SECTION 7.5 Sum and Difference Formulas 517 3 Use Sum and Difference Formulas Involving Inverse Trigonometric Functions CAUTION Be careful when using formulas (6) and (7). These formulas can be used only for angles α and β for which α tan and β tan are defined. That is, they can be used for all angles except odd integer multiples of π 2 . j Establishing an Identity Establish the identity: θ π θ ( ) + = − tan 2 cot EXAMPLE 8 Solution Formula (6) cannot be used because π tan 2 is not defined. Instead, proceed as follows: θ π θ π θ π θ π θ π θ π θ π θ θ θ θ θ θ θ ( ) ( ) ( ) ( ) ( ) ( ) ( ) + = + + = + − = ⋅ + ⋅ ⋅ − ⋅ = − = − tan 2 sin 2 cos 2 sin cos 2 cos sin 2 cos cos 2 sin sin 2 sin 0 cos 1 cos 0 sin 1 cos sin cot Finding the Exact Value of an Expression Involving Inverse Trigonometric Functions Find the exact value of: ( ) + − − sin cos 1 2 sin 3 5 1 1 EXAMPLE 9 Solution We want the sine of the sum of two angles, α = − cos 1 2 1 and β = − sin 3 5 . 1 Then cos 1 2 0 and sin 3 5 2 2 α α π β π β π = ≤ ≤ = − ≤ ≤ Use Pythagorean Identities to obtain α sin and β cos . Note that α ≥ sin 0 because α π ≤ ≤ 0 and β ≥ cos 0 because π β π − ≤ ≤ 2 2 . This means that α α β β = − = − = = = − = − = = sin 1 cos 1 1 4 3 4 3 2 cos 1 sin 1 9 25 16 25 4 5 2 2 As a result, α β α β α β ( ) ( ) + = + = + = ⋅ + ⋅ = + − − sin cos 1 2 sin 3 5 sin sin cos cos sin 3 2 4 5 1 2 3 5 4 3 3 10 1 1 Now Work PROBLEM 77 Writing a Trigonometric Expression as an Algebraic Expression Write ( ) + − − u v sin sin cos 1 1 as an algebraic expression containing u and v (that is, without trigonometric functions). State the restrictions on u and v. EXAMPLE 10 (continued) NOTE In Example 9, α sin also can be found by using α = = x r cos 1 2 , so = x 1 and = r 2. Then = y 3 and α = = y r sin 3 2 . Also, β cos can be found in a similar fashion. j Option 1 First, for − u sin , 1 the restriction on u is − ≤ ≤ u 1 1, and for − v cos , 1 the restriction on v is − ≤ ≤ v 1 1. Now let α = − u sin 1 and β = − v cos . 1 Then α π α π β β π = − ≤ ≤ − ≤ ≤ = ≤ ≤ − ≤ ≤ u u v v sin 2 2 1 1 cos 0 1 1 Because π α π − ≤ ≤ 2 2 , α ≥ cos 0. So, α α = − = −u cos 1 sin 1 2 2 Also, because β π ≤ ≤ 0 , β ≥ sin 0. Then β β = − = −v sin 1 cos 1 2 2 Solution
RkJQdWJsaXNoZXIy NjM5ODQ=