Question

Proving more Identities

Verify the identity${\mathbf{sin}}^{\mathbf{6}}\mathbf{\beta }\mathbf{+}{\mathbf{cos}}^{\mathbf{6}}\mathbf{\beta }\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{3}{\mathbf{sin}}^{\mathbf{2}}{\mathbf{\beta cos}}^{\mathbf{2}}\mathbf{\beta }$

Short answer

The expression${\mathbf{sin}}^{\mathbf{6}}\mathbf{\beta }\mathbf{+}{\mathbf{cos}}^{\mathbf{6}}\mathbf{\beta }\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{3}{\mathbf{sin}}^{\mathbf{2}}{\mathbf{\beta cos}}^{\mathbf{2}}\mathbf{\beta }$ is an identity.

Step-by-step solution

Step 1. Given information

An expression${\sin }^6\beta +{\cos }^6\beta =1-3{\sin }^2\beta {\cos }^2\beta$

Step 2. Concept used

There is a written expression in doubt. Divide it into two halves, such as LHS and RHS. Separately, simplify them. It's an identity if both results are the same.

Step 3. Calculation

Now, simplify LHS of ${\sin }^6\beta +{\cos }^6\beta =1-3{\sin }^2\beta {\cos }^2\beta$

$\begin{array}{l}\because a^3+b^3={(a+b)}^3-3ab(a+b)\\ =\left({\sin }^2\beta +{\cos }^2\beta \right)-3{\sin }^2\beta {\cos }^2\beta \cdot \left({\sin }^2\beta +{\cos }^2\beta \right)\\ \because \left({\sin }^2\beta +{\cos }^2\beta =1\right)\\ =1-3{\sin }^2\beta {\cos }^2\beta \end{array}$

RHS of the equation is$1-3{\sin }^2\beta {\cos }^2\beta$

Both RHS and LHS are equal. Hence it is an identity