
If \[\cos A\sin \left( {A - \dfrac{n}{6}} \right)\] is maximum, then the value of\[A\]is equal to
A. \[\dfrac{\pi }{3}\]
B. \[\dfrac{\pi }{4}\]
C. \[\dfrac{\pi }{2}\]
D. None of these
Answer
161.1k+ views
Hint: Trigonometry identities are useful when equations are much complicated to solve. They are the fundamental tools of trigonometry used in factoring, finding common denominators, and resolving trigonometric equations. In this case, since\[\cos A\sin \left( {A - \dfrac{\pi }{6}} \right)\], simply compute the trigonometry equation using trigonometry identities and formulas and then simply the obtained expression by finding common divisors.
Complete step by step solution: The given equation according to the question is:
\[\cos A\sin \left( {A - \dfrac{\pi }{6}} \right)\]
After finding, the equation becomes,
\[ = \dfrac{1}{2}\left[ {\sin \left( {2A - \dfrac{\pi }{6}} \right) - \sin \dfrac{\pi }{6}} \right]\]
Here, we will replace the value according to the trigonometry formula:
\[\cos \left( A \right)\left( {\dfrac{{\sqrt 3 }}{2}\sin \left( A \right) - \dfrac{1}{2}\cos \left( A \right)} \right)\]
Rewrite the obtained equation as:
\[ = \dfrac{1}{2}\left( {\dfrac{{\sqrt 3 }}{2}\sin \left( {2A} \right) - \dfrac{1}{2}\cos \left( {2A} \right) - \sin \left( {\dfrac{\pi }{6}} \right)} \right)\]
Simplify the denominator, we obtain:
\[ = \dfrac{{ - 1 + \sqrt 3 \sin \left( {2A} \right) - \cos \left( {2A} \right)}}{4}\]
But\[\sin \left( {2A - \dfrac{\pi }{6}} \right) - \dfrac{1}{2}\]reaches the maximum value at
\[2A - \dfrac{\pi }{6} = \dfrac{\pi }{2}\]
We must choose a side to resolve first.
Add \[\dfrac{\pi }{6}\]either side:
\[{\rm{2A}} - \dfrac{\pi }{6} + \dfrac{\pi }{6} = \dfrac{\pi }{2} + \dfrac{\pi }{6}\]
Simplify the above equation into less complicated form:
\[2A = \dfrac{{2\pi }}{3}\]
Divide either side by\[2\]:
\[\dfrac{{2A}}{2} = \dfrac{{\dfrac{{2\pi }}{3}}}{2}\]
Simplify to obtain the value of A:
Solve for A:
\[ \Rightarrow A = \dfrac{\pi }{3}\]
Hence, the value of \[A\]is equal to\[\dfrac{\pi }{3}\].
Option ‘A’ is correct
Note: Sometimes students build error in applying identities and aren't able to comprehend the way to make the opposite. Students from time to time apply identities incorrectly. Creating the error of simplifying either side at the same time, once applying pure trig identities could be a typical error. To unravel this kind of issues, there are numerous ways offered. There are totally different formulas offered to unravel these varieties of issues. To avoid this kind of downside, we've to settle on the right approach and therefore the right formulas and identities.
Complete step by step solution: The given equation according to the question is:
\[\cos A\sin \left( {A - \dfrac{\pi }{6}} \right)\]
After finding, the equation becomes,
\[ = \dfrac{1}{2}\left[ {\sin \left( {2A - \dfrac{\pi }{6}} \right) - \sin \dfrac{\pi }{6}} \right]\]
Here, we will replace the value according to the trigonometry formula:
\[\cos \left( A \right)\left( {\dfrac{{\sqrt 3 }}{2}\sin \left( A \right) - \dfrac{1}{2}\cos \left( A \right)} \right)\]
Rewrite the obtained equation as:
\[ = \dfrac{1}{2}\left( {\dfrac{{\sqrt 3 }}{2}\sin \left( {2A} \right) - \dfrac{1}{2}\cos \left( {2A} \right) - \sin \left( {\dfrac{\pi }{6}} \right)} \right)\]
Simplify the denominator, we obtain:
\[ = \dfrac{{ - 1 + \sqrt 3 \sin \left( {2A} \right) - \cos \left( {2A} \right)}}{4}\]
But\[\sin \left( {2A - \dfrac{\pi }{6}} \right) - \dfrac{1}{2}\]reaches the maximum value at
\[2A - \dfrac{\pi }{6} = \dfrac{\pi }{2}\]
We must choose a side to resolve first.
Add \[\dfrac{\pi }{6}\]either side:
\[{\rm{2A}} - \dfrac{\pi }{6} + \dfrac{\pi }{6} = \dfrac{\pi }{2} + \dfrac{\pi }{6}\]
Simplify the above equation into less complicated form:
\[2A = \dfrac{{2\pi }}{3}\]
Divide either side by\[2\]:
\[\dfrac{{2A}}{2} = \dfrac{{\dfrac{{2\pi }}{3}}}{2}\]
Simplify to obtain the value of A:
Solve for A:
\[ \Rightarrow A = \dfrac{\pi }{3}\]
Hence, the value of \[A\]is equal to\[\dfrac{\pi }{3}\].
Option ‘A’ is correct
Note: Sometimes students build error in applying identities and aren't able to comprehend the way to make the opposite. Students from time to time apply identities incorrectly. Creating the error of simplifying either side at the same time, once applying pure trig identities could be a typical error. To unravel this kind of issues, there are numerous ways offered. There are totally different formulas offered to unravel these varieties of issues. To avoid this kind of downside, we've to settle on the right approach and therefore the right formulas and identities.
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