Question

True or false sin6A+cos6A=1-3sin²A.cos²A

Answer 1

Hey there!

We're given this trigonometric equation, to prove either of the sides:

$\bf{sin^6(A) + cos^6(A) = 1 - 3 sin^2(A) \times cos^2(A)}$

For this query, you'll need to bracket out the exponential forms to form a common exponent rule of multiplication inside and outside the bracket. Here, squaring will be taken inside and the cubes to be transferred outside the Bracket. This allows us to apply the cubic equation for variables. That is:

$\bf{sin^6 (A) + cos^6 (A)}$

$\bf{\Big(sin^2 (A) \Big)^3 + \Big(cos^2 (A) \Big)^3}$

$\bf{\Big(sin^2 (A) + cos^2 (A) \Big)^3}$

Use this equation rule for variables cubed into a bracketing raised to a power of cube or a cubic equation, that'd be like this:

$\boxed{\bf{Cubic \: \: Equation: \: (a + b)^3 = a^3 + b^3 + 3 \times a \times b \times (a + b)}}$

Here, a = sin^2(A) and b = cos^2(A).

$\bf{\therefore \quad \Big(sin^2 (A) + cos^2 (A) \Big)^3 - 3 \times sin^2 (A)cos^2 (A) \times \Big(sin^2(A) + cos^2 (A) \Big)}$

Now, just apply the commonly known identity rule for sin and cos to be equalled with "1" (As they're squared), that is

$\bf{Identity \: \: Rule: \: sin^2 (A) + cos^2 (A) = 1}$

So:

$\bf{(1)^3 - 3 \times sin^2 (A) cos^2 (A) \times (1)}$

$\boxed{\bf{\underline{\therefore Hence \: \: Proved: \quad 1 - 3 \: sin^2 (A) cos^2(A)}}}$

Which is the required solution or the final proof for these types of queries.

Hope it helps you and clears your doubts for proving the two given trigonometric identities!!!!