Lecture 9 : Derivatives of Trigonometric Functions (Please review Trigonometry uner Algebra/Precalculus Review on the class webpage.) In this section we will look at the erivatives of the trigonometric functions sin, cos, tan, sec, csc, cot. Here the units use are raians an sin = sin( raians). Recall that sin an cos are efine an continuous everywhere an tan = sin cos, sec = cos, csc = sin, cos cot = sin, are continuous on their omains (all values of where the enominator is non-zero). The graphs of the above functions are shown at the en of this lecture to help refresh your memory: Before we calculate the erivatives of these functions, we will calculate two very important its. First Important Limit sin 0 =. See the en of this lecture for a geometric proof of the inequality, shown in the picture below for > 0, sin < < tan.... 0. 0. B D O 0. AD = tan! E! 0. sin! 0.5 0.5.5 0.! C A 0. 0. 0. From this we can easily erive that... cos < sin an we can use the squeeze theorem to prove that the it shown above is. <
From the above it, we can erive that : Another Important Limit cos 0 = 0 Eample Calculate the its: sin 5 0 sin 3, 0. Derivatives of Trigonometric Functions. From our trigonometric ientities, we can show that sin = cos : sin( + h) sin() sin = sin()[cos(h) ] + cos() sin(h) h 0 h = h 0 sin() cos(h) + cos() sin(h) sin() h [cos(h) ] sin(h) = sin() + cos() [cos(h) ] = sin() + cos() sin(h) = cos(). =. We can also show that cos = sin() : cos( + h) cos() cos = = h 0 cos() cos(h) sin() sin(h) cos() h =
cos()[cos(h) ] sin() sin(h) = [cos(h) ] sin(h) = cos() sin() = sin(). 3. Using the erivatives of sin() an cos() an the quotient rule, we can euce that tan = sec () : Eample Fin the erivative of the following function: g() = + cos + sin Higher Derivatives We see that the higher erivatives of sin an cos form a pattern in that they repeat with a cycle of four. For eample, if f() = sin, then f () = cos, f () = sin, f (3) () = cos, f () () = sin, f (5) () = cos,... (Note the erivatives follow a similar pattern for cos(). ) Eample Let f() = sin. What is f (0) ()? 3
A mass on a spring release at some point other than its equilibrium position will follow a pattern of simple harmonic motion ((t) = A sin(c + D) or equivalently (t) = A cos(c + D) ), when there is no friction or other forces to ampen the effect. The values of A, C an D epen on the elasticity of the spring, the mass an the point at which the mass is release. You will be able to prove this easily later when you learn about ifferential equations. Eample An object at the en of a vertical spring is stretche 5cm beyon its rest position an release at time t = 0. Its position at time t is given by (t) with the positive irection as shown in a ownwar irection, where (t) = 5 cos(t). (a) Fin the velocity an acceleration at time t. (b) Fin the position, velocity an acceleration of the mass at time t = π. In which irection is it moving at that time?
The following is a summary of the erivatives of the trigonometric functions. You shoul be able to verify all of the formulas easily. sin = cos, csc = csc cot, cos = sin, sec = sec tan, tan = sec cot = csc Eample The graph below shows the variations in ay length for various egrees of Lattitue. At 0 o North, at what times of the year is the length of the ay changing most rapily? Etras Eample (Preparation for Relate Rates) A police car is parke 0 feet from the roa at the point P in the iagram below. Your vehicle is approaching on the roa as in the iagram below an the police are pointing a raar gun at your car. Let enote the istance from your car to the police car an let be the angle between the line of sight of the raar gun an the roa. How fast is changing with respect to when = π? (Please attempt this problem before looking at the solution on the following page.) P 0ft! 5
Solution We have that the variables an are relate in the following way: Therefore an When = π, = π 0 = sin(). 0 sin() = [ cos() ] = 0 sin. () [ cos( π = 0 ) ] sin ( π) = 0 / / = 0 feet per raian. Graphs of Trigonometric functions 0 0 0 h( ) = tan( )! 3!!!!! 3!!! 3!!!!! 3!! f( ) = sin( ) g( ) = cos( )! 3!!!!! 3!! 0 0 0 0 0 s( ) = sin( ) 0 r( ) = cos( ) t( ) = tan( )! 3!!!!! 3!!! 3!!!!! 3!! 3!!!!! 3!! 0 0 0
Inequality Let be an angle close to 0, an between 0 an π. Note that since sin = sin( ), we have sin = sin( ) an sin 0 + = sin 0. Because of this, we nee only consier the right han it, sin 0 + with > 0. In the picture below, we see that, which is the length of the arc of the unit circle from A to B in larger than the length of the line segment from A to B. The line segment from A to B is larger than. sin since it is the hypotenuse of a right triangle with a sie of length sin.. 0. 0. B 0. sin! 0. O! 0.5 0.5 A 0. 0. 0. 0. From this we can conclue that sin < or.. sin <. Now consier the picture below. We can see intuitively that the length of the arc of the unit circle from A to B is smaller than the sum of the lengths of the line segments AE + EB. Because the line segment EB is a sie of a right triangle with hypotenuse ED, we see that EB < ED. Thus we have < AE + EB < AE + ED = AD Note now that AD = tan an AD = OA tan. = tan. OA. 0. 0. B D 0. sin!! 0. O! 0.5 0.5 0. C A E 0. 0. 0. We now have that. < tan = sin giving cos < sin cos since cos > 0 (when we multiply by positive numbers, inequalities are preserve). Putting both inequalities together we get. cos < sin < 7
. Calculate Etra Problems o.. Calculate 7 cot(3). 0 3. If g() = cos(), what is g () ()?. Fin f () if f() = cos() sin().
Etra Problems : Solutions. Calculate o. o = 0 = 0 3 0 = 0 = 0.. Calculate 3. If g() = cos(), what is g () ()? 7 cot(3). 0 7 cot(3) = 7cos(3) 0 0 sin(3) = 3 7 0 3 cos(3) sin(3) = 7 (3) 0 0 sin(3) cos(3) 0 3 = 7 0 = 0 g () = sin, g () = cos, g (3) () = sin, g () () = cos,... Therefore g (0) () = cos an g () () = cos.. Fin f () if f() = cos() sin(). Using the prouct rule, we get using the quotient rule a secon time, we get f () = (cos sin ) + (cos sin ) f () = (cos sin ) + (sin ( sin ) + cos cos ) = (cos sin ) + (cos sin ). In fact if we know our trig formulas very well, we see that f () = sin() + (cos()). 9