How do you find the exact functional value $\cos (\dfrac{{7\pi }}{{12}})$ using the cosine sum or difference identity?
Answer
626.7k+ views
Hint: Apply the formula of cosine sum $\cos (A + B) = \cos A\cos B - \sin A\sin B$.
In these questions try to break $\dfrac{{7\pi }}{{12}}$ in two terms so that you can use the above formula and denote A and B.
Complete step by step answer:
Firstly let's break $\dfrac{{7\pi }}{{12}}$ in two terms . We know $4 + 3 = 7$ , so we will break $\dfrac{{7\pi }}{{12}}$ in 4 and 3 . We cannot break it in other numbers because 12 is divisible by 4 and 3 and it is easier to calculate.
$\cos (\dfrac{{7\pi }}{{12}}) = \cos (\dfrac{{4\pi }}{{12}} + \dfrac{{3\pi }}{{12}})$
Simplifying the above
$ \Rightarrow \cos (\dfrac{\pi }{3} + \dfrac{\pi }{4})$
Now , let’s take A as $\dfrac{\pi }{4}$ and B as $\dfrac{\pi }{3}$
Put the values in the formula $\cos (A + B) = \cos A\cos B - \sin A\sin B$
Calculating cos and sin separately to avoid confusion
For cosA and cosB
$\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$
For sinA and sinB
$\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$
$ \Rightarrow \cos (\dfrac{\pi }{3} + \dfrac{\pi }{4}) = \cos \dfrac{\pi }{4}\cos \dfrac{\pi }{3} - \sin \dfrac{\pi }{4}\sin \dfrac{\pi }{3}$
Putting the values of cosine and sine
$ \Rightarrow \dfrac{1}{{\sqrt 2 }} \cdot \dfrac{1}{2} - \dfrac{1}{{\sqrt 2 }}\dfrac{{\sqrt 3 }}{2}$
Multiplying the above
$ \Rightarrow \dfrac{1}{{2\sqrt 2 }} - \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}$
Bothe have same denominators , subtract them
$ \Rightarrow \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$
Thus , value of $\cos (\dfrac{{7\pi }}{{12}})$ is $\dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$.
Additional information:
You can check from the calculator if the answer we obtained is correct or not.
For $\cos (\dfrac{{7\pi }}{{12}})$ , finding the value using calculator we get -0.259
For $\dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$ , finding the value using calculator we get -0.259
Therefore our obtained value is correct.
Note:
Finding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values.
Therefore in the above question, we have used the cosine sum formula, if you have tried to break it into different identity it would be difficult to know the complex angles and it will take a long time.
For example when you break $\cos (\dfrac{{7\pi }}{{12}})$ for difference identity , you will break 7 in 10 and 3 then you will get angles in $\dfrac{{10\pi }}{{12}}$ or $\dfrac{{5\pi }}{{12}}$ and $\dfrac{\pi }{4}$ . Here finding the value of $\dfrac{{5\pi }}{{12}}$ will be a long calculation.
So use the formulas accordingly.
In these questions try to break $\dfrac{{7\pi }}{{12}}$ in two terms so that you can use the above formula and denote A and B.
Complete step by step answer:
Firstly let's break $\dfrac{{7\pi }}{{12}}$ in two terms . We know $4 + 3 = 7$ , so we will break $\dfrac{{7\pi }}{{12}}$ in 4 and 3 . We cannot break it in other numbers because 12 is divisible by 4 and 3 and it is easier to calculate.
$\cos (\dfrac{{7\pi }}{{12}}) = \cos (\dfrac{{4\pi }}{{12}} + \dfrac{{3\pi }}{{12}})$
Simplifying the above
$ \Rightarrow \cos (\dfrac{\pi }{3} + \dfrac{\pi }{4})$
Now , let’s take A as $\dfrac{\pi }{4}$ and B as $\dfrac{\pi }{3}$
Put the values in the formula $\cos (A + B) = \cos A\cos B - \sin A\sin B$
Calculating cos and sin separately to avoid confusion
For cosA and cosB
$\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$
For sinA and sinB
$\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$
$ \Rightarrow \cos (\dfrac{\pi }{3} + \dfrac{\pi }{4}) = \cos \dfrac{\pi }{4}\cos \dfrac{\pi }{3} - \sin \dfrac{\pi }{4}\sin \dfrac{\pi }{3}$
Putting the values of cosine and sine
$ \Rightarrow \dfrac{1}{{\sqrt 2 }} \cdot \dfrac{1}{2} - \dfrac{1}{{\sqrt 2 }}\dfrac{{\sqrt 3 }}{2}$
Multiplying the above
$ \Rightarrow \dfrac{1}{{2\sqrt 2 }} - \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}$
Bothe have same denominators , subtract them
$ \Rightarrow \dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$
Thus , value of $\cos (\dfrac{{7\pi }}{{12}})$ is $\dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$.
Additional information:
You can check from the calculator if the answer we obtained is correct or not.
For $\cos (\dfrac{{7\pi }}{{12}})$ , finding the value using calculator we get -0.259
For $\dfrac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$ , finding the value using calculator we get -0.259
Therefore our obtained value is correct.
Note:
Finding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values.
Therefore in the above question, we have used the cosine sum formula, if you have tried to break it into different identity it would be difficult to know the complex angles and it will take a long time.
For example when you break $\cos (\dfrac{{7\pi }}{{12}})$ for difference identity , you will break 7 in 10 and 3 then you will get angles in $\dfrac{{10\pi }}{{12}}$ or $\dfrac{{5\pi }}{{12}}$ and $\dfrac{\pi }{4}$ . Here finding the value of $\dfrac{{5\pi }}{{12}}$ will be a long calculation.
So use the formulas accordingly.
Recently Updated Pages
10 examples of friction in our daily life

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Difference between physical and chemical change class 11 chemistry CBSE

What are derived physical quantities Give any two examples class 11 physics CBSE

Which of the following is a derived physical quantity class 11 physics CBSE

