In this section, we are going to give some of the most elementary trigonometric relations.
- $$\sin^2(x) +\cos^2(x)=1$$
- $$\tan(x)=\dfrac{\sin(x)}{\cos(x)}$$
- $$1+\tan^2(x)=\sec^2(x)$$
- $$1+\cot^2(x)=\csc^2(x)$$
- $$1-\cos(2x)=2\sin^2(x)$$
- $$1+\cos(2x)=2\cos^2(x)$$
- $$\sin(x+y)=\sin(x)\cdot\cos(x)+\cos(x)\cdot\sin(y)$$
- $$\sin(x-y)=\sin(x)\cdot\cos(x)-\cos(x)\cdot\sin(y)$$
- $$\cos(x+y)=\cos(x)\cdot\cos(y)-\sin(x)\cdot\sin(y)$$
- $$\cos(x-y)=\cos(x)\cdot\cos(y)+\sin(x)\cdot\sin(y)$$
- $$\tan(x+y)= \dfrac{\tan(x)+\tan(y)}{1-\tan(x)\cdot\tan(y)}$$
- $$\tan(x-y)= \dfrac{\tan(x)-\tan(y)}{1-\tan(x)\cdot\tan(y)}$$
- $$\sin(2x)=2\cdot\sin(x)\cdot\cos(x)$$
- $$\cos(2x)=\cos^2(x)-\sin^2(x)$$
- $$\tan(2x)=\dfrac{2\cdot\tan(x)}{1-\tan^2(x)}$$
Knowing that $$\cos(x)=\sin(x)$$ and $$\sin(y)=2\cos(y)$$, we are going to calculate the cosine and the sine of $$x+y$$:
$$\begin{array}{rl} \sin(x+y)=&\sin(x)\cos(y) + \cos(x)\sin(y)= \sin(x)\cos(y) + 2\sin(x)\cos(y) \\ =&3\sin(x)\cos(y) \end{array}$$
$$\begin{array}{rl} \cos(x+y)=&\cos(x)\cos(y)-\sin(x)\sin(y)= \sin(x)\cos(y)-2\sin(x)\cos(y) \\ =& -\sin(x)\cos(y) \end{array}$$
Proceeding in a similar way, we obtain that the sine and the cosine of $$x-y$$ is:
$$\sin(x-y)=-\sin(x)\cos(y)=\cos(x+y)$$
$$\cos(x-y)=3\sin(x)\cos(y)=\sin(x+y)$$
Knowing that $$\sin\alpha = \dfrac{3}{5}$$, and that $$90^\circ < \alpha < 180^\circ$$, calculate the remaining trigonometric ratios of the angle $$\alpha$$.
We are going to calculate other trigonometric ratios from the fact that we know the sine of the angle,
$$$\sin(\alpha)=\dfrac{3}{5} \Rightarrow \csc(\alpha)=\dfrac{5}{3}$$$
On the other hand, the cosine is: $$$\cos(\alpha)=-\sqrt{1-\sin^2(\alpha)}=-\sqrt{1-\dfrac{9}{25}}= -\sqrt{\dfrac{26}{25}}=-\dfrac{4}{5} \Rightarrow \sec(\alpha)=-\dfrac{5}{4}$$$
Finally, to calculate the tangent we use:
$$$\tan(\alpha)=\dfrac{\sin(\alpha)}{\cos(\alpha)}= -\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}=-\dfrac{3}{4} \Rightarrow \cot(\alpha)=-\dfrac{4}{3}$$$