COMEDK 2023 Shift 1 Paper

To find the correct expression for cos6A−sin‌6A in terms of cos‌2‌A and sin‌2A, we can use a factorization method or algebraic identities.
Firstly, we can recognize that cos6A−sin‌6A can be factorized using the difference of cubes formula:
a3−b3=(a−b)(a2+ab+b2)
Here, let a=cos2A and b=sin‌2A. Then,
We know that:
cos2A−sin‌2A=cos‌2‌A
Now, simplify (cos2A+sin‌2A)2=12=1 :
(cos2A)2+2cos2A‌sin‌2A+(sin‌2A)2=1
So, cos2Asin‌2A=‌

1

2

sin‌22A (since sin‌2A=2sin‌A‌cos‌A, hence sin‌22A=4sin‌2Acos2A ).
The expression becomes:
‌cos‌2‌A((cos2A)2+cos2A‌sin‌2A+(sin‌2A)2)=cos‌2‌A(1−cos2A‌sin‌2A)
From cos2A‌sin‌2A=‌

1

4

‌sin‌22A
‌cos‌2‌A(1−‌

1

4

sin‌22A)
So, the correct expression that matches the options provided is Option A: cos‌2‌A(1−‌

1

4

‌sin‌22A)