Prove that:-
\[\sin \left( \dfrac{4\pi }{9}+7 \right)\cdot \cos \left( \dfrac{\pi }{9}-7 \right)+\cos \left( \dfrac{4\pi }{9}+7 \right)\cdot \sin \left( \dfrac{\pi }{9}-7 \right)=\dfrac{\sqrt{3}}{2}\]
Answer
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Hint:In such questions, we prove them by either making the left hand side that is L.H.S. or by making the right hand side that is R.H.S. equal to the other in order to prove the proof that has been asked.
Now, the formula that would be used in solving this question is as follows
\[\sin A\cdot \cos B-\cos B\cdot \sin A=\sin (A-B)\]
Complete step-by-step answer:
Now, this is the result that would be used to prove the proof mentioned in this question as using this identity, we would convert the left hand side that is L.H.S. or the right hand side that is R.H.S. to make either of them equal to the other.
As mentioned in the question, we have to prove the given expression that is
\[\sin \left( \dfrac{4\pi }{9}+7 \right)\cdot \cos \left( \dfrac{\pi }{9}+7 \right)-\cos \left( \dfrac{4\pi }{9}+7 \right)\cdot \sin \left( \dfrac{\pi }{9}+7 \right)=\dfrac{\sqrt{3}}{2}\] .
Now, we will start with the left hand side that is L.H.S. as follows
\[=\sin \left( \dfrac{4\pi }{9}+7 \right)\cdot \cos \left( \dfrac{\pi }{9}+7 \right)-\cos \left( \dfrac{4\pi }{9}+7 \right)\cdot \sin \left( \dfrac{\pi }{9}+7 \right)\]
Now, we will use the relation that is mentioned the hint as follows
\[\begin{align}
& =\sin \left( \dfrac{4\pi }{9}+7 \right)\cdot \cos \left( \dfrac{\pi }{9}+7 \right)-\cos \left( \dfrac{4\pi }{9}+7 \right)\cdot \sin \left( \dfrac{\pi }{9}+7 \right) \\
& =\sin \left( \left( \dfrac{4\pi }{9}+7 \right)-\left( \dfrac{\pi }{9}+7 \right) \right) \\
& =\sin \left( \dfrac{4\pi }{9}-\dfrac{\pi }{9}+7-7 \right) \\
& =\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2} \\
\end{align}\]
Now, as the right hand side that is R.H.S. is equal to the left hand side that is L.H.S., hence, the expression has been proved.
Note:Another method of attempting this question is by converting the right hand side that is R.H.S. to the left hand side that is L.H.S. by using the relations that are given in the hint. Through this method also, we could get to the correct answer and hence, we would be able to prove the required proof but proving this proof from the right hand side is very difficult and lengthy.
Now, the formula that would be used in solving this question is as follows
\[\sin A\cdot \cos B-\cos B\cdot \sin A=\sin (A-B)\]
Complete step-by-step answer:
Now, this is the result that would be used to prove the proof mentioned in this question as using this identity, we would convert the left hand side that is L.H.S. or the right hand side that is R.H.S. to make either of them equal to the other.
As mentioned in the question, we have to prove the given expression that is
\[\sin \left( \dfrac{4\pi }{9}+7 \right)\cdot \cos \left( \dfrac{\pi }{9}+7 \right)-\cos \left( \dfrac{4\pi }{9}+7 \right)\cdot \sin \left( \dfrac{\pi }{9}+7 \right)=\dfrac{\sqrt{3}}{2}\] .
Now, we will start with the left hand side that is L.H.S. as follows
\[=\sin \left( \dfrac{4\pi }{9}+7 \right)\cdot \cos \left( \dfrac{\pi }{9}+7 \right)-\cos \left( \dfrac{4\pi }{9}+7 \right)\cdot \sin \left( \dfrac{\pi }{9}+7 \right)\]
Now, we will use the relation that is mentioned the hint as follows
\[\begin{align}
& =\sin \left( \dfrac{4\pi }{9}+7 \right)\cdot \cos \left( \dfrac{\pi }{9}+7 \right)-\cos \left( \dfrac{4\pi }{9}+7 \right)\cdot \sin \left( \dfrac{\pi }{9}+7 \right) \\
& =\sin \left( \left( \dfrac{4\pi }{9}+7 \right)-\left( \dfrac{\pi }{9}+7 \right) \right) \\
& =\sin \left( \dfrac{4\pi }{9}-\dfrac{\pi }{9}+7-7 \right) \\
& =\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2} \\
\end{align}\]
Now, as the right hand side that is R.H.S. is equal to the left hand side that is L.H.S., hence, the expression has been proved.
Note:Another method of attempting this question is by converting the right hand side that is R.H.S. to the left hand side that is L.H.S. by using the relations that are given in the hint. Through this method also, we could get to the correct answer and hence, we would be able to prove the required proof but proving this proof from the right hand side is very difficult and lengthy.
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