
(a)
To calculate: The exact value of function (f+g)(θ) for θ=−300° , where f(θ)=sinθ and g(θ)=cosθ .
(a)

Answer to Problem 114E
The value (f+g)(−300°) is √3+12 .
Explanation of Solution
Given:
The functions f(θ)=sinθ and g(θ)=cosθ .
Concept Used:
Use the property (f+g)(θ)=f(θ)+g(θ) , for the given value of θ .
Calculation:
Substitute the values of f(θ)=sinθ and g(θ)=cosθ in above identity
(f+g)(θ)=f(θ)+g(θ)=sinθ+cosθ , substitute θ=−300° in the above equation it gives
(f+g)(−300°)=sin(−300°)+cos(−300°) .
By the trigonometric identity sin(−θ)=−sinθ and cos(−θ)=cosθ , therefore sin(−300°)=−sin300° and cos(−300°)=cos300° .
Therefore, (f+g)(−300°)=−sin(300°)+cos(300°)
By the trigonometric property sin(360°−θ)=−sinθ and cos(360°−θ)=cosθ ,
Substitute θ=60° in both the properties sin(360°−60°)=−sin60° , it gives sin(300°)=−sin60° and also cos(360°−60°)=cos60° , it gives cos(300°)=cos60° .
Now substitute 12 for cos60° and √32 for sin60° in the above equation it gives,
sin(300°)=−√32 and cos(300°)=12 .
Now substitute the values in the formula (f+g)(−300°)=sin(−300°)+cos(−300°)=√32+12=√3+12
Therefore the value (f+g)(−300°) is √3+12 .
(b)
To calculate: The exact value of function (g−f)(θ) for θ=−300° , where f(θ)=sinθ and g(θ)=cosθ .
(b)

Answer to Problem 114E
The value (g−f)(−300°) is 1−√32 .
Explanation of Solution
Given:
The functions f(θ)=sinθ and g(θ)=cosθ .
Concept Used:
Use the property (g−f)(θ)=g(θ)−f(θ) , for the given value of θ .
Calculation:
Substitute the values of f(θ)=sinθ and g(θ)=cosθ in above identity
(g−f)(θ)=g(θ)−f(θ)=cosθ−sinθ , substitute θ=−300° in the above equation it gives
(g−f)(−300°)=cos(−300°)−sin(−300°) .
By the trigonometric identity sin(−θ)=−sinθ and cos(−θ)=cosθ , therefore sin(−300°)=−sin300° and cos(−300°)=cos300° .
Therefore, (g−f)(−300°)=cos(300°)+sin(300°)
By the trigonometric property sin(360°−θ)=−sinθ and cos(360°−θ)=cosθ ,
Substitute θ=60° in both the properties sin(360°−60°)=−sin60° , it gives sin(300°)=−sin60° and also cos(360°−60°)=cos60° , it gives cos(300°)=cos60° .
Now substitute 12 for cos60° and √32 for sin60° in the above equation it gives,
sin(300°)=−√32 and cos(300°)=12 .
Now substitute the values in the formula (g−f)(−300°)=cos(−300°)−sin(−300°)=12−√32=1−√32
Therefore the value (g−f)(−300°) is 1−√32 .
(c)
To calculate: The exact value of function [g(θ)]2 for θ=−300° , where g(θ)=cosθ .
(c)

Answer to Problem 114E
The value [g(−300°)]2 is 14 .
Explanation of Solution
Given:
The functions f(θ)=sinθ and g(θ)=cosθ .
Concept Used:
Use the value of g(θ)=cosθ in the required value [g(θ)]2 , for the given value of θ .
Calculation:
Substitute the values of g(θ)=cosθ in above identity
[g(θ)]2=(cosθ)2 , substitute θ=−300° in the above equation it gives
[g(−300°)]2=(cos(−300°))2 .
By the trigonometric identity cos(−θ)=cosθ , therefore cos(−300°)=cos300° .
Therefore, [g(−300°)]2=(cos(300°))2 .
By the trigonometric property cos(360°−θ)=cosθ ,
Substitute θ=60° in both the property cos(360°−60°)=cos60° , it gives cos(300°)=cos60° .
Now substitute 12 for cos60° in the above equation it gives cos(300°)=12 .
[g(−300°)]2=(cos300°)2=(12)2=14 .
Therefore the value [g(−300°)]2 is 14 .
(d)
To calculate: The exact value of function (fg)(θ) for θ=−300° , where f(θ)=sinθ and g(θ)=cosθ .
(d)

