Question

prove the following. que no. 4

Answer 1

Hey there!

Given trigonometric identity to prove :

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

Manipulating the left hand side to equal that of right hand side.

Apply the equation of factorial rule of two given set variables in cubes of "a" and "b" that is,

$\boxed{\bf{a^3 + b^3 = (a+ b)^3 - 3 ab \times (a + b)}}$

Now, take out the cubes outside the bracket to obtain "sin^2" and "cos^2" respectively to apply the principle of factorial rule of two cubical set variables:

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

Now apply the principles of factorial rule of two variables.

Here, a = "sin^2(theta)"  and   b = "cos^2(theta)", Just substitute the values into the factorial rule:

$\bf{(sin^2(\theta) + cos^2(\theta))^3 - 3 (sin^2(\theta) cos^2(\theta)) (sin^2(\theta) + cos^2(\theta))}$

Now, apply the basic pythagorean ruling trigonometric identity of "sin^2(theta) + cos^2(theta)" that is;

$\boxed{\bf{sin^2(\theta) + cos^2(\theta) = 1}}$

OR,

$\boxed{\bf{cos^2(\theta) + sin^2(\theta) = 1}}$    And vice versa.

Therefore,

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

$\boxed{\bf{\underline{\therefore \quad 1 - 3 sin^2(\theta) cos^2(\theta)}}}$

$\boxed{\large{\bf{\bigstar \: \: HENCE \: PROVED \: \: \bigstar}}}$

Hope this helps you and solves your doubts for proving two given trigonometric identities!!!!!