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Question: Prove that \({{\sin }^{6}}\theta +{{\cos }^{6}}\theta +3{{\sin }^{2}}\theta {{\cos }^{2}}\theta =1\)...

Prove that sin6θ+cos6θ+3sin2θcos2θ=1{{\sin }^{6}}\theta +{{\cos }^{6}}\theta +3{{\sin }^{2}}\theta {{\cos }^{2}}\theta =1

Explanation

Solution

Now to prove the given equation we will first write sin6θ=(sin2θ)3{{\sin }^{6}}\theta ={{\left( {{\sin }^{2}}\theta \right)}^{3}} by using the law of indices (am)n=amn{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}. Now we know that sin2θ+cos2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 using this we will write the LHS of the equation in the form of (a+b)3=a3+b3+3ab(a2+b2){{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( {{a}^{2}}+{{b}^{2}} \right) . Now again we will use the identity sin2θ+cos2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 to find the value of (a+b)3{{\left( a+b \right)}^{3}} . Hence we will get the required equation.

Complete step by step answer:
Now consider the given equation sin6θ+cos6θ+3sin2θcos2θ=1{{\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 (am)n=am×n{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} Hence using this we can write the LHS of the equation as,
(sin2θ)3+(cos2θ)3+3(sin2θ)(cos2θ){{\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 a3+b3+3ab{{a}^{3}}+{{b}^{3}}+3ab where a is sin2θ{{\sin }^{2}}\theta and b is cos2θ{{\cos }^{2}}\theta
Now we want to write the above expression as (a+b)3{{\left( a+b \right)}^{3}}
We know that sin2θ+cos2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 Hence we can write the above expression (sin2θ)3+(cos2θ)3+3(sin2θ)(cos2θ){{\left( {{\sin }^{2}}\theta \right)}^{3}}+{{\left( {{\cos }^{2}}\theta \right)}^{3}}+3\left( {{\sin }^{2}}\theta \right)\left( {{\cos }^{2}}\theta \right) as (sin2θ)3+(cos2θ)3+3(sin2θ)(cos2θ)(sin2θ+cos2θ){{\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 (a+b)3=a3+b3+3ab(a+b){{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)
Hence we get LHS of the given equation as,
(sin2θ+cos2θ)3\Rightarrow {{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}^{3}}
Now again using the identity sin2θ+cos2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 we get,
(sin2θ)3+(cos2θ)3+3(sin2θ)(cos2θ)=13\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 13=1{{1}^{3}}=1 . Hence we get the LHS of the equation as 1.
Hence we have sin6θ+cos6θ+3sin2θcos2θ=1{{\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 cosθ\cos \theta and sinθ\sin \theta gives us ratios of sides of the triangles in right angle triangle where sinθ=opposite sidehypotenuse\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}} and cosθ=adjacent sidehypotenuse\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}. Now we can easily obtained the identity sin2θ+cos2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 by using Pythagoras theorem in the triangle.