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Question: The exact value of \(\cos \dfrac{2\pi }{28}\csc \dfrac{3\pi }{28}+\cos \dfrac{6\pi }{28}\csc \dfrac{...

The exact value of cos⁡2π28csc⁡3π28+cos⁡6π28csc⁡9π28+cos⁡18π28csc⁡27π28\cos \dfrac{2\pi }{28}\csc \dfrac{3\pi }{28}+\cos \dfrac{6\pi }{28}\csc \dfrac{9\pi }{28}+\cos \dfrac{18\pi }{28}\csc \dfrac{27\pi }{28} is?

Explanation

Solution

Hint:For solving this question we will assume π28=θ\dfrac{\pi }{28}=\theta and simply the given term with the help of trigonometric formulas like 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right) , sin⁡(π2−θ)=cos⁡θ\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta , 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right) , sin⁡(π−θ)=sin⁡θ\sin \left( \pi -\theta \right)=\sin \theta . After that, we will try to analyse the result and try to get some numerical value of the given term with proper use of a suitable formula.

Complete step-by-step answer:
Given:
We have to find the exact value of cos⁡2π28csc⁡3π28+cos⁡6π28csc⁡9π28+cos⁡18π28csc⁡27π28\cos \dfrac{2\pi }{28}\csc \dfrac{3\pi }{28}+\cos \dfrac{6\pi }{28}\csc \dfrac{9\pi }{28}+\cos \dfrac{18\pi }{28}\csc \dfrac{27\pi }{28} .
Now, before we proceed we should know the following formulas:
sin⁡(π2−θ)=cos⁡θ.................(1) cos⁡(π2+θ)=−sin⁡θ..............(2) sin⁡(π−θ)=sin⁡θ...................(3) 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B).....................(4) 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B).......................(5) 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)........................(6) \begin{aligned} & \sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta .................\left( 1 \right) \\\ & \cos \left( \dfrac{\pi }{2}+\theta \right)=-\sin \theta ..............\left( 2 \right) \\\ & \sin \left( \pi -\theta \right)=\sin \theta ...................\left( 3 \right) \\\ & 2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right).....................\left( 4 \right) \\\ & 2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right).......................\left( 5 \right) \\\ & 2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)........................\left( 6 \right) \\\ \end{aligned}
Now, let π28=θ\dfrac{\pi }{28}=\theta . Then,
