# Arccos

### ARCCOS

ARCCOS Trigonometric Library Functions 7-2 September 3, 1996 DATAPLOT Reference Manual ARCCOS PURPOSE Compute the arccosine for a variable or parameter.

### Inverse Trigonometric Functions

We know that the reference angle is 45 degrees or */4. So the solutions are x = */4 and x = 7*/4. But the range of arccos x is given as [0, *]. So the answer will be x = */4.

### Section4.6: Inverse Trigonometric Functions

Note: arccos(x) is the angle in[0,*]whose cosine isx. Cancellation Equations: Recall f1 (f (x)) =xforxinthedomain off, and f (f 1 (x)) =xforxinthedomain off 1.

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### #11: Inverse trigonometric relations

The only problem is they probably don'thave A GAZILLION DOLLARS. 6 Inverse trig relations and problem-solving We can use arccos and arcsintosolve problems involving sine and cosine.

### MATH 150{TOPIC 15 INVERSE TRIGONOMETRIC FUNCTIONS

The derivations of the graphs ofarccos( x ) andarctan( x ) are similar to that ofarcsin( x ) : For arccos( x ) ; we rstconsidercos x where 0 x: cos x ,0 x 1 1 2 Fig. 15.4.

### Derivatives of inverse functions 1. Prove that

arccos(x)= −1 √ 1−x 2 . Proof: Set y = arccos(x) so that we want to show dy dx = −1 √ 1−x 2 . By the deﬁnition of arccosine we know x =cos(y) (⋆) and 0 ≤ y ≤ π (⋆⋆) Diﬀerentiating both sides of (⋆) with respect to x gives

### Spring'06/MAT 145/Worksheet 1

(6pts) Use the picture below to estimate (in degrees) the values of inverse trigonometric functions. Compare your answer with results you get with a calculator. estimate calculator arcsin0 : 8= arccos( ¡ 0 : 4) = arctan( ¡ 1 : 5) =

### INVERSE TRIGONOMETRIC FUNCTIONS PART 2

(3) arcsin(x) +arccos(x) =/2 for allx 2[1,1]. (4) cos is nonnegative on the range ofarcsin, and sin is nonnegative on the range of arccos. Last time, we graphed the arcsine function.

### Inverse Trig Functions

Graphs: x y sin =: x x y 1 sin arcsin − = =: x y cos =: x x y 1 cos arccos − = =: x y tan =: x x y 1 tan arctan − = =: Trig function Restricted domain Inverse trig function Principle value range x y sin = 2 2 π π ≤ ≤ − x x y arcsin = 2 2 π π ≤ ≤ − y x y cos = π ≤ ≤ x 0 x y arccos = π ≤ ≤ y 0 x y tan = 2 2 π π < < − x x y arctan = 2 2 π π < < − y 1 − 2 π 2 π − 1 2 π − ...