arcsin 5) 5𝜋 2 2. arccos (sin ) 7. sin [arccos (− 3)] 2 𝜋 12 4 3. 𝑐𝑜𝑠 −1 (sin 4 ) 8. sin (arcsin 13 + arcsin 5) 2𝜋 2 1 4. 𝑠𝑖𝑛−1 [cos (− )] 9. cos [𝑐𝑜𝑠 −1 3 − 𝑠𝑖𝑛−1 (− 3)] 3 −1 √2 2 1 5. cos (2 𝑠𝑖𝑛 2 ) 10. sin (𝑠𝑖𝑛−1 3 + 𝑐𝑜𝑠 −1 2) C. Graph each of the following equations below. 1. 𝑦 = 2 𝑐𝑜𝑠 −1 (3𝑥) 1 2. 𝑦 = 2 𝑐𝑜𝑠 −1 (𝑥 + 2) 3. 𝑦 = 2 𝑠𝑖𝑛−1 (𝑥 − 2) 4. 𝑦 = 2𝑠𝑖𝑛−1 (2𝑥) 5. 𝑦 = 2𝑠𝑖𝑛−1 (2𝑥 + 2) D. Solve for x. 5 𝜋 1. 𝑠𝑖𝑛−1 𝑥 = 𝑐𝑜𝑠 −1 13 6. 𝑠𝑖𝑛−1 (𝑥 − 2) = − 6 4 𝑥 4 𝜋 2. 𝑐𝑜𝑠 −1 (4) = 𝜋 7. 𝑠𝑖𝑛−1 𝑥 + 𝑐𝑜𝑠 −1 5 = 7 6 𝜋 3. 𝑠𝑖𝑛−1 (𝑥 − 1) = 8. arccos 2𝑥 = arcsin 𝑥 2 −1 3 √3 𝜋 4. −1 𝑠𝑖𝑛 5 + 𝑐𝑜𝑠 𝑥 = 𝜋 9. 𝑐𝑜𝑠 −1 𝑥 + 𝑠𝑖𝑛−1 = 2 2 √2 2𝜋 1 5. 𝑠𝑖𝑛−1 2 + 𝑐𝑜𝑠 −1 𝑥 = 3 10. arccos(2𝑥 − 1) = 2 arccos 2 2 7 Inverse Tangent Function The inverse tangent function is similarly defined as inverse sine and inverse cosine functions. 𝒚 = 𝒕𝒂𝒏−𝟏 𝒙 or 𝒚 = 𝐚𝐫𝐜𝐭𝐚𝐧 𝒙 means tan 𝑦 = 𝑥, 𝜋 𝜋 where 𝑥 ∈ ℝ and − 2 < 𝑦 < 2. Graph of Tangent Function Graph of Inverse Tangent Function Example 1. Find the exact value of each expression. 7𝜋 (c) 𝑡𝑎𝑛−1 (1) (d) 𝑡𝑎𝑛−1 (tan ) 6 19𝜋 (d) 𝑎𝑟𝑐𝑡𝑎𝑛(−√3) (e) 𝑎𝑟𝑐𝑡𝑎𝑛 (tan (− )) 6 𝜋 5 (e) 𝑡𝑎𝑛−1 (tan (− 6 )) (f) 𝑡𝑎𝑛−1 (tan (− 2)) 𝜋 𝜋 Solution: Note the range of arctan is the open interval (− 2 , 2 ) 𝜋 (a) 𝑡𝑎𝑛−1 (1) = 4 𝜋 (b) 𝑎𝑟𝑐𝑡𝑎𝑛(−√3) = − 3 𝜋 𝜋 𝜋 𝜋 𝜋 (c) 𝑡𝑎𝑛−1 (tan (− 6 )) = − 6 because − 6 𝜖 (− 2 , 2 ) 𝜋 𝜋 Note: If − 2 < 𝑥 < 2 , then 𝑡𝑎𝑛−1 (tan 𝑥) = 𝑥 7𝜋 (d) 𝑡𝑎𝑛−1 (tan ) 6 7𝜋 𝜋 𝜋 7𝜋 𝜋 Note that ∉ (− 2 , 2 ). Use the idea of reference angle, we know that tan = tan 6. 6 6 8  7𝜋 𝜋 𝜋 𝑡𝑎𝑛−1 (tan ) = 𝑡𝑎𝑛−1 (tan 6 ) = 6 6 19𝜋 (e) 𝑎𝑟𝑐𝑡𝑎𝑛 (tan (− )) 6 Here we cannot use the idea of reference angle, but the idea can help in a way. The 19𝜋 𝜋 number (or angle) − 6 is in QII, wherein tangent is negative, and its reference angle is 6. 