cos⁡2π28csc⁡3π28+cos⁡6π28csc⁡9π28+cos⁡18π28csc⁡27π28 ⇒cos⁡2θcsc⁡3θ+cos⁡6θcsc⁡9θ+cos⁡18θcsc⁡27θ \begin{aligned} & \cos \dfrac{2\pi }{28}\csc \dfrac{3\pi }{28}+\cos \dfrac{6\pi }{28}\csc \dfrac{9\pi }{28}+\cos \dfrac{18\pi }{28}\csc \dfrac{27\pi }{28} \\\ & \Rightarrow \cos 2\theta \csc 3\theta +\cos 6\theta \csc 9\theta +\cos 18\theta \csc 27\theta \\\ \end{aligned}
Now, we will simplify cos⁡2θcsc⁡3θ+cos⁡6θcsc⁡9θ+cos⁡18θcsc⁡27θ\cos 2\theta \csc 3\theta +\cos 6\theta \csc 9\theta +\cos 18\theta \csc 27\theta and as we know that csc⁡θ=1sin⁡θ\csc \theta =\dfrac{1}{\sin \theta } . Then,
cos⁡2θcsc⁡3θ+cos⁡6θcsc⁡9θ+cos⁡18θcsc⁡27θ ⇒cos⁡2θsin⁡3θ+cos⁡6θsin⁡9θ+cos⁡18θsin⁡27θ ⇒cos⁡2θsin⁡3θ+cos⁡6θsin⁡(14θ−5θ)+cos⁡(14θ+4θ)sin⁡(28θ−θ) \begin{aligned} & \cos 2\theta \csc 3\theta +\cos 6\theta \csc 9\theta +\cos 18\theta \csc 27\theta \\\ & \Rightarrow \dfrac{\cos 2\theta }{\sin 3\theta }+\dfrac{\cos 6\theta }{\sin 9\theta }+\dfrac{\cos 18\theta }{\sin 27\theta } \\\ & \Rightarrow \dfrac{\cos 2\theta }{\sin 3\theta }+\dfrac{\cos 6\theta }{\sin \left( 14\theta -5\theta \right)}+\dfrac{\cos \left( 14\theta +4\theta \right)}{\sin \left( 28\theta -\theta \right)} \\\ \end{aligned}
Now, as per our assumption π28=θ\dfrac{\pi }{28}=\theta we can write 14θ=π214\theta=\dfrac{\pi}{2} . Then,
cos⁡2θsin⁡3θ+cos⁡6θsin⁡(14θ−5θ)+cos⁡(14θ+4θ)sin⁡(28θ−θ) ⇒cos⁡2θsin⁡3θ+cos⁡6θsin⁡(π2−5θ)+cos⁡(π2+4θ)sin⁡(π−θ) \begin{aligned} & \dfrac{\cos 2\theta }{\sin 3\theta }+\dfrac{\cos 6\theta }{\sin \left( 14\theta -5\theta \right)}+\dfrac{\cos \left( 14\theta +4\theta \right)}{\sin \left( 28\theta -\theta \right)} \\\ & \Rightarrow \dfrac{\cos 2\theta }{\sin 3\theta }+\dfrac{\cos 6\theta }{\sin \left( \dfrac{\pi }{2}-5\theta \right)}+\dfrac{\cos \left( \dfrac{\pi }{2}+4\theta \right)}{\sin \left( \pi -\theta \right)} \\\ \end{aligned}
Now, using the formula from the equation (1) to write sin⁡(π2−5θ)=cos⁡5θ\sin \left( \dfrac{\pi }{2}-5\theta \right)=\cos 5\theta , formula from the equation (2) to write cos⁡(π2+4θ)=−sin⁡4θ\cos \left( \dfrac{\pi }{2}+4\theta \right)=-\sin 4\theta and formula from the equation (3) to write sin⁡(π−θ)=sin⁡θ\sin \left( \pi -\theta \right)=\sin \theta in the above. Then,
cos⁡2θsin⁡3θ+cos⁡6θsin⁡(π2−5θ)+cos⁡(π2+4θ)sin⁡(π−θ) ⇒cos⁡2θsin⁡3θ+cos⁡6θcos⁡5θ−sin⁡4θsin⁡θ ⇒sin⁡θcos⁡2θcos⁡5θ+cos⁡6θsin⁡3θsin⁡θ−cos⁡5θsin⁡3θsin⁡4θsin⁡3θcos⁡5θsin⁡θ=ND......................(7) \begin{aligned} & \dfrac{\cos 2\theta }{\sin 3\theta }+\dfrac{\cos 6\theta }{\sin \left( \dfrac{\pi }{2}-5\theta \right)}+\dfrac{\cos \left( \dfrac{\pi }{2}+4\theta \right)}{\sin \left( \pi -\theta \right)} \\\ & \Rightarrow \dfrac{\cos 2\theta }{\sin 3\theta }+\dfrac{\cos 6\theta }{\cos 5\theta }-\dfrac{\sin 4\theta }{\sin \theta } \\\ & \Rightarrow \dfrac{\sin \theta \cos 2\theta \cos 5\theta +\cos 6\theta \sin 3\theta \sin \theta -\cos 5\theta \sin 3\theta \sin 4\theta }{\sin 3\theta \cos 5\theta \sin \theta }=\dfrac{N}{D}......................\left( 7 \right) \\\ \end{aligned}
