
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
515.1k+ 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
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

State the laws of reflection of light

Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
