# Need help on a math exam question (1 Viewer)

#### Dashdorm24

##### New Member

I would be grateful for any ideas on this question , I know the use of the difference between cubes is used but how do you proceed from there?

#### quickoats

##### Well-Known Member
Use the fact that $\bg_white \sin^2 \theta + \cos^2 \theta = 1$. It might also be helpful to recognise that $\bg_white \sin 2 \theta = 2 \sin \theta \cos \theta$.

#### CM_Tutor

##### Moderator
Moderator
First, note the factorisation of the sum of two cubes: $\bg_white A^3 + B^3 = (A + B)(A^2 - AB + B^2)$. You can easily check this by expanding:

\bg_white \begin{align*} \text{RHS} &= (A + B)(A^2 - AB + B^2) \\ &= A(A^2 - AB + B^2) + B(A^2 - AB + B^2) \\ &= A^3 - A^2B + AB^2 + A^2B - AB^2 + B^3 \\ &= A^3 - A^2B + A^2B + AB^2 - AB^2 + B^3 \\ &= A^3 + B^3 \\ &= \text{RHS} \end{align*}

Now, taking $\bg_white A = \sin\theta$ and $\bg_white B = \cos\theta$, we get that $\bg_white \sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta)$

So,

\bg_white \begin{align*} \cfrac{\sin^3\theta + \cos^3\theta}{\sin\theta + \cos\theta} - 1 &= \cfrac{(\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta)}{\sin\theta + \cos\theta} - 1 \\ &= \sin^2\theta - \sin\theta\cos\theta + \cos^2\theta - 1 \\ &= 1 - \sin\theta\cos\theta - 1 \qquad \text{noting \sin^2\theta + \cos^2\theta = 1} \\ &= - \sin\theta\cos\theta \end{align*}

This could also be written as $\bg_white -\cfrac{1}{2}\sin{2\theta}$