
Find the value of
\[\cos \dfrac{{2\pi }}{3}\cos \dfrac{\pi }{4} - \sin \dfrac{{2\pi }}{3}\sin \dfrac{\pi }{4}\]
Answer
515.4k+ 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
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

