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

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

Last updated date: 21st Mar 2023

•

Total views: 306.3k

•

Views today: 7.85k

Answer

Verified

306.3k+ views

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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE