# How am I supposed to expand \sin^2 A+\sin^4 A=1 into

How am I supposed to expand $$\displaystyle{{\sin}^{{2}}{A}}+{{\sin}^{{4}}{A}}={1}$$ into $$\displaystyle{1}+{{\sin}^{{2}}{A}}={{\tan}^{{2}}{A}}$$?

## Want to know more about Trigonometry?

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Paul Mitchell
$$\displaystyle{{\sin}^{{2}}{A}}+{{\sin}^{{4}}{A}}={1}$$
$$\displaystyle{{\sin}^{{4}}{A}}={1}-{{\sin}^{{2}}{A}}$$
$$\displaystyle{{\sin}^{{2}}{A}}{{\sin}^{{2}}{A}}={{\cos}^{{2}}{A}}$$
$$\displaystyle{{\sin}^{{2}}{A}}{\left({1}-{{\cos}^{{2}}{A}}\right)}={{\cos}^{{2}}{A}}$$
$$\displaystyle{{\sin}^{{2}}{A}}-{{\sin}^{{2}}{A}}{{\cos}^{{2}}{A}}={{\cos}^{{2}}{A}}$$
$$\displaystyle{{\sin}^{{2}}{A}}={{\cos}^{{2}}{A}}+{{\sin}^{{2}}{A}}{{\cos}^{{2}}{A}}$$
$$\displaystyle{{\sin}^{{2}}{A}}={{\cos}^{{2}}{A}}{\left({1}+{{\sin}^{{2}}{A}}\right)}$$
$$\displaystyle{1}+{{\sin}^{{2}}{A}}={\frac{{{{\sin}^{{2}}{A}}}}{{{{\cos}^{{2}}{A}}}}}$$
$$\displaystyle{1}+{{\sin}^{{2}}{A}}={{\tan}^{{2}}{A}}$$
###### Not exactly what you’re looking for?
soanooooo40
Hint
If $$\displaystyle{{\sin}^{{2}}{A}}+{{\sin}^{{4}}{A}}={1}\Rightarrow{{\sin}^{{4}}{A}}={{\cos}^{{2}}{A}}$$
Can you proceed from here?
$$\displaystyle{{\sin}^{{2}}{A}}{\left({1}-{{\cos}^{{2}}{A}}\right)}={{\cos}^{{2}}{A}}$$
$$\displaystyle{{\sin}^{{2}}{A}}={{\cos}^{{2}}{A}}+{{\sin}^{{2}}{A}}{{\cos}^{{2}}{A}}$$
Divide by $$\displaystyle{{\cos}^{{2}}{A}}$$
Vasquez

$$\begin{array}{}\sin^2 A=1-\sin^4 A \\\sin^2 A=(1-\sin^2 A)(1+\sin^2 A) \\\sin^2 A=(\cos^2 A)(1+\sin^2 A) \\\therefore \tan^2 A=1+\sin^2 A \end{array}$$