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

## 1 Answer

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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