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Question: How do you use the sum and difference formula to simplify \(\cos \left( {\dfrac{{17\pi }}{{12}}} \ri...

How do you use the sum and difference formula to simplify cos(17π12)\cos \left( {\dfrac{{17\pi }}{{12}}} \right) ?

Explanation

Solution

In the question above, we have a function cos(17π12)\cos \left( {\dfrac{{17\pi }}{{12}}} \right) , and we have to solve it using the sum and difference formula, there are a lot of sine and cosine formulas with sum and difference.
One of the formulae is,
cos(A+B)=cosAcosBsinAsinB\cos (A + B) = \cos A\cos B - \sin A\sin B
We are going to solve this simplification with the help of this formula.

Complete step-by-step solution:
For a function cos(17π12)\cos \left( {\dfrac{{17\pi }}{{12}}} \right) , we can see that there exists a cos\cos and therefore we will be using cosine\cos ine formula that exists in the sum and difference formula.
cos(A+B)=cosAcosBsinAsinB\cos (A + B) = \cos A\cos B - \sin A\sin B
Parting the numbers in two halves, in order to divide it according to the requirement,
A=8π12=2π3A = \dfrac{{8\pi }}{{12}} = 2\dfrac{\pi }{3} and B=9π12=3π4B = \dfrac{{9\pi }}{{12}} = \dfrac{{3\pi }}{4},
Then, A+B=17π12A + B = \dfrac{{17\pi }}{{12}} since both the halves make a whole.
Now, substituting the values inside the formula for cos\cos ,
cosA=cos(2π3)=cos(ππ3)=cos(π3)=12\cos A = \cos (\dfrac{{2\pi }}{3}) = \cos \left( {\pi - \dfrac{\pi }{3}} \right) = - \cos \left( {\dfrac{\pi }{3}} \right) = - \dfrac{1}{2}
Also, substituting the values inside the formula for sin\sin ,
sinA=sin(2π3)=sin(ππ3)=sin(π3)=32\sin A = \sin (\dfrac{{2\pi }}{3}) = \sin \left( {\pi - \dfrac{\pi }{3}} \right) = \sin \left( {\dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}
Substituting the values for the part BB, in cos\cos ,
cosB=cos(3π)4=cos(ππ4)=cosπ4=12\cos B = \dfrac{{\cos (3\pi )}}{4} = \cos \left( {\pi - \dfrac{\pi }{4}} \right) = - \dfrac{{\cos \pi }}{4} = - \dfrac{1}{{\sqrt 2 }}
And,
Substituting the values for the part BB, in sin\sin ,
sinB=sin(3π4)=sin(ππ4)=sin(π4)=12\sin B = \sin \left( {\dfrac{{3\pi }}{4}} \right) = \sin \left( {\pi - \dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}
Now that we have the values that are required, we will substitute them in the formulas,
Hence, cos(17π12)=cos(A+B)\cos \left( {\dfrac{{17\pi }}{{12}}} \right) = \cos (A + B)
This clearly shows that the sum and difference method can be followed with the formula which is,
=cosAcosBsinAsinB= \cos A\cos B - \sin A\sin B
Substituting the values in the formula for simplifying the equation and multiplying the values,
=(12)×(12)(32)×(12)\Rightarrow = \left( { - \dfrac{1}{2}} \right) \times \left( { - \dfrac{1}{{\sqrt 2 }}} \right) - \left( {\dfrac{{\sqrt 3 }}{2}} \right) \times \left( {\dfrac{1}{{\sqrt 2 }}} \right)
Simplifying the equation further after multiplying the fractions,
=122322\Rightarrow = \dfrac{1}{{2\sqrt 2 }} - \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}
Subtracting the fractions in more detail,
=1322\Rightarrow = \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}

Therefore, after using the sum and difference formula, the simplification of cos(17π12)\cos \left( {\dfrac{{17\pi }}{{12}}} \right) will be forming the solution 1322\dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }} .

Note: The sum and difference method have formulas for a lot of sine, cosine, or tangent functions of two given angles. In order to use the formula, we have to first break it into numbers and find out what formula suits the numbers well. For example, in the above question the formula in use is cos(A+B)=cosAcosBsinAsinB\cos (A + B) = \cos A\cos B - \sin A\sin B . With the help of this identity, we can perform the sum and difference operation on the equation.