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Question: How do you show that \(2\sin x\cos x=\sin 2x\) ? Is it true for \(\dfrac{5\pi }{6}\) ?...

How do you show that 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x=\sin 2x ? Is it true for 5π6\dfrac{5\pi }{6} ?

Explanation

Solution

Here in this question we have been asked to show that 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x=\sin 2x and verify it for the value of xx equal to 5π6\dfrac{5\pi }{6} . For that sake we will use the formula sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B.

Complete step by step solution:
Now considering from the question we have been asked to show that 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x=\sin 2x and verify it for the value of xx equal to 5π6\dfrac{5\pi }{6} .
For that sake we will use the formula sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B which we have learnt during trigonometric basics and we also know that 5π6=150∘\dfrac{5\pi }{6}={{150}^{\circ }} .
By using sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B if we keep A=B=xA=B=x then we will have sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x .
From the trigonometric table which specifies values for different angles we have sin⁡30∘=12\sin {{30}^{\circ }}=\dfrac{1}{2} and cos⁡30∘=32\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} . We know that 150∘{{150}^{\circ }} lies in the second quadrant. In which only sine and cosecant functions are positive remaining are negative. We also know that sin⁡(180∘−x)=cos⁡x\sin \left( {{180}^{\circ }}-x \right)=\cos x and cos⁡(180∘−x)=−sin⁡x\cos \left( {{180}^{\circ }}-x \right)=-\sin x .
Now we can say that
⇒sin⁡150∘=sin⁡(180∘−30∘) ⇒cos⁡30∘=32 \begin{aligned} & \Rightarrow \sin {{150}^{\circ }}=\sin \left( {{180}^{\circ }}-{{30}^{\circ }} \right) \\\ & \Rightarrow \cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} \\\ \end{aligned}
Similarly we can also say that
⇒cos⁡150∘=cos⁡(180∘−30∘) ⇒−sin⁡30∘=−12 \begin{aligned} & \Rightarrow \cos {{150}^{\circ }}=\cos \left( {{180}^{\circ }}-{{30}^{\circ }} \right) \\\ & \Rightarrow -\sin {{30}^{\circ }}=\dfrac{-1}{2} \\\ \end{aligned}
Now we can say that
2sin⁡xcos⁡x=2(32)(−12) ⇒−32 \begin{aligned} & 2\sin x\cos x=2\left( \dfrac{\sqrt{3}}{2} \right)\left( \dfrac{-1}{2} \right) \\\ & \Rightarrow \dfrac{-\sqrt{3}}{2} \\\ \end{aligned}
For x=5π6x=\dfrac{5\pi }{6} and now we need to check for sin⁡2x\sin 2x by doing that we will have ⇒sin⁡300∘=sin⁡(360∘−60∘) ⇒−sin⁡60∘=−32 \begin{aligned} & \Rightarrow \sin {{300}^{\circ }}=\sin \left( {{360}^{\circ }}-{{60}^{\circ }} \right) \\\ & \Rightarrow -\sin {{60}^{\circ }}=\dfrac{-\sqrt{3}}{2} \\\ \end{aligned} .
Therefore we can conclude that the expression is verified and valid for any value.

Note: During the process of answering questions of this type we should be sure with our trigonometric concepts that we are going to apply in between. This is a very simple and easy question and can be answered accurately in a short span of time. It is completely theory based question and can be answered by applying accurate trigonometric concepts. We can also write
sin⁡150∘=sin⁡(90∘+60∘) ⇒sin⁡60∘=32 \begin{aligned} & \sin {{150}^{\circ }}=\sin \left( {{90}^{\circ }}+{{60}^{\circ }} \right) \\\ & \Rightarrow \sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2} \\\ \end{aligned}
And similarly the other simplifications can be done.