Now, we will simplify the numerator NN . Then,
N=sin⁡θcos⁡2θcos⁡5θ+cos⁡6θsin⁡3θsin⁡θ−cos⁡5θsin⁡3θsin⁡4θ ⇒2N=2sin⁡θcos⁡2θcos⁡5θ+2cos⁡6θsin⁡3θsin⁡θ−2cos⁡5θsin⁡4θsin⁡3θ ⇒2N=sin⁡θ(2cos⁡2θcos⁡5θ)+cos⁡6θ(2sin⁡3θsin⁡θ)−cos⁡5θ(2sin⁡4θsin⁡3θ) \begin{aligned} & N=\sin \theta \cos 2\theta \cos 5\theta +\cos 6\theta \sin 3\theta \sin \theta -\cos 5\theta \sin 3\theta \sin 4\theta \\\ & \Rightarrow 2N=2\sin \theta \cos 2\theta \cos 5\theta +2\cos 6\theta \sin 3\theta \sin \theta -2\cos 5\theta \sin 4\theta \sin 3\theta \\\ & \Rightarrow 2N=\sin \theta \left( 2\cos 2\theta \cos 5\theta \right)+\cos 6\theta \left( 2\sin 3\theta \sin \theta \right)-\cos 5\theta \left( 2\sin 4\theta \sin 3\theta \right) \\\ \end{aligned}
Now, use the formula form the equation (4) to write 2cos⁡2θcos⁡5θ=cos⁡7θ+cos⁡3θ2\cos 2\theta \cos 5\theta =\cos 7\theta +\cos 3\theta and formula from the equation (5) to write 2sin⁡3θsin⁡θ=cos⁡2θ−cos⁡4θ2\sin 3\theta \sin \theta =\cos 2\theta -\cos 4\theta and 2sin⁡4θsin⁡3θ=cos⁡θ−cos⁡7θ2\sin 4\theta \sin 3\theta =\cos \theta -\cos 7\theta in the above equation. Then,
2N=sin⁡θ(2cos⁡2θcos⁡5θ)+cos⁡6θ(2sin⁡3θsin⁡θ)−cos⁡5θ(2sin⁡4θsin⁡3θ) ⇒2N=sin⁡θ(cos⁡(5θ+2θ)+cos⁡(5θ−2θ))+cos⁡6θ(cos⁡(3θ−θ)−cos⁡(3θ+θ)) −cos⁡5θ(cos⁡(4θ−3θ)−cos⁡(4θ+3θ)) ⇒2N=sin⁡θ(cos⁡7θ+cos⁡3θ)+cos⁡6θ(cos⁡2θ−cos⁡4θ)−cos⁡5θ(cos⁡θ−cos⁡7θ) ⇒2N=sin⁡θcos⁡7θ+sin⁡θcos⁡3θ+cos⁡6θcos⁡2θ−cos⁡6θcos⁡4θ−cos⁡5θcos⁡θ+cos⁡5θcos⁡7θ ⇒4N=2sin⁡θcos⁡7θ+2sin⁡θcos⁡3θ+2cos⁡6θcos⁡2θ−2cos⁡6θcos⁡4θ−2cos⁡5θcos⁡θ+2cos⁡7θcos⁡5θ \begin{aligned} & 2N=\sin \theta \left( 2\cos 2\theta \cos 5\theta \right)+\cos 6\theta \left( 2\sin 3\theta \sin \theta \right)-\cos 5\theta \left( 2\sin 4\theta \sin 3\theta \right) \\\ & \Rightarrow 2N=\sin \theta \left( \cos \left( 5\theta +2\theta \right)+\cos \left( 5\theta -2\theta \right) \right)+\cos 6\theta \left( \cos \left( 3\theta -\theta \right)-\cos \left( 3\theta +\theta \right) \right) \\\ & -\cos 5\theta \left( \cos \left( 4\theta -3\theta \right)-\cos \left( 4\theta +3\theta \right) \right) \\\ & \Rightarrow 2N=\sin \theta \left( \cos 7\theta +\cos 3\theta \right)+\cos 6\theta \left( \cos 2\theta -\cos 4\theta \right)-\cos 5\theta \left( \cos \theta -\cos 7\theta \right) \\\ & \Rightarrow 2N=\sin \theta \cos 7\theta +\sin \theta \cos 3\theta +\cos 6\theta \cos 2\theta -\cos 6\theta \cos 4\theta -\cos 5\theta \cos \theta +\cos 5\theta \cos 7\theta \\\ & \Rightarrow 4N=2\sin \theta \cos 7\theta +2\sin \theta \cos 3\theta +2\cos 6\theta \cos 2\theta -2\cos 6\theta \cos 4\theta -2\cos 5\theta \cos \theta +2\cos 7\theta \cos 5\theta \\\ \end{aligned}
Now, use the formula from the equation (6) to write 2sin⁡θcos⁡7θ=sin⁡8θ−sin⁡6θ2\sin \theta \cos 7\theta =\sin 8\theta -\sin 6\theta , 2sin⁡θcos⁡3θ=sin⁡4θ−sin⁡2θ2\sin \theta \cos 3\theta =\sin 4\theta -\sin 2\theta and equation (4) to write 2cos⁡6θcos⁡2θ=cos⁡8θ+cos⁡4θ2\cos 6\theta \cos 2\theta =\cos 8\theta +\cos 4\theta , 2cos⁡6θcos⁡4θ=cos⁡10θ+cos⁡2θ2\cos 6\theta \cos 4\theta =\cos 10\theta +\cos 2\theta , 2cos⁡5θcos⁡θ=cos⁡6θ+cos⁡4θ2\cos 5\theta \cos \theta =\cos 6\theta +\cos 4\theta and 2cos⁡7θcos⁡5θ=cos⁡12θ+cos⁡2θ2\cos 7\theta \cos 5\theta =\cos 12\theta +\cos 2\theta in the above equation. Then,
4N=2sin⁡θcos⁡7θ+2sin⁡θcos⁡3θ+2cos⁡6θcos⁡2θ−2cos⁡6θcos⁡4θ−2cos⁡5θcos⁡θ+2cos⁡7θcos⁡5θ ⇒4N=sin⁡(θ+7θ)+sin⁡(θ−7θ)+sin⁡(θ+3θ)+sin⁡(θ−3θ)+cos⁡(6θ+2θ)+cos⁡(6θ−2θ) −cos⁡(6θ+4θ)−cos⁡(6θ−4θ)−cos⁡(5θ+θ)−cos⁡(5θ−θ)+cos⁡(7θ+5θ)+cos⁡(7θ−5θ) ⇒4N=sin⁡8θ+sin⁡(−6θ)+sin⁡4θ+sin⁡(−2θ)+cos⁡8θ+cos⁡4θ−cos⁡10θ−cos⁡2θ−cos⁡6θ−cos⁡4θ +cos⁡12θ+cos⁡2θ \begin{aligned} & 4N=2\sin \theta \cos 7\theta +2\sin \theta \cos 3\theta +2\cos 6\theta \cos 2\theta -2\cos 6\theta \cos 4\theta -2\cos 5\theta \cos \theta +2\cos 7\theta \cos 5\theta \\\ & \Rightarrow 4N=\sin \left( \theta +7\theta \right)+\sin \left( \theta -7\theta \right)+\sin \left( \theta +3\theta \right)+\sin \left( \theta -3\theta \right)+\cos \left( 6\theta +2\theta \right)+\cos \left( 6\theta -2\theta \right) \\\ & -\cos \left( 6\theta +4\theta \right)-\cos \left( 6\theta -4\theta \right)-\cos \left( 5\theta +\theta \right)-\cos \left( 5\theta -\theta \right)+\cos \left( 7\theta +5\theta \right)+\cos \left( 7\theta -5\theta \right) \\\ & \Rightarrow 4N=\sin 8\theta +\sin \left( -6\theta \right)+\sin 4\theta +\sin \left( -2\theta \right)+\cos 8\theta +\cos 4\theta -\cos 10\theta -\cos 2\theta -\cos 6\theta -\cos 4\theta \\\ & +\cos 12\theta +\cos 2\theta \\\ \end{aligned}
Now, as we know that sin⁡(−θ)=−sin⁡θ\sin \left( -\theta \right)=-\sin \theta . Then,
