3.5 Derivatives of Trig Functions

Transcription

1 3.5 Derivatives of Trig Functions Problem 1 (a) Suppose we re given the right triangle below. Epress sin( ) and cos( ) in terms of the sides of the triangle. sin( ) = B C = B and cos( ) = A C = A (b) Suppose we are given the triangle below. (i) Find the length of the sides A and B. This is an isosceles triangle. A = B Using the Pythagorean Theorem: A 2 + B 2 =1 2 2A 2 =1 A 2 = 1 r 2 p 1 2 A = 2 = 2 r p 1 2 B = 2 = 2 (ii) Epress sin and cos in terms of the sides of the triangle. 4 4 sin = B p 2 4 C = 2 and cos = A p 2 4 C = 2 1

2 (c) Suppose we are given the triangle below. (i) Find the length of the sides A and B. You might remember this as a 30/60/90 triangle. To find the lengths of the sides of the triangle, we create a triangle as in the figure below. All the angles in this triangle are of 60 degrees, therefore, this is an equilateral triangle Now that we have an equilateral triangle, we have C = C =2A. Thus,1=2A =) A = 1 2 To find B, we use the Pythagorean Theorem B 2 = B2 =1 B 2 = 3 r 4 p 3 3 B = 4 = 2 2

3 (ii) Epress sin and cos in terms of the sides of the triangle. 3 3 sin = B p 3 3 C = 2 and cos = A 3 C = 1 2 (d) Suppose we are given the triangle below. (i) Find the length of the sides A and B. r We ve p actually already found the lengths of the sides for this type of triangle in part 3 3 c. B = 4 = 2 and A = 1 2 (ii) Write sin and cos in terms of the sides of the triangle. 6 6 sin = A 6 C = 1 2 and cos = B p 3 6 C = 2 (e) For any point P(,y) on the unit circle, we can epress its coordinates in terms of sin( ) and cos( ). Here is the radian measure of the angle in standard position whose terminal side is the line through the origin and the point P (, y). (, y) = (cos( ), sin( )) (f) Use all of the above information to label the given points on the unit circle. That is, for each point on the unit circle, provide the angle measure in radians and degrees, and give the (,y) coordinate for the point. 3

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5 Problem 2 Find all real numbers which satisfy each of the equations. In the previous problem, we used to denote the radian measure of the angle. However, we can use any variable to represent the angle measure. For eample, in part a, is the variable representing the radian measure of the angle. (a) cos() =1 This is asking for the collection of all angles such that cosine of that angle equals 1. The unit circle shows that one such angle is 0 (since cos(0) = 1). There is a slight trick here: since cosine has period 2 we actually have cos(0 + 2 n) = 1 for every integer n. In summary, = 2 n, where n is any integer, gives all the solutions to this equation. (b) sin(3 ) = p 3/2 for 0 apple apple 2 5

6 Finding all numbers with 0 apple apple 2 that satisfy sin(3 ) = p 3/2 is a bit tricky. We first perform a useful trick from algebra variable substitution. Let =3. So, we are trying to find all numbers such that sin() = p 3/2 for 0 apple /3 apple 2. Then = 3 +2 n or = 2 +2 n for n any integer as long as 0 apple apple 2 Since =3, we can solve 3 for to obtain = /9 +(2 n)/3 or =(2 )/9 +(2 n)/3, wheren is again any integer as long as 0 apple apple 2. We are only looking for solutions of in [0, 2 ], and so our solutions are = 9, 2 9, 7 9, 8 9, 13 9, Problem 3 (a) Graph f( ) =sin( ) and g( ) = cos( ) from [ 8, ] (b) For 0 apple apple 2, find the following values or intervals. (i) Where does f 0 ( ) =0? f 0 ( ) =0at = 2, 3 2 (ii) Where is f 0 negative? Where is it positive? 6

7 (iii) Where is f 0 : f 0 is negative on i. negative and increasing? (, 3 2 ) ii. negative and decreasing? ( 2, ) iii. positive and increasing? ( 3 2, 2 ) iv. positive and decreasing? (0, 2 ) 2, 3. f 0 is positive on 0, 3, 2 2 2, 2 (iv) Estimate the slope of f at the points =0(i.e. f 0 (0)), = (i.e. f 0 ( )), =2 (i.e. f 0 (2 )). The slope of f at =0and =2 is 1. The slope of f at = is -1 (v) Where does g 0 ( ) =0? g 0 ( =0at =0,, 2 (vi) Where is g 0 negative? Where is it positive? g 0 is negative on (0, ) and positive on (, 2 ) (vii) Where is g 0 : i. negative and increasing? ( 2, ) ii. negative and decreasing? (0, 2 ) iii. positive and increasing? (, 3 2 ) iv. positive and decreasing? ( 3 2, 2 ) (viii) Estimate the slope of g at the points = 2 (i.e. g0 ( 2 )), = 3 2 (i.e. g0 ( 3 2 i)). The slope of g at = 2 is -1. The slope of g at = 3 2 is 1. (c) Use the above information to sketch a graph of f 0 and g 0 7

