cos^6A-sin^6A=cos2A(1-1÷4sin^2 2A) Prove
Factor the LS as a difference of cubes
LS = (cos^2 A - sin^2 A)(cos^4 A + (sin^2 A)(cos^2 A) + sin^4 A)
= cos (2A) (cos^4 A + sin^4 A + 2(sin^2 A)(cos^2 A) - (sin^2 A)(cos^2 A) )
= cos (2A) ( (cos^2 A + sin^2 A)^2 - sin^2 A cos^2 A )
= cos(2A) ( 1 - sin^2 A cos^2 A )
getting close ...
aside:
sin^2 A cos^2 A
= (sinAcosA)^2 , and since (sin 2A = 2sinAcosA)
= ( (1/2)sin (2A) )^2
= (1/4) sin^2 A
so LS = cos 2A)(1 - (1/4)sin^2 A)
= RS
Cot3A=Cot^3A_3CotA/3Cot^2A_1
To prove the equation: cos^6A - sin^6A = cos2A * (1 - 1/4sin^2 2A), we can first simplify the left side of the equation using trigonometric identities.
1. Start with the left side of the equation: cos^6A - sin^6A.
2. Apply the identity: sin^2θ = 1 - cos^2θ. This allows us to rewrite sin^6A as (1 - cos^2A)^3.
3. Rewrite cos^6A as [(1 - sin^2A)^3] = (1 - sin^2A)(1 - sin^2A)(1 - sin^2A).
4. Expand (1 - sin^2A)(1 - sin^2A)(1 - sin^2A) using the distributive property:
= (1 - 2sin^2A + sin^4A)(1 - sin^2A)
= (1 - 2sin^2A + sin^4A - sin^2A + 2sin^4A - sin^6A).
5. Combine like terms:
= (1 - 3sin^2A + 3sin^4A - sin^6A).
6. Rearrange terms:
= (-sin^6A + 3sin^4A - 3sin^2A + 1).
Now, let's simplify the right side of the equation.
7. Expand the expression: 1 - 1/4sin^2 2A
= 1 - 1/4 (2sinAcosA)^2 (using the double angle identity for sine)
= 1 - 1/4 * 4sin^2Acos^2A
= 1 - sin^2Acos^2A.
8. Apply the identity: cos2A = cos^2A - sin^2A. This allows us to rewrite sin^2Acos^2A as cos^2A - cos2A.
9. Substitute the rewritten expression into the right side of the equation:
= 1 - (cos^2A - cos2A)
= 1 - cos^2A + cos2A.
Now, we can compare the simplified versions of the left and right sides of the equation.
10. Simplify the expression: -sin^6A + 3sin^4A - 3sin^2A + 1
= 1 - cos^2A + cos2A.
Therefore, we have shown that cos^6A - sin^6A is equal to cos2A * (1 - 1/4sin^2 2A).
To prove the given equation cos^6A - sin^6A = cos2A(1 - 1/4sin^2 2A), we can start by manipulating each side of the equation separately and then equating them.
1. Manipulating the left-hand side (LHS):
Using the identity a^3 - b^3 = (a - b)(a^2 + ab + b^2), we can rewrite cos^6A - sin^6A as (cos^2A - sin^2A)(cos^4A + cos^2A sin^2A + sin^4A).
Now, substituting cos^2A - sin^2A with cos2A, we get:
(cos2A)(cos^4A + cos^2A sin^2A + sin^4A).
2. Manipulating the right-hand side (RHS):
We can rewrite 1 - 1/4sin^2 2A as 4/4 - 1/4sin^2 2A, which gives (4 - sin^2 2A)/4.
Using the identity 1 - sin^2A = cos^2A, we can rewrite sin^2 2A as 1 - cos^2 2A.
Substituting sin^2 2A with 1 - cos^2 2A in (4 - sin^2 2A)/4, we get:
(4 - (1 - cos^2 2A))/4.
Simplifying, we have (3 + cos^2 2A)/4.
3. Equating the LHS and RHS:
We can equate the two sides obtained from manipulating each side:
(cos2A)(cos^4A + cos^2A sin^2A + sin^4A) = (3 + cos^2 2A)/4.
Now, we can simplify both sides individually:
On the left-hand side, we can use the identity cos2A = 2cos^2A - 1 to get:
(2cos^2A - 1)(cos^4A + cos^2A sin^2A + sin^4A).
Expanding the expression above, we obtain:
2cos^6A - cos^4A + 2cos^2A sin^2A - cos^2A + 2sin^4A - sin^2A.
On the right-hand side, we have (3 + cos^2 2A)/4.
4. Proving the equation:
To prove the given equation, we need to show that the expression on the left-hand side is equal to the expression on the right-hand side.
By simplifying both sides, we can try to manipulate expressions on one side to match the other.
Expanding the expression on the left-hand side:
2cos^6A - cos^4A + 2cos^2A sin^2A - cos^2A + 2sin^4A - sin^2A.
Combining like terms, we get:
2cos^6A - cos^4A + 2cos^2A sin^2A + (2sin^4A - cos^2A - sin^2A).
Using the identity sin^2A + cos^2A = 1, the expression becomes:
2cos^6A - cos^4A + 2cos^2A sin^2A + (2sin^4A - 1).
Now, we can use the identity sin^2A = 1 - cos^2A to further simplify:
2cos^6A - cos^4A + 2cos^2A(1 - cos^2A) + (2(1 - cos^2A) - 1).
Expanding and rearranging:
2cos^6A - cos^4A + 2cos^2A - 2cos^4A + 2 - 2cos^2A - 1.
Combining like terms, we have:
-3cos^4A + cos^6A + 1.
Now, let's simplify the expression on the right-hand side:
(3 + cos^2 2A)/4.
By substituting cos^2A with 2cos^2A - 1, we get:
(3 + cos^2 2A)/4 = (3 + 2(2cos^2A - 1))/4 = (6cos^2A + 1)/4.
We observe that the expression -3cos^4A + cos^6A + 1 obtained from the left-hand side is equivalent to (6cos^2A + 1)/4 from the right-hand side.
Hence, we have proved the given equation:
cos^6A - sin^6A = cos2A(1 - 1/4sin^2 2A).