4N=sin⁡8θ+sin⁡(−6θ)+sin⁡4θ+sin⁡(−2θ)+cos⁡8θ+cos⁡4θ−cos⁡10θ−cos⁡2θ−cos⁡6θ−cos⁡4θ +cos⁡12θ+cos⁡2θ ⇒4N=sin⁡8θ−sin⁡6θ+sin⁡4θ−sin⁡2θ+cos⁡8θ+cos⁡4θ−cos⁡10θ−cos⁡2θ−cos⁡6θ−cos⁡4θ +cos⁡12θ+cos⁡2θ ⇒4N=(sin⁡8θ−cos⁡6θ)+(cos⁡8θ−sin⁡6θ)+(sin⁡4θ−cos⁡10θ)+(cos⁡12θ−sin⁡2θ)+(cos⁡4θ−cos⁡4θ) +(cos⁡2θ−cos⁡2θ) ⇒4N=(sin⁡(14θ−6θ)−cos⁡6θ)+(cos⁡8θ−sin⁡(14θ−8θ))+(sin⁡(14θ−10θ)−cos⁡10θ) +(cos⁡12θ−sin⁡(14θ−12θ)) \begin{aligned} & 4N=\sin 8\theta +\sin \left( -6\theta \right)+\sin 4\theta +\sin \left( -2\theta \right)+\cos 8\theta +\cos 4\theta -\cos 10\theta -\cos 2\theta -\cos 6\theta -\cos 4\theta \\\ & +\cos 12\theta +\cos 2\theta \\\ & \Rightarrow 4N=\sin 8\theta -\sin 6\theta +\sin 4\theta -\sin 2\theta +\cos 8\theta +\cos 4\theta -\cos 10\theta -\cos 2\theta -\cos 6\theta -\cos 4\theta \\\ & +\cos 12\theta +\cos 2\theta \\\ & \Rightarrow 4N=\left( \sin 8\theta -\cos 6\theta \right)+\left( \cos 8\theta -\sin 6\theta \right)+\left( \sin 4\theta -\cos 10\theta \right)+\left( \cos 12\theta -\sin 2\theta \right)+\left( \cos 4\theta -\cos 4\theta \right) \\\ & +\left( \cos 2\theta -\cos 2\theta \right) \\\ & \Rightarrow 4N=\left( \sin \left( 14\theta -6\theta \right)-\cos 6\theta \right)+\left( \cos 8\theta -\sin \left( 14\theta -8\theta \right) \right)+\left( \sin \left( 14\theta -10\theta \right)-\cos 10\theta \right) \\\ & +\left( \cos 12\theta -\sin \left( 14\theta -12\theta \right) \right) \\\ \end{aligned}
Now, as per our assumption π28=θ\dfrac{\pi }{28}=\theta . Then,
4N=(sin⁡(14θ−6θ)−cos⁡6θ)+(cos⁡8θ−sin⁡(14θ−8θ))+(sin⁡(14θ−10θ)−cos⁡10θ) +(cos⁡12θ−sin⁡(14θ−12θ)) ⇒4N=(sin⁡(π2−6θ)−cos⁡6θ)+(cos⁡8θ−sin⁡(π2−8θ))+(sin⁡(π2−10θ)−cos⁡10θ) +(cos⁡12θ−sin⁡(π2−12θ)) \begin{aligned} & 4N=\left( \sin \left( 14\theta -6\theta \right)-\cos 6\theta \right)+\left( \cos 8\theta -\sin \left( 14\theta -8\theta \right) \right)+\left( \sin \left( 14\theta -10\theta \right)-\cos 10\theta \right) \\\ & +\left( \cos 12\theta -\sin \left( 14\theta -12\theta \right) \right) \\\ & \Rightarrow 4N=\left( \sin \left( \dfrac{\pi }{2}-6\theta \right)-\cos 6\theta \right)+\left( \cos 8\theta -\sin \left( \dfrac{\pi }{2}-8\theta \right) \right)+\left( \sin \left( \dfrac{\pi }{2}-10\theta \right)-\cos 10\theta \right) \\\ & +\left( \cos 12\theta -\sin \left( \dfrac{\pi }{2}-12\theta \right) \right) \\\ \end{aligned}
Now, using the formula from the equation (1) to write sin⁡(π2−6θ)=cos⁡6θ\sin \left( \dfrac{\pi }{2}-6\theta \right)=\cos 6\theta , sin⁡(π2−8θ)=cos⁡8θ\sin \left( \dfrac{\pi }{2}-8\theta \right)=\cos 8\theta , sin⁡(π2−10θ)=cos⁡10θ\sin \left( \dfrac{\pi }{2}-10\theta \right)=\cos 10\theta and sin⁡(π2−12θ)=cos⁡12θ\sin \left( \dfrac{\pi }{2}-12\theta \right)=\cos 12\theta in the above equation. Then,
