Answer
Verified
479.7k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE