### Introduction to the number e

#### Recognise the definitions and notations of the number e and the natural logarithm

• If $$\frac{{\tan \theta }}{{1 - 3{{\sec }^2}\theta }} = \frac{2}{{11}}$$, where $$\frac{{3\pi }}{2} < \theta < 2\pi$$, find the values of $$\csc \theta$$. (Leave your answers in surd form if necessary.)
• $$\begin{array}{c}\frac{{\tan \theta }}{{1 - 3{{\sec }^2}\theta }} = \frac{2}{{11}}\\ 11\tan \theta = 2 - 6{\sec ^2}\theta \\ 11\tan \theta = 2 - 6(1 + {\tan ^2}\theta )\\ 11\tan \theta = - {\rm{ }}4 - 6{\tan ^2}\theta \\ 6{\tan ^2}\theta + 11\tan \theta + 4 = 0\\ (3\tan \theta + 4)(2\tan \theta + 1) = 0\end{array}$$
$$\tan \theta = - {\rm{ }}\frac{4}{3}$$ or $$\tan \theta = - {\rm{ }}\frac{1}{2}$$

When $$\tan \theta = - {\rm{ }}\frac{4}{3}$$,
[GraphMissing-M2E04 Q52]
$$\begin{array}{c}O{P^2} = O{N^2} + N{P^2}\\ = {3^2} + {4^2}\\ = 25\end{array}$$
∴ $$OP = 5$$
∴ $$\csc \theta = \underline{\underline { - {\rm{ }}\frac{5}{4}}}$$
When $$\tan \theta = - \frac{1}{2}$$,
[GraphMissing-M2E04 Q52]
$$\begin{array}{c}O{Q^2} = O{M^2} + M{Q^2}\\ = {2^2} + {1^2}\\ = 5\end{array}$$
∴ $$OQ = \sqrt 5$$
∴ $$\csc \theta = \underline{\underline { - {\rm{ }}\sqrt 5 }}$$

#### Understand compound angle formulae and double angle formulae for the functions sine, cosine and tangent, and product-to-sum and sum-to-product formulae for the functions sine and cosine

• [codemissing]
sin(A ± B) =
cos(A ± B) =
tan(A ± B) =
sin 2A = 2 sin A cos A
cos 2A =
tan 2A =
sin2A =
• Prove that $$\frac{{1 - \csc (\theta - 2\pi ) + \tan (\theta + {\textstyle{{3\pi } \over 2}})}}{{1 + \csc (\pi - \theta ) - \tan (\theta - {\textstyle{\pi \over 2}})}} = \frac{{1 + \sec ({\textstyle{\pi \over 2}} + \theta )}}{{\cot (\pi + \theta )}}$$.
• $$\begin{array}{l}\frac{{1 - \csc (\theta - 2\pi ) + \tan (\theta + {\textstyle{{3\pi } \over 2}})}}{{1 + \csc (\pi - \theta ) - \tan (\theta - {\textstyle{\pi \over 2}})}} - \frac{{1 + \sec ({\textstyle{\pi \over 2}} + \theta )}}{{\cot (\pi + \theta )}}\\ = \frac{{1 - \csc [ - {\rm{ }}(2\pi - \theta )] + \tan ({\textstyle{{3\pi } \over 2}} + \theta )}}{{1 + \csc (\pi - \theta ) - \tan [ - {\rm{ }}({\textstyle{\pi \over 2}} - \theta )]}} - {\frac{{1 + \sec ({\textstyle{\pi \over 2}} + \theta )}}{{\cot (\pi + \theta )}}^{}}\\ = \frac{{1 + \csc (2\pi - \theta ) + \tan ({\textstyle{{3\pi } \over 2}} + \theta )}}{{1 + \csc (\pi - \theta ) + \tan ({\textstyle{\pi \over 2}} - \theta )}} - {\frac{{1 + \sec ({\textstyle{\pi \over 2}} + \theta )}}{{\cot (\pi + \theta )}}^{}}\\ = \frac{{1 - \csc \theta - \cot \theta }}{{1 + \csc \theta + \cot \theta }} - {\frac{{1 - \csc \theta }}{{\cot \theta }}^{}}\\ = {\frac{{\cot \theta (1 - \csc \theta - \cot \theta ) - (1 - \csc \theta )(1 + \csc \theta + \cot \theta )}}{{(1 + \csc \theta + \cot \theta )\cot \theta }}^{^{}}}\\ = {\frac{{\cot \theta - \csc \theta \cot \theta - {{\cot }^2}\theta - 1 - \csc \theta - \cot \theta + \csc \theta + {{\csc }^2}\theta + \csc \theta \cot \theta }}{{(1 + \csc \theta + \cot \theta )\cot \theta }}^{}}\\ = {\frac{{ - {{\cot }^2}\theta - 1 + {{\csc }^2}\theta }}{{(1 + \csc \theta + \cot \theta )\cot \theta }}^{}}\\ = {\frac{{ - {{\csc }^2}\theta + {{\csc }^2}\theta }}{{(1 + \csc \theta + \cot \theta )\cot \theta }}^{}}\\ = {0^{^{}}}\end{array}$$
∴ $$\frac{{1 - \csc (\theta - 2\pi ) + \tan (\theta + {\textstyle{{3\pi } \over 2}})}}{{1 + \csc (\pi - \theta ) - \tan (\theta - {\textstyle{\pi \over 2}})}} = \frac{{1 + \sec ({\textstyle{\pi \over 2}} + \theta )}}{{\cot (\pi + \theta )}}$$