
Find the value of
\[\cos \dfrac{{2\pi }}{3}\cos \dfrac{\pi }{4} - \sin \dfrac{{2\pi }}{3}\sin \dfrac{\pi }{4}\]
Answer
500.4k+ views
Hint: In this question, we have to simplify the given expression using trigonometric values.
First, we have to simplify each term. For doing that we need to evaluate the quadrant where the angle lies for each term and find out the cosine and sine function accordingly. Putting all the values and by simplifying the expression we will get the required solution.
In the first quadrant (\[0to\dfrac{\pi }{2}\]) all trigonometric functions are positive, in the second quadrant (\[\dfrac{\pi }{2}to\pi \]) only sine function is positive, in third quadrant (\[\pi to\dfrac{{3\pi }}{2}\])tan function is positive, in the fourth quadrant
(\[\dfrac{{3\pi }}{2}to2\pi \]) cosine functions are positive.
Complete step by step answer:
It is given that,\[\cos \dfrac{{2\pi }}{3}\cos \dfrac{\pi }{4} - \sin \dfrac{{2\pi }}{3}\sin \dfrac{\pi }{4}\]
We need to find out the value of\[\cos \dfrac{{2\pi }}{3}\cos \dfrac{\pi }{4} - \sin \dfrac{{2\pi }}{3}\sin \dfrac{\pi }{4}\].
Now,\[\dfrac{{2\pi }}{3}\] can be written as\[\left( {\pi - \dfrac{\pi }{3}} \right)\].
To simplify the given expression, we need to simplify each term.
\[\dfrac{{2\pi }}{3}\]lies in the II quadrant where cosine is negative and sine is positive.
Thus, \[\cos \dfrac{{2\pi }}{3} = \cos \left( {\pi - \dfrac{\pi }{3}} \right) = - \cos \dfrac{\pi }{3} = - \dfrac{1}{2}\]
\[\sin \dfrac{{2\pi }}{3} = \sin \left( {\pi - \dfrac{\pi }{3}} \right) = \sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}\]
[we can the formula also,
\[
\cos (\pi - x) = - \cos x \\
\sin \left( {\pi - x} \right) = \sin x \\
\]
& the values
\[
\cos \dfrac{\pi }{3} = \dfrac{1}{2} \\
\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2} \\
\]]
Again, we know that,\[\cos \dfrac{\pi }{4} = \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
Putting the values, we get in the given expression we get,
\[\cos \dfrac{{2\pi }}{3}\cos \dfrac{\pi }{4} - \sin \dfrac{{2\pi }}{3}\sin \dfrac{\pi }{4}\]
\[ = - \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }} - \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }}\]
\[ = - \dfrac{1}{{2\sqrt 2 }} - \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}\]
\[ = - \dfrac{{\left( {1 + \sqrt 3 } \right)}}{{2\sqrt 2 }}\]
Hence simplifying,\[\cos \dfrac{{2\pi }}{3}\cos \dfrac{\pi }{4} - \sin \dfrac{{2\pi }}{3}\sin \dfrac{\pi }{4}\]\[ = - \dfrac{{\left( {1 + \sqrt 3 } \right)}}{{2\sqrt 2 }}\]
Note: Trigonometric expression:
Sin and cos formulas are calculated based on the sides of a right-angled triangle. The sine of an angle is equal to the ratio of the opposite side and the hypotenuse whereas the cosine of an angle is equal to the ratio of the adjacent side and the hypotenuse.
\[
\sin \theta = \dfrac{{Opposite Side}}{{Hypotenuse}} \\
\cos \theta = \dfrac{{Adjacent}}{{Hypotenuse}} \\
\]
In mathematics, trigonometric functions are real functions. We can find out that the most widely used trigonometric functions are the sine, cosine and tangent.
First, we have to simplify each term. For doing that we need to evaluate the quadrant where the angle lies for each term and find out the cosine and sine function accordingly. Putting all the values and by simplifying the expression we will get the required solution.
In the first quadrant (\[0to\dfrac{\pi }{2}\]) all trigonometric functions are positive, in the second quadrant (\[\dfrac{\pi }{2}to\pi \]) only sine function is positive, in third quadrant (\[\pi to\dfrac{{3\pi }}{2}\])tan function is positive, in the fourth quadrant
(\[\dfrac{{3\pi }}{2}to2\pi \]) cosine functions are positive.
Complete step by step answer:
It is given that,\[\cos \dfrac{{2\pi }}{3}\cos \dfrac{\pi }{4} - \sin \dfrac{{2\pi }}{3}\sin \dfrac{\pi }{4}\]
We need to find out the value of\[\cos \dfrac{{2\pi }}{3}\cos \dfrac{\pi }{4} - \sin \dfrac{{2\pi }}{3}\sin \dfrac{\pi }{4}\].
Now,\[\dfrac{{2\pi }}{3}\] can be written as\[\left( {\pi - \dfrac{\pi }{3}} \right)\].
To simplify the given expression, we need to simplify each term.
\[\dfrac{{2\pi }}{3}\]lies in the II quadrant where cosine is negative and sine is positive.
Thus, \[\cos \dfrac{{2\pi }}{3} = \cos \left( {\pi - \dfrac{\pi }{3}} \right) = - \cos \dfrac{\pi }{3} = - \dfrac{1}{2}\]
\[\sin \dfrac{{2\pi }}{3} = \sin \left( {\pi - \dfrac{\pi }{3}} \right) = \sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}\]
[we can the formula also,
\[
\cos (\pi - x) = - \cos x \\
\sin \left( {\pi - x} \right) = \sin x \\
\]
& the values
\[
\cos \dfrac{\pi }{3} = \dfrac{1}{2} \\
\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2} \\
\]]
Again, we know that,\[\cos \dfrac{\pi }{4} = \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
Putting the values, we get in the given expression we get,
\[\cos \dfrac{{2\pi }}{3}\cos \dfrac{\pi }{4} - \sin \dfrac{{2\pi }}{3}\sin \dfrac{\pi }{4}\]
\[ = - \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }} - \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }}\]
\[ = - \dfrac{1}{{2\sqrt 2 }} - \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}\]
\[ = - \dfrac{{\left( {1 + \sqrt 3 } \right)}}{{2\sqrt 2 }}\]
Hence simplifying,\[\cos \dfrac{{2\pi }}{3}\cos \dfrac{\pi }{4} - \sin \dfrac{{2\pi }}{3}\sin \dfrac{\pi }{4}\]\[ = - \dfrac{{\left( {1 + \sqrt 3 } \right)}}{{2\sqrt 2 }}\]
Note: Trigonometric expression:
Sin and cos formulas are calculated based on the sides of a right-angled triangle. The sine of an angle is equal to the ratio of the opposite side and the hypotenuse whereas the cosine of an angle is equal to the ratio of the adjacent side and the hypotenuse.
\[
\sin \theta = \dfrac{{Opposite Side}}{{Hypotenuse}} \\
\cos \theta = \dfrac{{Adjacent}}{{Hypotenuse}} \\
\]
In mathematics, trigonometric functions are real functions. We can find out that the most widely used trigonometric functions are the sine, cosine and tangent.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

