Question Statement Prove that:
Cos β‘ ( x + h ) = Cos β‘ x β h Sin β‘ x β h 2 2 ! Cos β‘ x + h 3 3 ! Sin β‘ x + β¦ \operatorname{Cos}(x + h) = \operatorname{Cos}x - h\operatorname{Sin}x - \frac{h^2}{2!}\operatorname{Cos}x + \frac{h^3}{3!}\operatorname{Sin}x + \ldots Cos ( x + h ) = Cos x β h Sin x β 2 ! h 2 β Cos x + 3 ! h 3 β Sin x + β¦ and evaluate Cos β‘ ( 6 1 β ) \operatorname{Cos}(61^\circ) Cos ( 6 1 β )
Background and Explanation The problem involves the Taylor series expansion of a function, specifically Cos β‘ ( x + h ) \operatorname{Cos}(x + h) Cos ( x + h ) x x x h h h f ( x ) f(x) f ( x ) x x x
f ( x + h ) = f ( x ) + f β² ( x ) 1 ! h + f β² β² ( x ) 2 ! h 2 + f β² β² β² ( x ) 3 ! h 3 + β¦ f(x + h) = f(x) + \frac{f'(x)}{1!}h + \frac{f''(x)}{2!}h^2 + \frac{f'''(x)}{3!}h^3 + \ldots f ( x + h ) = f ( x ) + 1 ! f β² ( x ) β h + 2 ! f β²β² ( x ) β h 2 + 3 ! f β²β²β² ( x ) β h 3 + β¦ For trigonometric functions like Cos β‘ x \operatorname{Cos}x Cos x
f ( x ) = Cos β‘ x f(x) = \operatorname{Cos}x f ( x ) = Cos x f β² ( x ) = β Sin β‘ x f'(x) = -\operatorname{Sin}x f β² ( x ) = β Sin x f β² β² ( x ) = β Cos β‘ x f''(x) = -\operatorname{Cos}x f β²β² ( x ) = β Cos x f β² β² β² ( x ) = Sin β‘ x f'''(x) = \operatorname{Sin}x f β²β²β² ( x ) = Sin x
Solution Step 1: Taylor Series Expansion for Cos β‘ ( x + h ) \operatorname{Cos}(x + h) Cos ( x + h )
Using the formula:
f ( x + h ) = f ( x ) + f β² ( x ) 1 ! h + f β² β² ( x ) 2 ! h 2 + f β² β² β² ( x ) 3 ! h 3 + β¦ f(x + h) = f(x) + \frac{f'(x)}{1!}h + \frac{f''(x)}{2!}h^2 + \frac{f'''(x)}{3!}h^3 + \ldots f ( x + h ) = f ( x ) + 1 ! f β² ( x ) β h + 2 ! f β²β² ( x ) β h 2 + 3 ! f β²β²β² ( x ) β h 3 + β¦ Substitute the derivatives for f ( x ) = Cos β‘ x f(x) = \operatorname{Cos}x f ( x ) = Cos x
Cos β‘ ( x + h ) = Cos β‘ x β Sin β‘ x β
h β Cos β‘ x 2 ! h 2 + Sin β‘ x 3 ! h 3 + β¦ \operatorname{Cos}(x + h) = \operatorname{Cos}x - \operatorname{Sin}x \cdot h - \frac{\operatorname{Cos}x}{2!}h^2 + \frac{\operatorname{Sin}x}{3!}h^3 + \ldots Cos ( x + h ) = Cos x β Sin x β
h β 2 ! Cos x β h 2 + 3 ! Sin x β h 3 + β¦ Step 2: Evaluate Cos β‘ ( 6 1 β ) \operatorname{Cos}(61^\circ) Cos ( 6 1 β )
Here, x = 6 0 β x = 60^\circ x = 6 0 β h = 1 β h = 1^\circ h = 1 β
x = Ο 3 x = \frac{\pi}{3} x = 3 Ο β h = Ο 180 h = \frac{\pi}{180} h = 180 Ο β
Substitute values into the expanded series:
Cos β‘ ( 6 1 β ) = Cos β‘ Ο 3 β Sin β‘ Ο 3 β
Ο 180 β Cos β‘ Ο 3 2 ! ( Ο 180 ) 2 + Sin β‘ Ο 3 3 ! ( Ο 180 ) 3 + β¦ \operatorname{Cos}(61^\circ) = \operatorname{Cos}\frac{\pi}{3} - \operatorname{Sin}\frac{\pi}{3} \cdot \frac{\pi}{180} - \frac{\operatorname{Cos}\frac{\pi}{3}}{2!} \left(\frac{\pi}{180}\right)^2 + \frac{\operatorname{Sin}\frac{\pi}{3}}{3!} \left(\frac{\pi}{180}\right)^3 + \ldots Cos ( 6 1 β ) = Cos 3 Ο β β Sin 3 Ο β β
