Question

Prove that ${{\sin }^{6}}\theta +{{\cos }^{6}}\theta +3{{\sin }^{2}}\theta {{\cos }^{2}}\theta =1$

Answer 1

Hint: Now to prove the given equation we will first write ${{\sin }^{6}}\theta ={{\left( {{\sin }^{2}}\theta \right)}^{3}}$ by using the law of indices ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$. Now we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ using this we will write the LHS of the equation in the form of ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( {{a}^{2}}+{{b}^{2}} \right)$ . Now again we will use the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to find the value of ${{\left( a+b \right)}^{3}}$ . Hence we will get the required equation.

Complete step by step answer:
Now consider the given equation ${{\sin }^{6}}\theta +{{\cos }^{6}}\theta +3{{\sin }^{2}}\theta {{\cos }^{2}}\theta =1$
To prove the given equation we will first consider the LHS of the equation.
Now we know by the law of indices that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ Hence using this we can write the LHS of the equation as,
${{\left( {{\sin }^{2}}\theta \right)}^{3}}+{{\left( {{\cos }^{2}}\theta \right)}^{3}}+3\left( {{\sin }^{2}}\theta \right)\left( {{\cos }^{2}}\theta \right)$
The LHS of the equation is now in the form ${{a}^{3}}+{{b}^{3}}+3ab$ where a is ${{\sin }^{2}}\theta $ and b is ${{\cos }^{2}}\theta $
Now we want to write the above expression as ${{\left( a+b \right)}^{3}}$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ Hence we can write the above expression ${{\left( {{\sin }^{2}}\theta \right)}^{3}}+{{\left( {{\cos }^{2}}\theta \right)}^{3}}+3\left( {{\sin }^{2}}\theta \right)\left( {{\cos }^{2}}\theta \right)$ as ${{\left( {{\sin }^{2}}\theta \right)}^{3}}+{{\left( {{\cos }^{2}}\theta \right)}^{3}}+3\left( {{\sin }^{2}}\theta \right)\left( {{\cos }^{2}}\theta \right)\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)$
Now we know that ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$
Hence we get LHS of the given equation as,
$\Rightarrow {{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}^{3}}$
Now again using the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ we get,
$\Rightarrow {{\left( {{\sin }^{2}}\theta \right)}^{3}}+{{\left( {{\cos }^{2}}\theta \right)}^{3}}+3\left( {{\sin }^{2}}\theta \right)\left( {{\cos }^{2}}\theta \right)={{1}^{3}}$
Now we also know that ${{1}^{3}}=1$ . Hence we get the LHS of the equation as 1.
Hence we have ${{\sin }^{6}}\theta +{{\cos }^{6}}\theta +3{{\sin }^{2}}\theta {{\cos }^{2}}\theta =1$
Hence the given equation is proved.

Note: Now note that the trigonometric functions $\cos \theta $ and $\sin \theta $ gives us ratios of sides of the triangles in right angle triangle where $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$. Now we can easily obtained the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ by using Pythagoras theorem in the triangle.