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Question: Prove that \[{\sin ^8}\theta - {\cos ^8}\theta = ({\sin ^2}\theta - {\cos ^2}\theta )(1 - 2{\sin ^2}...

Prove that sin8θcos8θ=(sin2θcos2θ)(12sin2θcos2θ){\sin ^8}\theta - {\cos ^8}\theta = ({\sin ^2}\theta - {\cos ^2}\theta )(1 - 2{\sin ^2}\theta {\cos ^2}\theta )

Explanation

Solution

Here in this question some trigonometric identities and algebraic identities will get used. They are mentioned below: -
sin2θ+cos2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
(a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab
a2+b2=(a+b)22ab{a^2} + {b^2} = {(a + b)^2} - 2ab
a4b4=(a2b2)(a2+b2){a^4} - {b^4} = ({a^2} - {b^2})({a^2} + {b^2})
a2b2=(ab)(a+b){a^2} - {b^2} = (a - b)(a + b)

Complete step-by-step answer:
We will convert the L.H.S (left hand side) part equal to R.H.S (right hand side) using some trigonometric identities and algebraic identities.
First of all we will split the power 8 like this a8b8=(a4)2(b4)2{a^8} - {b^8} = {({a^4})^2} - {({b^4})^2} so that we can apply further identity.
(sin4θ)2(cos4θ)2\Rightarrow {({\sin ^4}\theta )^2} - {({\cos ^4}\theta )^2}
Now applying identity a2b2=(ab)(a+b){a^2} - {b^2} = (a - b)(a + b) we will get
(sin4θcos4θ)(sin4θ+cos4θ)\Rightarrow ({\sin ^4}\theta - {\cos ^4}\theta )({\sin ^4}\theta + {\cos ^4}\theta )
Now we will split power 4 so that we can apply identity.
(sin4θcos4θ)[(sin2θ)2+(cos2θ)2]\Rightarrow ({\sin ^4}\theta - {\cos ^4}\theta )[{({\sin ^2}\theta )^2} + {({\cos ^2}\theta )^2}]
Now we will apply identitya2+b2=(a+b)22ab{a^2} + {b^2} = {(a + b)^2} - 2abwe will get
(sin4θcos4θ)[(sin2θ+cos2θ)22sin2θcos2θ]\Rightarrow ({\sin ^4}\theta - {\cos ^4}\theta )[{({\sin ^2}\theta + {\cos ^2}\theta )^2} - 2{\sin ^2}\theta {\cos ^2}\theta ]
Now we will apply identitya4b4=(a2b2)(a2+b2){a^4} - {b^4} = ({a^2} - {b^2})({a^2} + {b^2})we will get
(sin2θcos2θ)(sin2θ+cos2θ)[(sin2θ+cos2θ)22sin2θcos2θ]\Rightarrow ({\sin ^2}\theta - {\cos ^2}\theta )({\sin ^2}\theta + {\cos ^2}\theta )[{({\sin ^2}\theta + {\cos ^2}\theta )^2} - 2{\sin ^2}\theta {\cos ^2}\theta ]
Now we will apply trigonometric identity sin2θ+cos2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1and we will get
(sin2θcos2θ)[(sin2θ+cos2θ)22sin2θcos2θ]\Rightarrow ({\sin ^2}\theta - {\cos ^2}\theta )[{({\sin ^2}\theta + {\cos ^2}\theta )^2} - 2{\sin ^2}\theta {\cos ^2}\theta ]
Now we will apply identity sin2θ+cos2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1again and we will get
(sin2θcos2θ)[(1)22sin2θcos2θ]\Rightarrow ({\sin ^2}\theta - {\cos ^2}\theta )[{(1)^2} - 2{\sin ^2}\theta {\cos ^2}\theta ]
(sin2θcos2θ)[12sin2θcos2θ]\therefore ({\sin ^2}\theta - {\cos ^2}\theta )[1 - 2{\sin ^2}\theta {\cos ^2}\theta ]
Hence it is proved that sin8θcos8θ=(sin2θcos2θ)(12sin2θcos2θ){\sin ^8}\theta - {\cos ^8}\theta = ({\sin ^2}\theta - {\cos ^2}\theta )(1 - 2{\sin ^2}\theta {\cos ^2}\theta )

Note: Students are likely to make mistakes in relating algebraic identities with trigonometric functions but they can be applied easily by only carefully observing the degrees. With the use of algebraic identities questions can be solved with very less complexity. Further this question can be solved alternatively by proofing the R.H.S (right hand side) part equal to L.H.S (left hand side). Same steps will take place but in reverse. In the previous case we eliminate the term as if to simplify the equation. Now in an alternate method we will just add more terms so that we can make power 8. That method will be more complex than the method we have used so the mentioned method must be followed to avoid any discrepancies.