4N=(sin⁡(π2−6θ)−cos⁡6θ)+(cos⁡8θ−sin⁡(π2−8θ))+(sin⁡(π2−10θ)−cos⁡10θ) +(cos⁡12θ−sin⁡(π2−12θ)) ⇒4N=(cos⁡6θ−cos⁡6θ)+(cos⁡8θ−sin⁡8θ)+(cos⁡10θ−cos⁡10θ)+(cos⁡12θ−cos⁡12θ) ⇒4N=0 ⇒N=0 \begin{aligned} & 4N=\left( \sin \left( \dfrac{\pi }{2}-6\theta \right)-\cos 6\theta \right)+\left( \cos 8\theta -\sin \left( \dfrac{\pi }{2}-8\theta \right) \right)+\left( \sin \left( \dfrac{\pi }{2}-10\theta \right)-\cos 10\theta \right) \\\ & +\left( \cos 12\theta -\sin \left( \dfrac{\pi }{2}-12\theta \right) \right) \\\ & \Rightarrow 4N=\left( \cos 6\theta -\cos 6\theta \right)+\left( \cos 8\theta -\sin 8\theta \right)+\left( \cos 10\theta -\cos 10\theta \right)+\left( \cos 12\theta -\cos 12\theta \right) \\\ & \Rightarrow 4N=0 \\\ & \Rightarrow N=0 \\\ \end{aligned}
Now, put N=0N=0 from the above equation in the equation (7). Then,
sin⁡θcos⁡2θcos⁡5θ+cos⁡6θsin⁡3θsin⁡θ−cos⁡5θsin⁡3θsin⁡4θsin⁡3θcos⁡5θsin⁡θ=ND ⇒sin⁡θcos⁡2θcos⁡5θ+cos⁡6θsin⁡3θsin⁡θ−cos⁡5θsin⁡3θsin⁡4θsin⁡3θcos⁡5θsin⁡θ=0 \begin{aligned} & \dfrac{\sin \theta \cos 2\theta \cos 5\theta +\cos 6\theta \sin 3\theta \sin \theta -\cos 5\theta \sin 3\theta \sin 4\theta }{\sin 3\theta \cos 5\theta \sin \theta }=\dfrac{N}{D} \\\ & \Rightarrow \dfrac{\sin \theta \cos 2\theta \cos 5\theta +\cos 6\theta \sin 3\theta \sin \theta -\cos 5\theta \sin 3\theta \sin 4\theta }{\sin 3\theta \cos 5\theta \sin \theta }=0 \\\ \end{aligned}
Now, as we have simplified the given term into sin⁡θcos⁡2θcos⁡5θ+cos⁡6θsin⁡3θsin⁡θ−cos⁡5θsin⁡3θsin⁡4θsin⁡3θcos⁡5θsin⁡θ\dfrac{\sin \theta \cos 2\theta \cos 5\theta +\cos 6\theta \sin 3\theta \sin \theta -\cos 5\theta \sin 3\theta \sin 4\theta }{\sin 3\theta \cos 5\theta \sin \theta } and its value is zero. Then,
cos⁡2π28csc⁡3π28+cos⁡6π28csc⁡9π28+cos⁡18π28csc⁡27π28=0\cos \dfrac{2\pi }{28}\csc \dfrac{3\pi }{28}+\cos \dfrac{6\pi }{28}\csc \dfrac{9\pi }{28}+\cos \dfrac{18\pi }{28}\csc \dfrac{27\pi }{28}=0
Thus, the exact value of the given term will be 0.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer. And we should proceed in a stepwise manner and use each formula with proper values and make assumptions like θ=π28\theta =\dfrac{\pi }{28} for smooth calculation. Moreover, we should avoid calculation mistakes while solving to get the correct answer.Students should remember important trigonometric identities ,formulas and standard trigonometric angles for solving these types of questions.