19𝜋 𝜋 𝜋 𝑎𝑟𝑐𝑡𝑎𝑛 (tan (− )) = 𝑎𝑟𝑐𝑡𝑎𝑛 (tan (− )) = − 6 6 6 5 5 (f) 𝑡𝑎𝑛−1 (tan (− 2)) = − 2 Note: If 𝑥 is any real number, then tan(𝑡𝑎𝑛−1 𝑥) = 𝑥 Example 2: Find the exact value of each expression. 8 (a) sin (2 𝑡𝑎𝑛−1 (− 3)) 3 1 (b) tan (𝑠𝑖𝑛−1 5 − 𝑡𝑎𝑛−1 4) Solution. 8 8 (a) Let 𝜃 = 𝑡𝑎𝑛−1 (− 3). Then tan 𝜃 = − 3, and 𝜃 lies in QIV, and 𝑥 = 3 and 𝑦 = −8. We get, 𝒓 = √𝒙𝟐 + 𝒚𝟐 = √𝟑𝟐 + (−𝟖)𝟐 = √𝟕𝟑 Applying the Sine Double-Angle Identity, 8 sin (2 𝑡𝑎𝑛−1 (− )) = sin 2𝜃 3 = 2 sin 𝜃 cos 𝜃 𝑦 𝑥 =2 ⋅𝑟⋅𝑟 −8 3 =2⋅ ⋅ √73 √73 48 = − 73 (b) Using the Tangent Difference Identity, 3 1 tan (𝑠𝑖𝑛−1 − 𝑡𝑎𝑛−1 ) 5 4 −1 3 1 tan (𝑠𝑖𝑛 ) − tan (𝑡𝑎𝑛−1 4) = 5 3 1 1 + tan (𝑠𝑖𝑛−1 ) tan (𝑡𝑎𝑛−1 4) 5 3 1 tan (𝑠𝑖𝑛−1 ) − 4 = 5 3 1 1 + tan (𝑠𝑖𝑛−1 ) (4) 5 −1 3 3 Let 𝜃 = 𝑠𝑖𝑛 5. Then sin 𝜃 = 5 and 𝜃 lies in QI, and 𝑦 = 3 and 𝑟 = 5. We get, 𝒙 = √𝒓𝟐 − 𝒚𝟐 = √𝟓𝟐 − (𝟑)𝟐 = 𝟒 9  3 so that, tan 𝜃 = 4. 3 1 3 1 tan (𝑠𝑖𝑛−1 ) − 4 tan (𝑠𝑖𝑛−1 − 𝑡𝑎𝑛−1 ) = 5 5 4 3 1 1 + tan (𝑠𝑖𝑛−1 ) (4) 5 3 1 − 4 4 = 3 1 1+( )( ) 4 4 8 = 19 Inverse Cotangent Function Tangent and cotangent have a slightly different relationship. The graph of cotangent differs 𝜋 from tangent by a reflection over the y-axis and shift of 2. As an equation, this can be written 𝜋 as cot 𝑥 = − tan (𝑥 − 2 ). Taking the inverse of this function will show the inverse reciprocal between 𝑎𝑟𝑐𝑐𝑜𝑡𝑎𝑛𝑔𝑒𝑛𝑡 and 𝑎𝑟𝑐𝑡𝑎𝑛𝑔𝑒𝑛𝑡. 𝑦 = 𝑐𝑜𝑡 −1 𝑥 𝜋 𝑦 = −𝑡𝑎𝑛−1 (𝑥 − 2 ) 𝜋 𝑥 = −𝑡𝑎𝑛 (𝑦 − 2 ) 𝜋 −𝑥 = 𝑡𝑎𝑛 (𝑦 − 2 ) 𝜋 𝑡𝑎𝑛−1 (−𝑥) = 𝑦 − 2 𝜋 + 𝑡𝑎𝑛−1 (−𝑥) = 𝑦 2 𝜋 − 𝑡𝑎𝑛−1 𝑥 = 𝑦 2 Definition 𝜋 𝑐𝑜𝑡 −1 𝑥 = 2 − 𝑡𝑎𝑛−1 𝑥 where the domain of 𝑦 = 𝑐𝑜𝑡 −1 𝑥 is ℝ and its range is (0, 𝜋). Graph of Cotangent Function Graph of Inverse Cotangent Function Inverse Secant Function 𝑦 = 𝑠𝑒𝑐 −1 𝑥 or 𝑦 = arcsec 𝑥 means sec 𝑦 = 𝑥, 𝜋 𝜋 where 𝑥 ≤ −1 or 𝑥 ≥ 1 or the domain is (−∞, −1] ∪ [1, ∞) and the range is [0, 2 ) ∪ ( 2 , 𝜋] or 0 ≤ 𝜋 𝑦 ≤ 𝜋, 𝑦 ≠ 2. 