180 Ο β β 2 ! Cos 3 Ο β β ( 180 Ο β ) 2 + 3 ! Sin 3 Ο β β ( 180 Ο β ) 3 + β¦ Step 3: Substitute Trigonometric Values
From trigonometric identities:
Cos β‘ Ο 3 = 1 2 \operatorname{Cos}\frac{\pi}{3} = \frac{1}{2} Cos 3 Ο β = 2 1 β Sin β‘ Ο 3 = 3 2 \operatorname{Sin}\frac{\pi}{3} = \frac{\sqrt{3}}{2} Sin 3 Ο β = 2 3 β β
Substitute these values:
Cos β‘ ( 6 1 β ) = 1 2 β 3 2 β
Ο 180 β 1 2 2 ! ( Ο 180 ) 2 + 3 2 3 ! ( Ο 180 ) 3 \operatorname{Cos}(61^\circ) = \frac{1}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\pi}{180} - \frac{\frac{1}{2}}{2!} \left(\frac{\pi}{180}\right)^2 + \frac{\frac{\sqrt{3}}{2}}{3!} \left(\frac{\pi}{180}\right)^3 Cos ( 6 1 β ) = 2 1 β β 2 3 β β β
180 Ο β β 2 ! 2 1 β β ( 180 Ο β ) 2 + 3 ! 2 3 β β β ( 180 Ο β ) 3 Step 4: Approximate Terms
Let Ο 180 β 0.017453 \frac{\pi}{180} \approx 0.017453 180 Ο β β 0.017453
First term: 1 2 = 0.5 \frac{1}{2} = 0.5 2 1 β = 0.5
Second term: β 3 2 β
0.017453 β β 0.0151 -\frac{\sqrt{3}}{2} \cdot 0.017453 \approx -0.0151 β 2 3 β β β
0.017453 β β 0.0151
Third term: β 1 2 2 ! β
( 0.017453 ) 2 β β 0.000076 -\frac{\frac{1}{2}}{2!} \cdot (0.017453)^2 \approx -0.000076 β 2 ! 2 1 β β β
( 0.017453 ) 2 β β 0.000076
Fourth term: 3 2 3 ! β
( 0.017453 ) 3 β 0.0000005 \frac{\frac{\sqrt{3}}{2}}{3!} \cdot (0.017453)^3 \approx 0.0000005 3 ! 2 3 β β β β
( 0.017453 ) 3 β 0.0000005
Add the terms:
Cos β‘ ( 6 1 β ) β 0.5 β 0.0151 β 0.000076 + 0.0000005 β 0.4848 \operatorname{Cos}(61^\circ) \approx 0.5 - 0.0151 - 0.000076 + 0.0000005 \approx 0.4848 Cos ( 6 1 β ) β 0.5 β 0.0151 β 0.000076 + 0.0000005 β 0.4848
Taylor Series Expansion :f ( x + h ) = f ( x ) + f β² ( x ) 1 ! h + f β² β² ( x ) 2 ! h 2 + β¦ f(x + h) = f(x) + \frac{f'(x)}{1!}h + \frac{f''(x)}{2!}h^2 + \ldots f ( x + h ) = f ( x ) + 1 ! f β² ( x ) β h + 2 ! f β²β² ( x ) β h 2 + β¦ Conversion to radians:h = 1 β = Ο 180 β 0.017453 h = 1^\circ = \frac{\pi}{180} \approx 0.017453 h = 1 β = 180 Ο β β 0.017453
Trigonometric values:Cos β‘ Ο 3 = 1 2 , Sin β‘ Ο 3 = 3 2 \operatorname{Cos}\frac{\pi}{3} = \frac{1}{2}, \quad \operatorname{Sin}\frac{\pi}{3} = \frac{\sqrt{3}}{2} Cos 3 Ο β = 2 1 β , Sin 3 Ο β = 2 3 β β
Summary of Steps
Use the Taylor series expansion for Cos β‘ ( x + h ) \operatorname{Cos}(x + h) Cos ( x + h )
Substitute f ( x ) = Cos β‘ x f(x) = \operatorname{Cos}x f ( x ) = Cos x
Convert x x x h h h
Substitute known values for Cos β‘ Ο 3 \operatorname{Cos}\frac{\pi}{3} Cos 3 Ο β Sin β‘ Ο 3 \operatorname{Sin}\frac{\pi}{3} Sin 3 Ο β
Approximate terms and sum them up to get the result.
Final answer:
Cos β‘ ( 6 1 β ) β 0.4848 \operatorname{Cos}(61^\circ) \approx 0.4848 Cos ( 6 1 β ) β 0.4848 
