Need help on a math exam question (1 Viewer)

Dashdorm24 · Jun 24, 2021

Screen Shot 2021-06-24 at 11.40.11 am.png

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 · Jun 24, 2021

Use the fact that $\sin^2 \theta + \cos^2 \theta = 1$ . It might also be helpful to recognise that $\sin 2 \theta = 2 \sin \theta \cos \theta$ .

CM_Tutor · Jun 27, 2021

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

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

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

So,

$\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 $-\cfrac{1}{2}\sin{2\theta}$

Need help on a math exam question (1 Viewer)

Dashdorm24

New Member

quickoats

Well-Known Member

CM_Tutor

Moderator

Users Who Are Viewing This Thread (Users: 0, Guests: 1)