Pre-Calc Exam Notes 127

Pre-Calc Exam Notes 127 - Inverse Trigonometric Functions...

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Inverse Trigonometric Functions Section 5.3 127 x y 0 π 2 π 2 1 1 y = csc 1 x (a) Graph of y = csc 1 x x y 0 π 2 π 1 1 y = sec 1 x (b) Graph of y = sec 1 x Figure 5.3.12 Example 5.21 Prove the identity tan 1 x + cot 1 x = π 2 . Solution: Let θ = cot 1 x . Using relations from Section 1.5, we have tan ( π 2 θ ) = − tan ( θ π 2 ) = cot θ = cot (cot 1 x ) = x , by formula (5.9). So since tan (tan 1 x ) = x for all x , this means that tan (tan 1 x ) = tan ( π 2 θ ) . Thus, tan (tan 1 x ) = tan ( π 2 cot 1 x ) . Now, we know that 0 < cot 1 x < π , so π 2 < π 2 cot 1 x < π 2 , i.e. π 2 cot 1 x is in the restricted subset on which the tangent function is one-to-one. Hence, tan (tan 1 x ) = tan ( π 2 cot 1 x ) implies that tan 1 x = π 2 cot 1 x , which proves the identity. Example 5.22 Is tan 1 a + tan 1 b = tan 1 p a + b 1 ab P an identity? Solution: In the tangent addition formula tan ( A + B ) = tan A + tan B 1 tan A tan B , let A = tan 1 a and B = tan 1 b . Then tan (tan 1 a + tan 1 b ) = tan (tan 1 a ) + tan (tan 1 b ) 1 tan (tan 1 a ) tan (tan 1
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