# Grade 11 General Trigonometry questions, source [Mind Action Series Mathematics Grade 11 New Edition]

Given that $(\sin \theta + \cos \theta)² = p$,  express the following in terms of $p$:

1. $\sin \theta - \cos \theta$
2. $\sin^3 \theta + \cos^ 3 \theta$

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Lets write
$p=(\sin \theta +\cos \theta)^2=\sin^2 \theta +\cos^2 \theta +2\sin \theta \cos \theta=1+2\sin \theta \cos \theta.$
So
$2\sin \theta \cos \theta=p-1. (1)$

1. We have
$(\sin \theta -\cos \theta)^2=\sin^2 \theta +\cos^2 \theta +2\sin \theta \cos \theta$
$=1-2\sin \theta \cos \theta=1-(p-1)=2-p.$
Thus
$\sin \theta -\cos \theta = \pm \sqrt{2-p}.$

2. We have
$(\sin \theta +\cos \theta)^3=\sin^3 \theta +\cos^3\theta+ 3\sin^2 \theta \cos \theta+3\sin\theta \cos^2 \theta$
$=\sin^3 \theta +\cos^3\theta+ 3\sin \theta \cos \theta(\sin \theta +\cos \theta ).$
Hence
$\sin^3 \theta +\cos^3\theta =(\sin \theta +\cos \theta)^3-3\sin \theta \cos \theta(\sin \theta +\cos \theta )$
$=(\sin \theta +\cos \theta) [(\sin \theta +\cos \theta)^2-3\sin \theta \cos \theta]$
$=(\sin \theta +\cos \theta) [(\sin \theta +\cos \theta)^2-\frac{3}{2} (2\sin \theta \cos \theta)]$
$=\pm \sqrt{p}(p-\frac{3}{2}(p-1))=\pm \sqrt{p}(-\frac{1}{2}p+\frac{3}{2})).$
Thus
$\sin^3 \theta +\cos^3\theta=\pm \sqrt{p}(-\frac{1}{2}p+\frac{3}{2}).$

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