# If \[\sin \theta +{{\sin }^{2}}\theta =1\], find the value of

\[{{\cos }^{12}}\theta +3{{\cos }^{10}}\theta +3{{\cos }^{8}}\theta +{{\cos }^{6}}\theta +2{{\cos }^{4}}\theta +2{{\cos }^{2}}\theta -2\]

Answer

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Hint: First of all, use the formula \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] in the given equation to get \[\sin \theta ={{\cos }^{2}}\theta \]. Now use \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\] in the given expression to express it in terms of \[\left( \sin \theta +{{\sin }^{2}}\theta \right)\] whose value is 1.

Complete step-by-step answer:

Here, we are given that \[\sin \theta +{{\sin }^{2}}\theta =1\] and we have to find the value of \[{{\cos }^{12}}\theta +3{{\cos }^{10}}\theta +3{{\cos }^{8}}\theta +{{\cos }^{6}}\theta +2{{\cos }^{4}}\theta +2{{\cos }^{2}}\theta -2\]

First of all let us take the given equation, that is, \[\sin \theta +{{\sin }^{2}}\theta =1\]

We know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Or, \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]

By putting the value of \[{{\sin }^{2}}\theta \] in the given equation, we get,

\[\sin \theta +1-{{\cos }^{2}}\theta =1\]

By cancelling 1 from both sides, we get,

\[\sin \theta -{{\cos }^{2}}\theta =0\]

Or, \[\sin \theta ={{\cos }^{2}}\theta ....\left( i \right)\]

Now, let us consider the expression whose value is to be found as,

\[A={{\cos }^{12}}\theta +3{{\cos }^{10}}\theta +3{{\cos }^{8}}\theta +{{\cos }^{6}}\theta +2{{\cos }^{4}}\theta +2{{\cos }^{2}}\theta -2\]

We can also write the above expression as,

\[A={{\left( {{\cos }^{4}}\theta \right)}^{3}}+3{{\cos }^{6}}\theta \left( {{\cos }^{4}}\theta +{{\cos }^{2}}\theta \right)+{{\left( {{\cos }^{2}}\theta \right)}^{3}}+2\left( {{\cos }^{4}}\theta -{{\cos }^{2}}\theta -1 \right)\]

Now by writing, \[{{\cos }^{6}}\theta ={{\cos }^{2}}\theta .{{\cos }^{4}}\theta \], we get

\[A={{\left( {{\cos }^{4}}\theta \right)}^{3}}+3{{\cos }^{2}}\theta .{{\cos }^{4}}\theta \left( {{\cos }^{4}}\theta +{{\cos }^{2}}\theta \right)+{{\left( {{\cos }^{2}}\theta \right)}^{3}}+2\left( {{\cos }^{4}}\theta -{{\cos }^{2}}\theta -1 \right)\]

Now, we know that \[{{a}^{3}}+3ab\left( a+b \right)+{{b}^{3}}={{\left( a+b \right)}^{3}}\]

By taking \[a={{\cos }^{4}}\theta \] and \[b={{\cos }^{2}}\theta \] in the above expression, we get,

\[A={{\left( {{\cos }^{4}}\theta +{{\cos }^{2}}\theta \right)}^{3}}+2\left( {{\cos }^{4}}\theta +{{\cos }^{2}}\theta -1 \right)\]

By writing \[{{\cos }^{4}}\theta ={{\left( {{\cos }^{2}}\theta \right)}^{2}}\] in the above expression, we get,

\[A={{\left[ {{\left( {{\cos }^{2}}\theta \right)}^{2}}+\left( {{\cos }^{2}}\theta \right) \right]}^{3}}+2\left[ {{\left( {{\cos }^{2}}\theta \right)}^{2}}+{{\cos }^{2}}\theta -1 \right]\]

From equation (i), we know that \[{{\cos }^{2}}\theta =\sin \theta \]. By applying it in the above expression, we get

\[A={{\left[ {{\left( \sin \theta \right)}^{2}}+\sin \theta \right]}^{3}}+2\left( {{\left( \sin \theta \right)}^{2}}+\sin \theta -1 \right)\]

As we are given that \[\sin \theta +{{\sin }^{2}}\theta =1\], therefore by applying it in the above equation, we get,

\[\begin{align}

& A={{\left[ 1 \right]}^{3}}+2\left[ 1-1 \right] \\

& A=1+2\left( 0 \right) \\

& A=1 \\

\end{align}\]

Therefore, we get the value of \[{{\cos }^{12}}\theta +3{{\cos }^{10}}\theta +3{{\cos }^{8}}\theta +{{\cos }^{6}}\theta +2{{\cos }^{4}}\theta +2{{\cos }^{2}}\theta -2=1\].

