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ratelena [41]
3 years ago
11

Please solve 5 f (Trigonometric Equations) #salute u if u solved it

Mathematics
1 answer:
Zanzabum3 years ago
4 0

Answer:

\beta=45\degree\:\:or\:\:\beta=135\degree

Step-by-step explanation:

We want to solve \tan \beta \sec \beta=\sqrt{2}, where 0\le \beta \le360\degree.

We rewrite in terms of sine and cosine.

\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}

\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}

Use the Pythagorean identity: \cos^2\beta=1-\sin^2\beta.

\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}

\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)

\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta

\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0

This is a quadratic equation in \sin \beta.

By the quadratic formula, we have:

\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }

\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }

\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }

\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }

\sin \beta=\frac{2}{2\cdot \sqrt{2} } or \sin \beta=\frac{-4}{2\cdot \sqrt{2} }

\sin \beta=\frac{1}{\sqrt{2} } or \sin \beta=-\frac{2}{\sqrt{2} }

\sin \beta=\frac{\sqrt{2}}{2} or \sin \beta=-\sqrt{2}

When \sin \beta=\frac{\sqrt{2}}{2} , \beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )

\implies \beta=45\degree\:\:or\:\:\beta=135\degree on the interval 0\le \beta \le360\degree.

When  \sin \beta=-\sqrt{2}, \beta is not defined because -1\le \sin \beta \le1

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