Question

If $\sin \theta +\cos \theta =m$, then prove that ${{\sin }^{6}}\theta +{{\cos }^{6}}\theta =\dfrac{1}{4}\left( 4-3{{\left( {{m}^{2}}-1 \right)}^{2}} \right)$

Answer 1

Hint: Square both sides of the equation $\left( \sin \theta +\cos \theta \right)=m$ and expand using the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$.Use identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Hence find the value of $\sin \theta \cos \theta $.
Factorise ${{\sin }^{6}}\theta +{{\cos }^{6}}\theta $ using the identity ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$. Finally, use ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$ and hence prove that ${{\sin }^{6}}\theta +{{\cos }^{6}}\theta =\dfrac{4-3{{\left( {{m}^{2}}-1 \right)}^{2}}}{4}$

Complete step-by-step answer:
We have $\sin \theta +\cos \theta =m$
Squaring both sides of the equation, we get
${{\left( \sin \theta +\cos \theta \right)}^{2}}={{m}^{2}}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Using the above algebraic identity, we get
${{\sin }^{2}}\theta +2\sin \theta \cos \theta +{{\cos }^{2}}\theta ={{m}^{2}}$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Using the above identity, we get
$1+2\sin \theta \cos \theta ={{m}^{2}}$
Subtracting 1 from both sides, we get
$2\sin \theta \cos \theta ={{m}^{2}}-1$
Now, we have
${{\sin }^{6}}\theta +{{\cos }^{6}}\theta ={{\left( {{\sin }^{2}}\theta \right)}^{3}}+{{\left( {{\cos }^{2}}\theta \right)}^{3}}$
We know that ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$
Using the above algebraic identity, we get
${{\sin }^{6}}\theta +{{\cos }^{6}}\theta =\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)\left( {{\sin }^{4}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta +{{\cos }^{4}}\theta \right)$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,$ we get
${{\sin }^{6}}\theta +{{\cos }^{6}}\theta ={{\sin }^{4}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta +{{\cos }^{4}}\theta $
Adding and subtracting $2{{\sin }^{2}}\theta {{\cos }^{2}}\theta ,$ we get
${{\sin }^{6}}\theta +{{\cos }^{6}}\theta ={{\sin }^{4}}\theta +2{{\sin }^{2}}\theta {{\cos }^{2}}\theta +{{\cos }^{4}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta -2{{\sin }^{2}}\theta {{\cos }^{2}}\theta $
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}},$ we get
${{\sin }^{6}}\theta +{{\cos }^{6}}\theta ={{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}^{2}}-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta $
Using ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,$ we get
${{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-3{{\left( \sin \theta \cos \theta \right)}^{2}}$
Substituting the value of $\sin \theta \cos \theta ,$ we get
${{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-3{{\left( \dfrac{{{m}^{2}}-1}{2} \right)}^{2}}$
Hence, we have
${{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-3\dfrac{{{\left( {{m}^{2}}-1 \right)}^{2}}}{4}=\dfrac{4-3{{\left( {{m}^{2}}-1 \right)}^{2}}}{4}$
Hence proved.

Note: Alternative solution 1:
We know from the binomial theorem that ${{\sin }^{6}}x={{\left( \dfrac{{{e}^{ix}}-{{e}^{-ix}}}{2i} \right)}^{6}}=\dfrac{-1}{{{2}^{6}}}\left( {{e}^{i6x}}-6{{e}^{i4x}}+15{{e}^{i2x}}-20+15{{e}^{-i2x}}-6{{e}^{-i4x}}+{{e}^{-i6x}} \right)$
Similarly ${{\cos }^{6}}x=\dfrac{1}{{{2}^{6}}}\left( {{e}^{i6x}}+6{{e}^{i4x}}+15{{e}^{i2x}}+20+15{{e}^{-i2x}}+6{{e}^{-i4x}}+{{e}^{-i6x}} \right)$
Hence, we have
${{\cos }^{6}}x+{{\sin }^{6}}x=\dfrac{1}{{{2}^{6}}}\left( 12{{e}^{i4x}}+40+12{{e}^{-i4x}} \right)=\dfrac{1}{16}\left( 3\left( 2\cos 4x \right)+10 \right)=\dfrac{1}{8}\left( 3\cos 4x+5 \right)$
Now, we have $\cos 4x=1-2{{\sin }^{2}}2x=1-2{{\left( {{m}^{2}}-1 \right)}^{2}}$
Hence, we have ${{\cos }^{6}}x+{{\sin }^{6}}x=\dfrac{1}{8}\left( 3\left( 1-2{{\left( {{m}^{2}}-1 \right)}^{2}} \right)+5 \right)=\dfrac{1}{8}\left( 8-6{{\left( {{m}^{2}}-1 \right)}^{2}} \right)=\dfrac{1}{4}\left( 4-3{{\left( {{m}^{2}}-1 \right)}^{2}} \right)$
Alternative solution:
Use the fact that $\sin \theta +\cos \theta =m,\sin \theta \cos \theta =\dfrac{{{m}^{2}}-1}{2}$ to prove that $\sin \theta $ and $\cos \theta $ are the roots of the equation ${{x}^{2}}-mx+\dfrac{{{m}^{2}}-1}{2}=0$.
Hence use the recurrence relation ${{P}_{n}}-m{{P}_{n-1}}+\dfrac{{{m}^{2}}-1}{2}{{P}_{n-2}}=0$, where ${{P}_{n}}={{\sin }^{n}}\theta +{{\cos }^{n}}\theta ,n\ge 3$ to prove the above relation.
Start by substituting n = 3 to n = 6 and hence find the value of ${{P}_{6}}={{\sin }^{6}}\theta +{{\cos }^{6}}\theta $