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Question: If \(y = \cos x,{y_n} = \dfrac{{{d^n}\left( {\cos x} \right)}}{{d{x^n}}}\), then \(\left| {\begin{ar...

If y=cos⁡x,yn=dn(cos⁡x)dxny = \cos x,{y_n} = \dfrac{{{d^n}\left( {\cos x} \right)}}{{d{x^n}}}, then \left| {\begin{array}{*{20}{c}} {{y_4}}&{{y_5}}&{{y_6}} \\\ {{y_7}}&{{y_8}}&{{y_9}} \\\ {{y_{10}}}&{{y_{11}}}&{{y_{12}}} \end{array}} \right|=....
(a)0\left( a \right)0
(b)−cos⁡x\left( b \right) - \cos x
(c)cos⁡x\left( c \right)\cos x
(d)sin⁡x\left( d \right)\sin x

Explanation

Solution

In this particular question use the concept that the differentiation of cos x is – sin x and the differentiation of sin x is cos x, later on use the concept of expansion of determinant so use these concepts to reach the solution of the question.

Complete step-by-step answer :
Given equation
y=cos⁡xy = \cos x.............. (1)
yn=dn(cos⁡x)dxn{y_n} = \dfrac{{{d^n}\left( {\cos x} \right)}}{{d{x^n}}}
Now as we know that the differentiation of cos x is – sin x, and the differentiation of sin x is cos x so differentiate equation (1) w.r.t x we have,
⇒y1=d(cos⁡x)dx=−sin⁡x\Rightarrow {y_1} = \dfrac{{d\left( {\cos x} \right)}}{{dx}} = - \sin x
Now again differentiate w.r.t x we have,
⇒y2=−ddxsin⁡x=−cos⁡x\Rightarrow {y_2} = - \dfrac{d}{{dx}}\sin x = - \cos x
Now again differentiate w.r.t x we have,
⇒y3=−ddxcos⁡x=sin⁡x\Rightarrow {y_3} = - \dfrac{d}{{dx}}\cos x = \sin x
Now again differentiate w.r.t x we have,
⇒y4=ddxsin⁡x=cos⁡x\Rightarrow {y_4} = \dfrac{d}{{dx}}\sin x = \cos x
So as we see that y4{y_4} is the same as y, so the value will be repeated.
⇒y5=y=y9−sin⁡x\Rightarrow {y_5} = y = {y_9} - \sin x
⇒y6=y2=y10=−cos⁡x\Rightarrow {y_6} = {y_2} = {y_{10}} = - \cos x
⇒y7=y3=y11=sin⁡x\Rightarrow {y_7} = {y_3} = {y_{11}} = \sin x
⇒y8=y4=y12=cos⁡x\Rightarrow {y_8} = {y_4} = {y_{12}} = \cos x
Now substitute these values in the given determinant we have,

{\cos x}&{ - \sin x}&{ - \cos x} \\\ {\sin x}&{\cos x}&{ - \sin x} \\\ { - \cos x}&{\sin x}&{\cos x} \end{array}} \right|$$ Now expand this determinant we have, $$ \Rightarrow \cos x\left| {\begin{array}{*{20}{c}} {\cos x}&{ - \sin x} \\\ {\sin x}&{\cos x} \end{array}} \right| - \left( { - \sin x} \right)\left| {\begin{array}{*{20}{c}} {\sin x}&{ - \sin x} \\\ { - \cos x}&{\cos x} \end{array}} \right| - \cos x\left| {\begin{array}{*{20}{c}} {\sin x}&{\cos x} \\\ { - \cos x}&{\sin x} \end{array}} \right|$$ Now expand the mini determinant we have, $$ \Rightarrow \cos x\left( {{{\cos }^2}x - \left( { - {{\sin }^2}x} \right)} \right) - \left( { - \sin x} \right)\left( {\sin x\cos x - \left( { - \sin x} \right)\left( { - \cos x} \right)} \right) - \cos x\left( {{{\sin }^2}x - \left( { - {{\cos }^2}x} \right)} \right)$$ Now simplify it we have, $$ \Rightarrow \cos x\left( {{{\cos }^2}x + {{\sin }^2}x} \right) + \sin x\left( {\sin x\cos x - \sin x\cos x} \right) - \cos x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)$$ $$ \Rightarrow \cos x\left( {{{\cos }^2}x + {{\sin }^2}x} \right) + 0 - \cos x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)$$ $$ \Rightarrow \cos x\left( {{{\cos }^2}x + {{\sin }^2}x} \right) - \cos x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)$$ $$ \Rightarrow 0$$ **Hence option (a) is the correct answer.** **Note** : Whenever we face such types of questions the key concept we have to remember is that always recall the basic differentiation of sin x as well as cos x which is stated above, and always recall how to expand the determinant so expand it as above and simplify we will get the required answer.