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

Mary Reyes 2022-01-01 Answered
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?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Paul Mitchell
Answered 2022-01-02 Author has 201 answers
\(\displaystyle{{\sin}^{{2}}{A}}+{{\sin}^{{4}}{A}}={1}\)
\(\displaystyle{{\sin}^{{4}}{A}}={1}-{{\sin}^{{2}}{A}}\)
\(\displaystyle{{\sin}^{{2}}{A}}{{\sin}^{{2}}{A}}={{\cos}^{{2}}{A}}\)
\(\displaystyle{{\sin}^{{2}}{A}}{\left({1}-{{\cos}^{{2}}{A}}\right)}={{\cos}^{{2}}{A}}\)
\(\displaystyle{{\sin}^{{2}}{A}}-{{\sin}^{{2}}{A}}{{\cos}^{{2}}{A}}={{\cos}^{{2}}{A}}\)
\(\displaystyle{{\sin}^{{2}}{A}}={{\cos}^{{2}}{A}}+{{\sin}^{{2}}{A}}{{\cos}^{{2}}{A}}\)
\(\displaystyle{{\sin}^{{2}}{A}}={{\cos}^{{2}}{A}}{\left({1}+{{\sin}^{{2}}{A}}\right)}\)
\(\displaystyle{1}+{{\sin}^{{2}}{A}}={\frac{{{{\sin}^{{2}}{A}}}}{{{{\cos}^{{2}}{A}}}}}\)
\(\displaystyle{1}+{{\sin}^{{2}}{A}}={{\tan}^{{2}}{A}}\)
Not exactly what you’re looking for?
Ask My Question
0
 
soanooooo40
Answered 2022-01-03 Author has 714 answers
Hint
If \(\displaystyle{{\sin}^{{2}}{A}}+{{\sin}^{{4}}{A}}={1}\Rightarrow{{\sin}^{{4}}{A}}={{\cos}^{{2}}{A}}\)
Can you proceed from here?
Additional Info
\(\displaystyle{{\sin}^{{2}}{A}}{\left({1}-{{\cos}^{{2}}{A}}\right)}={{\cos}^{{2}}{A}}\)
\(\displaystyle{{\sin}^{{2}}{A}}={{\cos}^{{2}}{A}}+{{\sin}^{{2}}{A}}{{\cos}^{{2}}{A}}\)
Divide by \(\displaystyle{{\cos}^{{2}}{A}}\)
0
Vasquez
Answered 2022-01-08 Author has 8850 answers

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

0

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2022-01-02
I can't figure out how to simplify:
\(\displaystyle{\frac{{12}}{=}}{\frac{{{{\sin}^{{2}}{x}}}}{{{\tan{{x}}}}}}\)
What would be my next step to get to \(\displaystyle{x}={45}^{{\circ}}\)?
asked 2021-12-30
Evaluating \(\displaystyle\lim_{{{x}\to{0}}}{\left({\frac{{{1}}}{{{\sin{{\left({x}\right)}}}}}}-{\frac{{{1}}}{{{x}}}}\right)}\) without L'Hôpital's rule
The only thing I can think of using is the basic identity
\(\displaystyle\lim_{{{x}\to{0}}}{\left({\frac{{{\sin{{x}}}}}{{{x}}}}\right)}={1}\)
but I can't reduce the original problem down to a point where I can apply this identity.
asked 2021-12-30
Why does \(\displaystyle{\sum_{{{i}={1}}}^{{k}}}{\sin{{\left({i}{\frac{{{2}\pi}}{{{k}}}}\right)}}}={0}\) for integers k
asked 2021-12-31
How does \(\displaystyle{\frac{{{4}\pi{x}+{\cos{{\left({4}\pi{x}\right)}}}{\sin{{\left({4}\pi{x}\right)}}}}}{{{8}\pi}}}+{C}\) become \(\displaystyle{\frac{{{\sin{{\left({8}\pi{x}\right)}}}+{8}\pi{x}}}{{{16}\pi}}}+{C}\)?
asked 2022-01-01
I was trying to find the least value of \(\displaystyle{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}\) and applied the above inequality as :
\(\displaystyle{\left|{\sin{{x}}}+{\cos{{x}}}\right|}\leq{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}\)
\(\displaystyle\Rightarrow\sqrt{{2}}\leq{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}\)
But the range of the given function is \(\displaystyle{\left[{1},\sqrt{{2}}\right]}\).
asked 2022-01-02
How could I approach
\(\displaystyle\lim_{{{n}\to\infty}}{\left({\frac{{{1}+{\cos{{\left({\frac{{{1}}}{{{2}^{{n}}}}}\right)}}}}}{{{2}}}}\right)}^{{n}}\)
I tried several things to evaluate it, namely looking at it as \(\displaystyle{{\cos{{\left({\frac{{{1}}}{{{2}^{{{n}+{1}}}}}}\right)}}}^{{{2}{n}}}}\) instead or as \(\displaystyle{\exp{{\left({2}{n}\cdot{\ln{{\left({\cos{{\left({\frac{{{1}}}{{{2}^{{{n}+{1}}}}}}\right)}}}\right)}}}\right.}}}\) and then trying to show that the limit of \(\displaystyle{n}\cdot{\ln{{\left({\cos{{\left({\frac{{{1}}}{{{2}^{{{n}+{1}}}}}}\right)}}}\right)}}}\) is 0 (for example using L'Hopital's rule), but I haven't been very successful
asked 2021-12-31
How to integrate hyperbolic function \(\displaystyle{\frac{{{1}-{\sin{{\left({x}\right)}}}}}{{{1}+{\sin{{\left({x}\right)}}}}}}\)
...