10 Graph of Secant Function Graph of Inverse Secant Function Note: Keep in mind that the domain restrictions are conventions we set. Other books and sources might have different domain restrictions. The restrictions made aim to make calculus computations easier in the future. Inverse Cosecant Function 𝑦 = 𝑐𝑠𝑐 −1 𝑥 or 𝑦 = arccsc 𝑥 means csc 𝑦 = 𝑥, 𝜋 𝜋 where 𝑥 ≤ −1 or 𝑥 ≥ 1 or the domain is (−∞, 1] ∪ [1, ∞) and the range is [− 2 , 0) ∪ (0, 2 ] or 𝜋 𝜋 −2 ≤ 𝑦 ≤ 2,𝑦 ≠ 0 Graph of Cosecant Function Graph of Inverse Cosecant Function The functions 𝑦 = 𝑠𝑒𝑐 −1 𝑥, 𝑦 = 𝑐𝑠𝑐 −1 𝑥, and 𝑦 = 𝑐𝑜𝑡 −1 𝑥 can be evaluated by noting their relationship to 𝑦 = 𝑐𝑜𝑠 −1 𝑥, 𝑦 = 𝑠𝑖𝑛−1 𝑥, and 𝑦 = 𝑡𝑎𝑛−1 𝑥, respectively. For 𝑦 = 𝑠𝑒𝑐 −1 𝑥, we have 𝑦 = sec 𝑥 𝑥 = sec 𝑦 1 1 = sec 𝑦 𝑥 1 cos 𝑦 = 𝑥 1 𝑦 = 𝑐𝑜𝑠 −1 ( ) 𝑥 1 𝑠𝑒𝑐 −1 𝑥 = 𝑐𝑜𝑠 −1 (𝑥) 11  1 Similarly, the expression 𝑐𝑠𝑐 −1 𝑥 is the same as 𝑠𝑖𝑛−1 (𝑥). The expression 𝑐𝑜𝑡 −1 𝑥 can likewise be 1 the same as 𝑡𝑎𝑛−1 (𝑥). Summary for Inverse Tangent, Secant, Cosecant, and Cotangent Functions 1. For tan 𝑥 and 𝑡𝑎𝑛−1 𝑥, tan(𝑡𝑎𝑛−1 𝑥) = 𝑥, for any real number x 𝜋 𝜋 𝑡𝑎𝑛−1 (tan 𝑥) = 𝑥, for any x in the interval (− 2 , 2 ) 1 2. To evaluate 𝑠𝑒𝑐 −1 𝑥, use 𝑐𝑜𝑠 −1 (𝑥) , |𝑥| ≥ 1, 1 𝑐𝑠𝑐 −1 𝑥, use 𝑠𝑖𝑛−1 (𝑥) , |𝑥| ≥ 1 𝜋 𝑐𝑜𝑡 −1 𝑥, use 2 − 𝑡𝑎𝑛−1 𝑥, for all real numbers x Example 1. Find the exact value of each expression. 2√3 (c) 𝑐𝑜𝑡 −1 (−√3) (d) 𝑐𝑠𝑐 −1 (− ) 3 (d) 𝑎𝑟𝑐𝑐𝑜𝑡(0) (e) arccsc(−1) 3 2√3 (e) 𝑠𝑒𝑐 −1 (2) (f) sin (𝑠𝑒𝑐 −1 (− 2) − 𝑐𝑠𝑐 −1 (− )) 3 Solution. 