5.3 Identities
Note 5.9. Two identities are worth memorising, because the rest follow from them: \[\sin ^2\theta +\cos ^2\theta =1\] \[\sin (A+B)=\sin A\cos B+\cos A\sin B.\]
Dividing the first by \(\cos ^2\theta \) gives \(\tan ^2\theta +1=\sec ^2\theta \), and dividing it by \(\sin ^2\theta \) gives \(1+\cot ^2\theta =\csc ^2\theta \). These three are the Pythagorean identities.
5.3.1 Compound and double angle formulae
\[\sin (A\pm B)=\sin A\cos B\pm \cos A\sin B\] \[\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B\] \[\tan (A\pm B)=\frac {\tan A\pm \tan B}{1\mp \tan A\tan B}\]
Note 5.10. The signs in the cosine formula are opposite to the sign on the left: \(\cos (A+B)\) has a minus in the middle. This is the single most-forgotten detail in the topic.
Putting \(B=A\) gives the double angle formulae: \[\sin 2A=2\sin A\cos A,\qquad \cos 2A=\cos ^2A-\sin ^2A=1-2\sin ^2A=2\cos ^2A-1,\qquad \tan 2A=\frac {2\tan A}{1-\tan ^2A}.\]
Solution. Write \(75^\circ \) as a sum of two angles whose ratios are known exactly: \(75^\circ =45^\circ +30^\circ \). Then \begin {align*} \sin 75^\circ &=\sin (45^\circ +30^\circ )\\ &=\sin 45^\circ \cos 30^\circ +\cos 45^\circ \sin 30^\circ \\ &=\left (\frac {\sqrt {2}}{2}\right )\left (\frac {\sqrt {3}}{2}\right ) +\left (\frac {\sqrt {2}}{2}\right )\left (\frac {1}{2}\right )\\ &=\frac {\sqrt {6}}{4}+\frac {\sqrt {2}}{4}\\ &=\frac {\sqrt {6}+\sqrt {2}}{4}. \end {align*}
Check. \(\frac {\sqrt 6+\sqrt 2}{4}\approx \frac {2.449+1.414}{4}\approx 0.966\), and \(\sin 75^\circ \approx 0.966\). A decimal check like this takes seconds and catches a wrong sign or a swapped ratio immediately.
Solution. Work on one side only — here the left — and turn everything into sines and cosines, which is almost always the way in. \begin {align*} \cot \theta +\tan \theta &=\frac {\cos \theta }{\sin \theta }+\frac {\sin \theta }{\cos \theta }\\ &=\frac {\cos ^2\theta +\sin ^2\theta }{\sin \theta \cos \theta } &&\text {(common denominator)}\\ &=\frac {1}{\sin \theta \cos \theta } &&\text {(Pythagorean identity)}\\ &=\frac {1}{\sin \theta }\cdot \frac {1}{\cos \theta }\\ &=\csc \theta \sec \theta . \end {align*}
which is the right-hand side, as required.
Note 5.13. To prove an identity, start from one side and work to the other. Do not move terms across the \(\equiv \) sign as though solving an equation — that assumes the very thing being proved.
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