Prove the following statement using the basic trigonometry identities and formulas and the value of different trigonometry function at different arguments $\sin \dfrac{8\pi }{3}\cos \dfrac{23\pi }{6}+\cos \dfrac{13\pi }{3}\sin \dfrac{35\pi }{6}=\dfrac{1}{2}$.

Hint: $\sin x$ is positive in both the first and second quadrant whereas $\cos x$ is positive in the first quadrant and it is negative in the second quadrant. Mathematically, we deduce $\sin \left( 90+\theta \right)=\cos \theta $ and $\cos \left( 90+\theta \right)=\sin \theta $. Using these formulas we will prove $\sin \dfrac{8\pi }{3}\cos \dfrac{23\pi }{6}+\cos \dfrac{13\pi }{3}\sin \dfrac{35\pi }{6}=\dfrac{1}{2}$.

Complete step-by-step answer:

It is given in the question to show that $\sin \dfrac{8\pi }{3}\cos \dfrac{23\pi }{6}+\cos \dfrac{13\pi }{3}\sin \dfrac{35\pi }{6}=\dfrac{1}{2}$. We will use some basic formulas of trigonometry as follows $\sin \left( 90+\theta \right)=\cos \theta $ and $\cos \left( 90+\theta \right)=\sin \theta $. Also we know that

\[\sin x\] is positive in both the first and second quadrant whereas $\cos x$ is positive in the first quadrant and it is negative in the second quadrant.

Now, we have been given expression in the LHS side as $\sin \dfrac{8\pi }{3}\cos \dfrac{23\pi }{6}+\cos \dfrac{13\pi }{3}\sin \dfrac{35\pi }{6}$, we can write it in degree form as $\sin \dfrac{8\times {{180}^{{}^\circ }}}{3}\cos \dfrac{23\times {{180}^{{}^\circ }}}{6}+\cos \dfrac{13\times {{180}^{{}^\circ }}}{3}\sin \dfrac{35\times {{180}^{{}^\circ }}}{6}$ solving the angle values we get $\sin {{480}^{{}^\circ }}\cos {{690}^{{}^\circ }}+\cos {{780}^{{}^\circ }}\sin {{1050}^{{}^\circ }}$. Now we can write $\sin \left( {{480}^{{}^\circ }} \right)$ as $\sin \left( {{90}^{{}^\circ }}\times 5+{{30}^{{}^\circ }} \right)$ and $\cos \left( {{690}^{{}^\circ }} \right)$ as $\cos \left( {{90}^{{}^\circ }}\times 7+{{60}^{{}^\circ }} \right)$, similarly we can write $\cos \left( {{780}^{{}^\circ }} \right)$ as $\cos \left( {{90}^{{}^\circ }}\times 8+{{60}^{{}^\circ }} \right)$ and $\sin \left( {{1050}^{{}^\circ }} \right)$ as $\sin \left( {{90}^{{}^\circ }}\times 11+{{60}^{{}^\circ }} \right)$, therefore we get the expression as $\sin \left( {{90}^{{}^\circ }}\times 5+{{30}^{{}^\circ }} \right)\cos \left( {{90}^{{}^\circ }}\times 7+{{60}^{{}^\circ }} \right)+\cos \left( {{90}^{{}^\circ }}\times 8+{{60}^{{}^\circ }} \right)\sin \left( {{90}^{{}^\circ }}\times 11+{{60}^{{}^\circ }} \right)$.

We know that $\sin \left( 90+\theta \right)=\cos \theta $,$\cos \left( 90+\theta \right)=\sin \theta $ and also $\sin \left( {{90}^{{}^\circ }}\times 11+{{60}^{{}^\circ }} \right)=-\cos \left( {{60}^{{}^\circ }} \right)$ because $\cos \theta $ is negative in the fourth quadrant. Therefore using the above relations we get our expression reduced to $\cos \left( {{30}^{{}^\circ }} \right)\sin \left( {{60}^{{}^\circ }} \right)+\cos \left( {{60}^{{}^\circ }} \right)\left[ -\cos \left( {{60}^{{}^\circ }} \right) \right]$= $\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \left( \dfrac{-1}{2} \right)$ which on solving results in $\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}=RHS$ of the given equation.

Hence Proved $\sin \dfrac{8\pi }{3}\cos \dfrac{23\pi }{6}+\cos \dfrac{13\pi }{3}\sin \dfrac{35\pi }{6}=\dfrac{1}{2}$.


Note: Usually students forget the nature of the trigonometric functions in the various quadrants, $\sin x$ is positive in both first and second quadrant whereas $\cos x$ is positive in first quadrant and it is negative in second quadrant. Student may take $\sin \left( {{90}^{{}^\circ }}\times 11+{{60}^{{}^\circ }} \right)=\cos \left( {{60}^{{}^\circ }} \right)$ instead of $\sin \left( {{90}^{{}^\circ }}\times 11+{{60}^{{}^\circ }} \right)=-\cos \left( {{60}^{{}^\circ }} \right)$ which will give us wrong answer and we cannot reach to prove the given statement. Since sine and cosine functions are complementary, therefore we have such a behaviour possible.