# High school trigonometry questions and answers

Recent questions in Trigonometry

### Why is $$\displaystyle{\tan{\theta}}\approx{\frac{{{1}}}{{{\frac{{\pi}}{{{2}}}}-\theta}}}$$ for $$\displaystyle\theta$$ close to $$\displaystyle{\frac{{\pi}}{{{2}}}}$$? I wanted to see what the behaviour of the steep part of the $$\displaystyle{\tan{}}$$ curve was like, i.e. the behaviour of $$\displaystyle{\tan{{\left({x}\right)}}}\ {a}{s}\ {x}\to{\left({\frac{{\pi}}{{{2}}}}\right)}^{{-{}}}$$. So by thinking about a shift of the graph of $$\displaystyle{\tan{{\left({x}\right)}}}\ {b}{y}{\frac{{\pi}}{{{2}}}}$$ to the left, I put some small (positive and negative) values of $$\displaystyle\theta$$ into my calculator for the function $$\displaystyle{\tan{{\left(\theta+{\frac{{\pi}}{{{2}}}}\right)}}}$$ $$\displaystyle{\tan{\theta}}\approx{\frac{{{1}}}{{{\frac{{\pi}}{{{2}}}}-\theta}}}$$ for $$\displaystyle\theta$$ close to $$\displaystyle{\frac{{\pi}}{{{2}}}}$$? or, in more colloquial terms, The steep part of $$\displaystyle{\tan{{x}}}$$ is just like the steep part of $$\displaystyle{\frac{{1}}{{x}}}$$ But why is this the case? I couldn't deduce it easily using the Maclaurin expansion of $$\displaystyle{\tan{{\left({x}\right)}}}$$. Is there a more intuitive explanation? I couldn't think of any explanations analogous to those explaining small angle approximations.

Carla Murphy 2021-12-27

### Find the general solution of the equation $$\displaystyle{\sin{{x}}}+{\sin{{2}}}{x}+{\sin{{3}}}{x}={0}$$. I have started doing this problem by applying the formula of $$\displaystyle{\sin{{A}}}+{\sin{{B}}}$$ but couldn't generalise it. Please solve it for me.

hvacwk 2021-12-26