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Question: Prove that \(\sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -...

Prove that sin⁡(−420∘)cos⁡(390∘)+cos⁡(−660∘)sin⁡(330∘)=−1\sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -{{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right)=-1 .

Explanation

Solution

Hint:For solving this question, we will use the formulas like sin⁡(−θ)=−sin⁡θ\sin \left( -\theta \right)=-\sin \theta to write sin⁡(−420∘)=−sin⁡(420∘)\sin \left( -{{420}^{\circ }} \right)=-\sin \left( {{420}^{\circ }} \right) and cos⁡(−θ)=cos⁡θ\cos \left( -\theta \right)=\cos \theta to write cos⁡(−660∘)=cos⁡(660∘)\cos \left( -{{660}^{\circ }} \right)=\cos \left( {{660}^{\circ }} \right) . After that, we will write the given trigonometric ratios into the form of sin⁡420∘=sin⁡(2π+60∘)\sin {{420}^{\circ }}=\sin \left( 2\pi +{{60}^{\circ }} \right) , cos⁡390∘=cos⁡(2π+30∘)\cos {{390}^{\circ }}=\cos \left( 2\pi +{{30}^{\circ }} \right) , cos⁡660∘=cos⁡(4π−60∘)\cos {{660}^{\circ }}=\cos \left( 4\pi -{{60}^{\circ }} \right) and sin⁡330∘=sin⁡(2π−30∘)\sin {{330}^{\circ }}=\sin \left( 2\pi -{{30}^{\circ }} \right) . Then, we will use the formulas like sin⁡(2π+θ)=sin⁡θ\sin \left( 2\pi +\theta \right)=\sin \theta , cos⁡(2π+θ)=cos⁡θ\cos \left( 2\pi +\theta \right)=\cos \theta , cos⁡(4π−θ)=cos⁡θ\cos \left( 4\pi -\theta \right)=\cos \theta and sin⁡(2π−θ)=−sin⁡θ\sin \left( 2\pi -\theta \right)=-\sin \theta to simplify further. And finally, we will use the result of trigonometric ratios like sin⁡60∘=cos⁡30∘=32\sin {{60}^{\circ }}=\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} and cos⁡60∘=sin⁡30∘=12\cos {{60}^{\circ }}=\sin {{30}^{\circ }}=\dfrac{1}{2} to solve further and prove the desired result easily.