Note: Take special care of powers of each term while writing them and always cross-check after writing each equation. Whenever you get the higher powers, always try using the formulas like \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\]or \[{{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)\] according to the question. Here, while solving the given equation, some students finally write it as \[\sin \theta =\cos \theta \] or \[{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \] instead of \[\sin \theta ={{\cos }^{2}}\theta \]. So this mistake must be avoided.

Complete step-by-step answer:

Here, we are given that \[\sin \theta +{{\sin }^{2}}\theta =1\] and we have to find the value of \[{{\cos }^{12}}\theta +3{{\cos }^{10}}\theta +3{{\cos }^{8}}\theta +{{\cos }^{6}}\theta +2{{\cos }^{4}}\theta +2{{\cos }^{2}}\theta -2\]

First of all let us take the given equation, that is, \[\sin \theta +{{\sin }^{2}}\theta =1\]

We know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Or, \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]

By putting the value of \[{{\sin }^{2}}\theta \] in the given equation, we get,

\[\sin \theta +1-{{\cos }^{2}}\theta =1\]

By cancelling 1 from both sides, we get,

\[\sin \theta -{{\cos }^{2}}\theta =0\]

Or, \[\sin \theta ={{\cos }^{2}}\theta ....\left( i \right)\]

Now, let us consider the expression whose value is to be found as,

\[A={{\cos }^{12}}\theta +3{{\cos }^{10}}\theta +3{{\cos }^{8}}\theta +{{\cos }^{6}}\theta +2{{\cos }^{4}}\theta +2{{\cos }^{2}}\theta -2\]

We can also write the above expression as,

\[A={{\left( {{\cos }^{4}}\theta \right)}^{3}}+3{{\cos }^{6}}\theta \left( {{\cos }^{4}}\theta +{{\cos }^{2}}\theta \right)+{{\left( {{\cos }^{2}}\theta \right)}^{3}}+2\left( {{\cos }^{4}}\theta -{{\cos }^{2}}\theta -1 \right)\]

Now by writing, \[{{\cos }^{6}}\theta ={{\cos }^{2}}\theta .{{\cos }^{4}}\theta \], we get

\[A={{\left( {{\cos }^{4}}\theta \right)}^{3}}+3{{\cos }^{2}}\theta .{{\cos }^{4}}\theta \left( {{\cos }^{4}}\theta +{{\cos }^{2}}\theta \right)+{{\left( {{\cos }^{2}}\theta \right)}^{3}}+2\left( {{\cos }^{4}}\theta -{{\cos }^{2}}\theta -1 \right)\]

Now, we know that \[{{a}^{3}}+3ab\left( a+b \right)+{{b}^{3}}={{\left( a+b \right)}^{3}}\]

By taking \[a={{\cos }^{4}}\theta \] and \[b={{\cos }^{2}}\theta \] in the above expression, we get,

\[A={{\left( {{\cos }^{4}}\theta +{{\cos }^{2}}\theta \right)}^{3}}+2\left( {{\cos }^{4}}\theta +{{\cos }^{2}}\theta -1 \right)\]

By writing \[{{\cos }^{4}}\theta ={{\left( {{\cos }^{2}}\theta \right)}^{2}}\] in the above expression, we get,

\[A={{\left[ {{\left( {{\cos }^{2}}\theta \right)}^{2}}+\left( {{\cos }^{2}}\theta \right) \right]}^{3}}+2\left[ {{\left( {{\cos }^{2}}\theta \right)}^{2}}+{{\cos }^{2}}\theta -1 \right]\]

From equation (i), we know that \[{{\cos }^{2}}\theta =\sin \theta \]. By applying it in the above expression, we get

\[A={{\left[ {{\left( \sin \theta \right)}^{2}}+\sin \theta \right]}^{3}}+2\left( {{\left( \sin \theta \right)}^{2}}+\sin \theta -1 \right)\]

As we are given that \[\sin \theta +{{\sin }^{2}}\theta =1\], therefore by applying it in the above equation, we get,

\[\begin{align}

& A={{\left[ 1 \right]}^{3}}+2\left[ 1-1 \right] \\

& A=1+2\left( 0 \right) \\

& A=1 \\

\end{align}\]

Therefore, we get the value of \[{{\cos }^{12}}\theta +3{{\cos }^{10}}\theta +3{{\cos }^{8}}\theta +{{\cos }^{6}}\theta +2{{\cos }^{4}}\theta +2{{\cos }^{2}}\theta -2=1\].

Note: Take special care of powers of each term while writing them and always cross-check after writing each equation. Whenever you get the higher powers, always try using the formulas like \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\]or \[{{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)\] according to the question. Here, while solving the given equation, some students finally write it as \[\sin \theta =\cos \theta \] or \[{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \] instead of \[\sin \theta ={{\cos }^{2}}\theta \]. So this mistake must be avoided.

Last updated date: 17th Sep 2023

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