Find the value of
\[\cos \dfrac{{2\pi }}{3}\cos \dfrac{\pi }{4} - \sin \dfrac{{2\pi }}{3}\sin \dfrac{\pi }{4}\]
Answer
545.4k+ 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.
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