Complete step-by-step answer:
Given:
We have to prove the following equation:
sin⁡(−420∘)cos⁡(390∘)+cos⁡(−660∘)sin⁡(330∘)=−1\sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -{{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right)=-1
Now, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side.
Now, before we proceed we should know the following formulas:
sin⁡(−θ)=−sin⁡θ...............(1) cos⁡(−θ)=cos⁡θ.................(2) sin⁡(2nπ+θ)=sin⁡θ...........(3) cos⁡(2nπ+θ)=cos⁡θ..........(4) cos⁡(2nπ−θ)=cos⁡θ...........(5) sin⁡(2nπ−θ)=−sin⁡θ..........(6) \begin{aligned} & \sin \left( -\theta \right)=-\sin \theta ...............\left( 1 \right) \\\ & \cos \left( -\theta \right)=\cos \theta .................\left( 2 \right) \\\ & \sin \left( 2n\pi +\theta \right)=\sin \theta ...........\left( 3 \right) \\\ & \cos \left( 2n\pi +\theta \right)=\cos \theta ..........\left( 4 \right) \\\ & \cos \left( 2n\pi -\theta \right)=\cos \theta ...........\left( 5 \right) \\\ & \sin \left( 2n\pi -\theta \right)=-\sin \theta ..........\left( 6 \right) \\\ \end{aligned}
Now, we will use the above six formulas to simplify the term on the left-hand side.
On the left-hand side, we have sin⁡(−420∘)cos⁡(390∘)+cos⁡(−660∘)sin⁡(330∘)\sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -{{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right) .
Now, we will use the formula from the equation (1) to write sin⁡(−420∘)=−sin⁡(420∘)\sin \left( -{{420}^{\circ }} \right)=-\sin \left( {{420}^{\circ }} \right) and formula from the equation (2) to write cos⁡(−660∘)=cos⁡(660∘)\cos \left( -{{660}^{\circ }} \right)=\cos \left( {{660}^{\circ }} \right) in the term on the left-hand side. Then,
sin⁡(−420∘)cos⁡(390∘)+cos⁡(−660∘)sin⁡(330∘) ⇒−sin⁡(420∘)cos⁡(390∘)+cos⁡(660∘)sin⁡(330∘) \begin{aligned} & \sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -{{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right) \\\ & \Rightarrow -\sin \left( {{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( {{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right) \\\ \end{aligned}
Now, we will write sin⁡(420∘)=sin⁡(360∘+60∘)\sin \left( {{420}^{\circ }} \right)=\sin \left( {{360}^{\circ }}+{{60}^{\circ }} \right) , cos⁡(390∘)=cos⁡(360∘+30∘)\cos \left( {{390}^{\circ }} \right)=\cos \left( {{360}^{\circ }}+{{30}^{\circ }} \right) , cos⁡(660∘)=cos⁡(720∘−60∘)\cos \left( {{660}^{\circ }} \right)=\cos \left( {{720}^{\circ }}-{{60}^{\circ }} \right) and sin⁡(330∘)=sin⁡(360∘−30∘)\sin \left( {{330}^{\circ }} \right)=\sin \left( {{360}^{\circ }}-{{30}^{\circ }} \right) in the above term. Then,
−sin⁡(420∘)cos⁡(390∘)+cos⁡(660∘)sin⁡(330∘) ⇒−sin⁡(360∘+60∘)cos⁡(360∘+30∘)+cos⁡(720∘−60∘)sin⁡(360∘−30∘) \begin{aligned} & -\sin \left( {{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( {{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right) \\\ & \Rightarrow -\sin \left( {{360}^{\circ }}+{{60}^{\circ }} \right)\cos \left( {{360}^{\circ }}+{{30}^{\circ }} \right)+\cos \left( {{720}^{\circ }}-{{60}^{\circ }} \right)\sin \left( {{360}^{\circ }}-{{30}^{\circ }} \right) \\\ \end{aligned}
Now, as we know that πc=180∘{{\pi }^{c}}={{180}^{\circ }} so, we can write 360∘=2πc{{360}^{\circ }}=2{{\pi }^{c}} and 720∘=4πc{{720}^{\circ }}=4{{\pi }^{c}} in the above term. Then,
−sin⁡(360∘+60∘)cos⁡(360∘+30∘)+cos⁡(720∘−60∘)sin⁡(360∘−30∘) ⇒−sin⁡(2π+60∘)cos⁡(2π+30∘)+cos⁡(4π−60∘)sin⁡(2π−30∘) \begin{aligned} & -\sin \left( {{360}^{\circ }}+{{60}^{\circ }} \right)\cos \left( {{360}^{\circ }}+{{30}^{\circ }} \right)+\cos \left( {{720}^{\circ }}-{{60}^{\circ }} \right)\sin \left( {{360}^{\circ }}-{{30}^{\circ }} \right) \\\ & \Rightarrow -\sin \left( 2\pi +{{60}^{\circ }} \right)\cos \left( 2\pi +{{30}^{\circ }} \right)+\cos \left( 4\pi -{{60}^{\circ }} \right)\sin \left( 2\pi -{{30}^{\circ }} \right) \\\ \end{aligned}
Now, when we put n=1n=1 in equation (3), then sin⁡(2π+θ)=sin⁡θ\sin \left( 2\pi +\theta \right)=\sin \theta so, we can write sin⁡(2π+60∘)=sin⁡60∘\sin \left( 2\pi +{{60}^{\circ }} \right)=\sin {{60}^{\circ }} in the above equation. Then,
−sin⁡(2π+60∘)cos⁡(2π+30∘)+cos⁡(4π−60∘)sin⁡(2π−30∘) ⇒−sin⁡60∘cos⁡(2π+30∘)+cos⁡(4π−60∘)sin⁡(2π−30∘) \begin{aligned} & -\sin \left( 2\pi +{{60}^{\circ }} \right)\cos \left( 2\pi +{{30}^{\circ }} \right)+\cos \left( 4\pi -{{60}^{\circ }} \right)\sin \left( 2\pi -{{30}^{\circ }} \right) \\\ & \Rightarrow -\sin {{60}^{\circ }}\cos \left( 2\pi +{{30}^{\circ }} \right)+\cos \left( 4\pi -{{60}^{\circ }} \right)\sin \left( 2\pi -{{30}^{\circ }} \right) \\\ \end{aligned}