8 (d) Based on the graphs of f 0 and g 0, what might the equation for f 0 and g 0 be? It appears f 0 ( ) = cos( ) and g 0 ( ) = sin( ) Problem 4 Find the following its: (a) sin(8) where u =8. (b) tan(5) sin(8) sin(8) = 8 8 =8 sin(8) 8 =8 u!0 sin(u) u =8 1=8 8

9 tan(5) = sin(5) cos(5) = 1 cos(5) sin(5) = cos(5) sin(5) 5 5 = 1 5 cos(5) 5 sin(5) = 1 5 (1)(1) = 1 5 Problem 5 Find the derivative of the following functions: (a) f() = cot() (b) f() =sin() cos() f 0 () = (76 + cot())(1) ( + 5)(42 5 csc 2 ()) (7 6 + cot()) 2 = 76 + cot() ( + 5)(42 5 csc 2 ()) (7 6 + cot()) 2. f 0 () = (cos())(cos()) + (sin())( sin()) = cos 2 () sin 2 (). (c) f() = e tan() sec()+2 (d) f() =sin() cos()e 3 f 0 () = (sec() + 2)(e tan()+e sec 2 ()) e tan()(sec() tan()) (sec() + 2) 2 = e [(sec() + 2)(tan()+sec 2 ()) sec() tan 2 ()] (sec() + 2) 2. f 0 () = d d [sin() cos()]e3 +(sin() cos()) d d (e3 ) = (cos 2 () sin 2 ())e 3 +3e 3 sin() cos() = e 3 (cos 2 ()+3sin() cos() sin 2 ()). 9

10 Problem 6 Let f() =sin() and g() = cos(). Can you compute f (48) (), g (42) (), and g (39) ()? Let s find the first several derivatives of f() =sin(). f (0) () =sin() f (1) () = cos() f (2) () = sin() f (3) () = cos() f (4) () =sin() Recall that, for n a nonnegative integer, f (4n) () =f() =sin() and g (4n) () =g() = cos(). Thus: f (48) () =f() =sin() g (42) () =g (40+2) () =g (40)(2) () =g (2) () =g 00 () = cos() g (39) () =g (36+3) () =g (36)(3) () =g (3) () =g 000 () =sin() Problem 7 5 (a) Let f() = ( + 5) sin(). Compute f 0 () 2 f 0 () = (2 5)( + 5) sin() (2 5)(sin()+( + 5) cos()) ( + 5) 2 sin 2 () (b) Evaluate the it.!5 2 5 sin( 5) 10

11 !5 2 sin( 5 5) = ( 5)!5 sin( 5) =!5 ()!5 5 sin( 5) 1 =()!5!5 sin( 5) 5 =!5 ()!5 1 sin( 5) 5 if we let u = =!5 () u!0 5 we have: 1 sin(u) u = 5(1) = 5 Problem 8 Find values for a, b, and c so that the following function is di erentiable everywhere. a sin()+bcos() if <0 f() = a 2 + b + c if 0 The function f is di erentiable for < 0 since f is a combination of trigonometric functions. f is also di erentiable >0 since f is a polynomial on that interval. We need to focus on =0.Sincefis di erentiable everywhere, f must also be di erentiable at =0and therefore continuous at =0. Therefore: We need that f() = f() = f() =f(0). Observe that + f() = (a sin()+b cos()) = b(1) = b. f() = + b + c) =c. + +(a2 Thus, we must have that b = c Net, if f 0 (0) eists, then: f 0 (0) = f() f(0) 0 since f(0) = a b 0+c = c f() f(0) 0 = f() c 11

12 Since this it eists, the left and right its must be equal. f() c a 2 + b + c c = + + a 2 + b = + = +(a + b) = b f() c = = a sin()+bcos() c a sin() b cos() + c we already found b = c so we have a sin() = a + sin() sin() since a sin() c cos() + c c cos() 1 cos() 1 =1and + c This means in order for the derivative to eist, a = b = c. cos() 1 = a =0we have 12