Question

If y = sin x, ${{y}_{n}}=\dfrac{{{d}^{n}}\left( \sin x \right)}{d{{x}^{n}}}$ then find the value of $\left| \begin{matrix} y & {{y}_{1}} & {{y}_{2}} \\\ {{y}_{3}} & {{y}_{4}} & {{y}_{5}} \\\ {{y}_{6}} & {{y}_{7}} & {{y}_{8}} \\\ \end{matrix} \right|=?$
(a) – sin x
(b) 0
(c) sin x
(d) cos x

Answer 1

Use the formula for ${{n}^{th}}$ differentiation of sin x given by $\dfrac{{{d}^{n}}\left( \sin x \right)}{d{{x}^{n}}}=\sin \left( x+\dfrac{1}{2}nx \right),$ where n = 1, 2, …..n, to evaluate the values of ${{y}_{1}},{{y}_{2}},{{y}_{3}}.....{{y}_{8}}.$ Then substitute these values in the determinant and expand it to get the answer.

Complete step by step answer:
We have been given that y = sin x and ${{y}_{n}}=\dfrac{{{d}^{n}}\left( \sin x \right)}{d{{x}^{n}}}.$ Here, $'{{y}_{n}}'$ denotes the ${{n}^{th}}$ differentiation of sin x. We know that the ${{n}^{th}}$ differentiation of sin x is given by the formula $\dfrac{{{d}^{n}}\left( \sin x \right)}{d{{x}^{n}}}=\sin \left( x+\dfrac{1}{2}nx \right),$ where n = 1, 2, 3…..n. Therefore, substituting the values of n = 1, 2, 3, ….8 one by one to determine the values of ${{y}_{1}},{{y}_{2}},{{y}_{3}},...{{y}_{n}},$ we get,
$\left( i \right){{y}_{1}}=\sin \left( x+\dfrac{\pi }{2} \right)=\cos x$
$\left( ii \right){{y}_{2}}=\sin \left( x+\dfrac{2\pi }{2} \right)$
$\Rightarrow {{y}_{2}}=\sin \left( x+\pi \right)$
$\Rightarrow {{y}_{2}}=-\sin x$
$\left( iii \right){{y}_{3}}=\sin \left( x+\dfrac{3\pi }{2} \right)=-\cos x$
$\left( iv \right){{y}_{4}}=\sin \left( x+\dfrac{4\pi }{2} \right)$
$\Rightarrow {{y}_{4}}=\sin \left( x+2\pi \right)$
$\Rightarrow {{y}_{4}}=\sin x$
$\left( v \right){{y}_{5}}=\sin \left( x+\dfrac{5\pi }{2} \right)=\cos x$
$\left( vi \right){{y}_{6}}=\sin \left( x+\dfrac{6\pi }{2} \right)$
$\Rightarrow {{y}_{6}}=\sin \left( x+3\pi \right)$
$\Rightarrow {{y}_{6}}=-\sin x$
$\left( vii \right){{y}_{7}}=\sin \left( x+\dfrac{7\pi }{2} \right)=-\cos x$
$\left( viii \right){{y}_{8}}=\sin \left( x+\dfrac{8\pi }{2} \right)$
$\Rightarrow {{y}_{8}}=\sin \left( x+4\pi \right)$
$\Rightarrow {{y}_{8}}=\sin x$
Now, let us assume the value of the determinant $\left| \begin{matrix} y & {{y}_{1}} & {{y}_{2}} \\\ {{y}_{3}} & {{y}_{4}} & {{y}_{5}} \\\ {{y}_{6}} & {{y}_{7}} & {{y}_{8}} \\\ \end{matrix} \right|$ as A. Therefore, substituting the values of ${{y}_{1}},{{y}_{2}},{{y}_{3}},....{{y}_{8}},$ we get,

\sin x & \cos x & -\sin x \\\ -\cos x & \sin x & \cos x \\\ -\sin x & -\cos x & \sin x \\\ \end{matrix} \right|$$ Expanding the determinant, we get, $$\begin{aligned} & \Rightarrow A=\sin x\left[ \sin x\times \sin x-\cos x\times \left( -\cos x \right) \right]-\cos x\left[ \left( -\cos x \right)\times \sin x-\left( -\sin x \right)\times \cos x \right] \\\ & +\left( -\sin x \right)\left[ \left( -\cos x \right)\times \left( -\cos x \right)-\left( -\sin x \right)\left( \sin x \right) \right] \\\ \end{aligned}$$ Here, we have expanded the determinant using the first row. $$\Rightarrow A=\sin x\left[ {{\sin }^{2}}x+{{\cos }^{2}}x \right]-\cos x\left[ -\sin x\cos x+\sin x\cos x \right]-\sin x\left[ {{\cos }^{2}}x+{{\sin }^{2}}x \right]$$ Cancelling the like terms and using the identity $${{\cos }^{2}}x+{{\sin }^{2}}x=1,$$ we get, $$\Rightarrow A=\sin x\times 1-\cos x\times 0-\sin x\times 1$$ $$\Rightarrow A=\sin x-\sin x$$ $$\Rightarrow A=0$$ **So, the correct answer is “Option b”.** **Note:** One may note that we need not require to find the $${{n}^{th}}$$ differentiation using the given formula. If we do not remember the formula, then also we can find all the differentiation values step by step. We can write the expression of each step of differentiation in terms of y and $${{y}_{1}}$$ and then we can easily find the determinant value. Note that there is a very easy method to determine the answer if the options are given just like in the above question. We can substitute x = 0 or $$x=\dfrac{\pi }{2}$$ in $$y,{{y}_{1}}{{y}_{2}},...{{y}_{8}}.$$ It will make our expression of the determinant easier as we already know the numerical values of $$\sin {{0}^{\circ }},\cos {{0}^{\circ }},\sin \dfrac{\pi }{2},\cos \dfrac{\pi }{2}.$$