Now, when we put n=1n=1 in equation (4), then cos⁡(2π+θ)=cos⁡θ\cos \left( 2\pi +\theta \right)=\cos \theta so, we can write cos⁡(2π+30∘)=cos⁡30∘\cos \left( 2\pi +{{30}^{\circ }} \right)=\cos {{30}^{\circ }} in the above term. Then,
−sin⁡60∘cos⁡(2π+30∘)+cos⁡(4π−60∘)sin⁡(2π−30∘) ⇒−sin⁡60∘cos⁡30∘+cos⁡(4π−60∘)sin⁡(2π−30∘) \begin{aligned} & -\sin {{60}^{\circ }}\cos \left( 2\pi +{{30}^{\circ }} \right)+\cos \left( 4\pi -{{60}^{\circ }} \right)\sin \left( 2\pi -{{30}^{\circ }} \right) \\\ & \Rightarrow -\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos \left( 4\pi -{{60}^{\circ }} \right)\sin \left( 2\pi -{{30}^{\circ }} \right) \\\ \end{aligned}
Now, when we put n=2n=2 in equation (5), then cos⁡(4π−θ)=cos⁡θ\cos \left( 4\pi -\theta \right)=\cos \theta so, we can write cos⁡(4π−60∘)=cos⁡60∘\cos \left( 4\pi -{{60}^{\circ }} \right)=\cos {{60}^{\circ }} in the above term. Then,
−sin⁡60∘cos⁡30∘+cos⁡(4π−60∘)sin⁡(2π−30∘) ⇒−sin⁡60∘cos⁡30∘+cos⁡60∘sin⁡(2π−30∘) \begin{aligned} & -\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos \left( 4\pi -{{60}^{\circ }} \right)\sin \left( 2\pi -{{30}^{\circ }} \right) \\\ & \Rightarrow -\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin \left( 2\pi -{{30}^{\circ }} \right) \\\ \end{aligned}
Now, when we put n=1n=1 in equation (6), then sin⁡(2π−θ)=−sin⁡θ\sin \left( 2\pi -\theta \right)=-\sin \theta so, we can write sin⁡(2π−30∘)=−sin⁡30∘\sin \left( 2\pi -{{30}^{\circ }} \right)=-\sin {{30}^{\circ }} in the above term. Then,
−sin⁡60∘cos⁡30∘+cos⁡60∘sin⁡(2π−30∘) ⇒−sin⁡60∘cos⁡30∘−cos⁡60∘sin⁡30∘ ⇒−(sin⁡60∘cos⁡30∘+cos⁡60∘sin⁡30∘) \begin{aligned} & -\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin \left( 2\pi -{{30}^{\circ }} \right) \\\ & \Rightarrow -\sin {{60}^{\circ }}\cos {{30}^{\circ }}-\cos {{60}^{\circ }}\sin {{30}^{\circ }} \\\ & \Rightarrow -\left( \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin {{30}^{\circ }} \right) \\\ \end{aligned}
Now, as we know that, sin⁡60∘=cos⁡30∘=32\sin {{60}^{\circ }}=\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} and cos⁡60∘=sin⁡30∘=12\cos {{60}^{\circ }}=\sin {{30}^{\circ }}=\dfrac{1}{2} . Then,
−(sin⁡60∘cos⁡30∘+cos⁡60∘sin⁡30∘) ⇒−(32×32+12×12) ⇒−(34+14) ⇒−1 \begin{aligned} & -\left( \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin {{30}^{\circ }} \right) \\\ & \Rightarrow -\left( \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \dfrac{1}{2} \right) \\\ & \Rightarrow -\left( \dfrac{3}{4}+\dfrac{1}{4} \right) \\\ & \Rightarrow -1 \\\ \end{aligned}
Now, from the above result, we conclude that the value of the term sin⁡(−420∘)cos⁡(390∘)+cos⁡(−660∘)sin⁡(330∘)\sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -{{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right) will be equal to −1-1 . Then,
sin⁡(−420∘)cos⁡(390∘)+cos⁡(−660∘)sin⁡(330∘)=−1\sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -{{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right)=-1
Now, from the above result, we conclude that the term on the left-hand side is equal to the term on the right-hand side.
Thus, sin⁡(−420∘)cos⁡(390∘)+cos⁡(−660∘)sin⁡(330∘)=−1\sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -{{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right)=-1 .
Hence, proved.

Note: Here, the student should first understand what we have to prove in the question. After that, we should proceed in a stepwise manner and apply trigonometric formulas like sin⁡(2π+θ)=sin⁡θ\sin \left( 2\pi +\theta \right)=\sin \theta , cos⁡(2π+θ)=cos⁡θ\cos \left( 2\pi +\theta \right)=\cos \theta , cos⁡(4π−θ)=cos⁡θ\cos \left( 4\pi -\theta \right)=\cos \theta and sin⁡(2π−θ)=−sin⁡θ\sin \left( 2\pi -\theta \right)=-\sin \theta correctly. Moreover, we could have also used the formula sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A to write −(sin⁡60∘cos⁡30∘+cos⁡60∘sin⁡30∘)=−sin⁡(60∘+30∘)=−sin⁡90∘-\left( \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin {{30}^{\circ }} \right)=-\sin \left( {{60}^{\circ }}+{{30}^{\circ }} \right)=-\sin {{90}^{\circ }} directly and then, used the result sin⁡90∘=1\sin {{90}^{\circ }}=1 to prove the desired result.