5𝜋 5𝜋 (a) 𝑐𝑜𝑡 −1 (−√3) = 6 , because 6 ∈ (0, 𝜋) 𝜋 (b) 𝑎𝑟𝑐𝑐𝑜𝑡(0) = 2 1 𝜋 𝜋 𝜋 (c) 𝑠𝑒𝑐 −1 (2) = 𝑐𝑜𝑠 −1 (2) = 3 , because ∈ [0, 2 ) 3 2√3 √3 𝜋 𝜋 𝜋 (d) 𝑐𝑠𝑐 −1 (− ) = 𝑠𝑖𝑛−1 (− ) = − 3 , because − 3 ∈ [− 2 , 0) 3 2 𝜋 (e) arccsc(−1) = arcsin(−1) = − 2 3 2√3 (f) sin (𝑠𝑒𝑐 −1 (− 2) − 𝑐𝑠𝑐 −1 (− )) 3 2√3 𝜋 3 3 From (d), we know that 𝑐𝑠𝑐 −1 (− ) = − 3. Let 𝜃 = 𝑠𝑒𝑐 −1 (− 2), then sec 𝜃 = − 2. From 3 the defined range of the inverse secant function and the notations, 𝜃 lies in QII, and 𝑟 = 3 and 𝑥 = −2. Solving for y, we get, 𝑦 = √32 − (−2)2 = √5 √5 2 It follows that sin 𝜃 = and cos 𝜃 = − 3 3 Using the Sine Sum Identity, 3 2√3 𝜋 sin (𝑠𝑒𝑐 −1 (− ) − 𝑐𝑠𝑐 −1 (− )) = sin (𝜃 − (− )) 2 3 3 𝜋 = sin (𝜃 + 3 ) 𝜋 𝜋 = sin 𝜃 cos 3 + cos 𝜃 sin 3 √5 1 2 √3 = ( 3 ) (2) + (− 3) ( 2 ) 12  √5−2√3 = 6 Practice Exercises: A. Find the exact value of each expression. 6. 𝑠𝑒𝑐 −1 (−1) 6. cot(𝑐𝑜𝑡 −1 (−10)) 3𝜋 7. arctan(−1) 7. csc −1 (− tan 4 ) 5𝜋 3 8. arccsc (csc ) 8. csc (𝑐𝑠𝑐 −1 (− 8)) 2 𝜋 9. sin(arctan 2) 9. 𝑐𝑠𝑐 −1 (cos 3 ) 𝜋 10. tan(arcsin 0) 10. 𝑐𝑠𝑐 −1 (tan 6 ) B. Simplify each expression 5 1 √3 6. cos (arcsec 2 − arccot 3) 6. sin (2𝑠𝑖𝑛−1 2 − 3𝑡𝑎𝑛−1 ) 3 1 5 3 15 7. tan (𝑡𝑎𝑛−1 (− 2) + 𝑡𝑎𝑛−1 3) 7. tan (arctan 4 + arccot 8 ) √3 4 12 8. cos (𝑡𝑎𝑛−1 √3 + 𝑠𝑖𝑛−1 (− )) 8. cos (arctan (− 3) +arcsin 13) 2 3 5 9. tan(2𝑡𝑎𝑛−1 (−1)) 9. tan (arcsin (− ) − arccos ) 5 13 4 5 4 12 10. cos (𝑡𝑎𝑛−1 + 𝑐𝑜𝑠 −1 ) 10. tan (2 arcsin + arccos ) 3 13 5 13 D. Solve for x. 𝜋 𝜋 6. 𝑡𝑎𝑛−1 (4𝑥 2 + 5𝑥 − 7) = − 6. 𝑠𝑖𝑛−1 (𝑥 − 2) = − 6 4 4 4 𝜋 7. arctan 𝑥 + arctan(1 − 𝑥) = arctan 3 7. 𝑠𝑖𝑛−1 𝑥 + 𝑐𝑜𝑠 −1 5 = 6 𝜋 8. arctan 2𝑥 + arctan 𝑥 = 8. arccos 2𝑥 = arcsin 𝑥 4 𝜋 √3 𝜋 9. arcsin 𝑥 + arctan 𝑥 = 9. 𝑐𝑜𝑠 −1 𝑥 + 𝑠𝑖𝑛−1 = 2 2 2 𝜋 1 10. arccos 𝑥 + arctan 𝑥 = 10. arccos(2𝑥 − 1) = 2 arccos 2 2 2 13