Question
Question: Prove that \({{\sin }^{6}}\theta +{{\cos }^{6}}\theta +3{{\sin }^{2}}\theta {{\cos }^{2}}\theta =1\)...
Prove that sin6θ+cos6θ+3sin2θcos2θ=1
Solution
Now to prove the given equation we will first write sin6θ=(sin2θ)3 by using the law of indices (am)n=amn. Now we know that sin2θ+cos2θ=1 using this we will write the LHS of the equation in the form of (a+b)3=a3+b3+3ab(a2+b2) . Now again we will use the identity sin2θ+cos2θ=1 to find the value of (a+b)3 . Hence we will get the required equation.
Complete step by step answer:
Now consider the given equation sin6θ+cos6θ+3sin2θcos2θ=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 Hence using this we can write the LHS of the equation as,
(sin2θ)3+(cos2θ)3+3(sin2θ)(cos2θ)
The LHS of the equation is now in the form a3+b3+3ab where a is sin2θ and b is cos2θ
Now we want to write the above expression as (a+b)3
We know that sin2θ+cos2θ=1 Hence we can write the above expression (sin2θ)3+(cos2θ)3+3(sin2θ)(cos2θ) as (sin2θ)3+(cos2θ)3+3(sin2θ)(cos2θ)(sin2θ+cos2θ)
Now we know that (a+b)3=a3+b3+3ab(a+b)
Hence we get LHS of the given equation as,
⇒(sin2θ+cos2θ)3
Now again using the identity sin2θ+cos2θ=1 we get,
⇒(sin2θ)3+(cos2θ)3+3(sin2θ)(cos2θ)=13
Now we also know that 13=1 . Hence we get the LHS of the equation as 1.
Hence we have sin6θ+cos6θ+3sin2θcos2θ=1
Hence the given equation is proved.
Note: Now note that the trigonometric functions cosθ and sinθ gives us ratios of sides of the triangles in right angle triangle where sinθ=hypotenuseopposite side and cosθ=hypotenuseadjacent side. Now we can easily obtained the identity sin2θ+cos2θ=1 by using Pythagoras theorem in the triangle.