Answer to Problem 114E
The value (fg)(−300°) is √34 .
Explanation of Solution
Given:
The functions f(θ)=sinθ and g(θ)=cosθ .
Concept Used:
Use the property (fg)(θ)=f(θ)g(θ) , for the given value of θ .
Calculation:
Substitute the values of f(θ)=sinθ and g(θ)=cosθ in above identity
(fg)(θ)=f(θ)g(θ)=sinθcosθ , substitute θ=−300° in the above equation it gives
(fg)(−300°)=sin(−300°)cos(−300°) .
By the trigonometric identity sin(−θ)=−sinθ and cos(−θ)=cosθ , therefore sin(−300°)=−sin300° and cos(−300°)=cos300° .
Therefore, (fg)(−300°)=−sin(300°)cos(300°) .
By the trigonometric property sin(360°−θ)=−sinθ and cos(360°−θ)=cosθ ,
Substitute θ=60° in both the properties sin(360°−60°)=−sin60° , it gives sin(300°)=−sin60° and also cos(360°−60°)=cos60° , it gives cos(300°)=cos60° .
Now substitute 12 for cos60° and √32 for sin60° in the above equation it gives,
sin(300°)=−√32 and cos(300°)=12 .
Now substitute the above values in the formula
(fg)(−300°)=−sin(300°)cos(300°)=−(−√32)×12=√34 .
Therefore the value (fg)(−300°) for f(θ)=sinθ and g(θ)=cosθ is √34 .
(e)
To calculate: The exact value of function f(2θ) for θ=−300° , where f(θ)=sinθ .
(e)

Answer to Problem 114E
The value f(2θ) for θ=−300° is −√32 .
Explanation of Solution
Given:
The functions f(θ)=sinθ and g(θ)=cosθ .
Concept Used:
Firstly calculate the angle 2θ for given θ and then calculate f(2θ) .
Calculation:
As θ=−300° , so 2θ=−600° .
Substitute the value of 2θ in f(2θ)=sin(2θ)
f(−600°)=sin(−600°) ,
By the trigonometric property sin(−θ)=−sinθ , therefore sin(−600°)=−sin600° .
Therefore, f(−600°)=−sin(600°)
As sin(360°+θ)=sinθ , substitute θ=240° in the equation it gives sin(360°+240°)=sin240° , it gives sin600°=sin240° .
By the trigonometric identity sin(180°+θ)=−sinθ , substitute θ=60° it gives sin(240°)=−sin60° .
Now substitute √32 for sin60° it gives sin(600°)=−√32
Therefore value f(2θ) for θ=−300° is −√32 .
(f)
To calculate: The exact value of function g(−θ) for θ=−300° , where g(θ)=cosθ .
(f)

Answer to Problem 114E
The value g(−300°) is 12 .
Explanation of Solution
Given:
The functions f(θ)=sinθ and g(θ)=cosθ .
Concept Used:
Use the property g(−θ)=g(θ) for the given even function g(θ)=cosθ , for the given value of θ .
Calculation:
Substitute θ=−300° in the equation it gives g(−θ)=g(300°)=cos(300°) .
As per trigonometric identity cos(360°−θ)=cosθ , where θ is any acute angle.
Now substitute θ=60° in the above equation it gives cos(360°−60°)=cos60° , which gives cos(300°)=cos60°
Substitute 12 for cos(60°) it gives
cos(300°)=12 .
Therefore the value of cos(−θ) for θ=−300° is 12 .
Chapter 4 Solutions
Precalculus with Limits: A Graphing Approach
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