Step by step solution to "What is tan⁡α if sin⁡α+cos⁡α=3−12 and α∈(90∘,135∘)"

Dylan Yoder

Answered question

2022-04-19

What is

\tan α

\sin α + \cos α = \frac{\sqrt{3} - 1}{2}

and

α \in (90^{\circ}, 135^{\circ})

Juan Goodwin

Beginner2022-04-20Added 10 answers

\sin α + \cos α = \sqrt{2} \sin (α + \frac{π}{4}) = \frac{\sqrt{3} - 1}{2} \Rightarrow α

= \arcsin (\frac{\sqrt{3} - 1}{2 \sqrt{2}} - \frac{π}{4}

Now:

\tan α = \tan (\arcsin (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) - \frac{π}{4})

= \frac{\tan \arcsin (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) - \tan \frac{π}{4}}{1 + \tan \arcsin (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) \tan \frac{π}{4}}

Now:

\tan \arcsin (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) = \frac{\sin \arcsin (\frac{\sqrt{3} - 1}{2 \sqrt{2}})}{\cos \arcsin (\frac{\sqrt{3} - 1}{2 \sqrt{2}})}

= \frac{\sin \arcsin (\frac{\sqrt{3} - 1}{2 \sqrt{2}})}{\sqrt{1 - \sin^{2} \arcsin (\frac{\sqrt{3} - 1}{2 \sqrt{2}})}}

= \frac{\frac{\sqrt{3} - 1}{2 \sqrt{2}}}{\sqrt{1 - {(\frac{\sqrt{3} - 1}{2 \sqrt{2}})}^{2}}} = 2 - \sqrt{3}

Now answer is:

\tan α = \frac{2 - \sqrt{3} - 1}{1 + 2 - \sqrt{3}} = \frac{1 - \sqrt{3}}{3 - \sqrt{3}} = - \frac{1}{\sqrt{3}}

Nico Buck

Beginner2022-04-21Added 9 answers

Let

\sin A = \frac{\sqrt{3}}{2}, \cos A = - \frac{12}{}

Clearly,

A = 120^{\circ}

is a solution
So, we can write

\sin α - \sin A = \cos A - \cos α

Using Prosthaphaeresis Formulas,

2 \sin \frac{α - A}{2} \cos \frac{α + A}{2} = 2 \sin \frac{α - A}{2} \sin \frac{α + A}{2}

\Leftrightarrow \sin \frac{α - A}{2} [\cos \frac{α + A}{2} - \sin \frac{α + A}{2}] = 0

\sin \frac{α - A}{2} = 0, \frac{α - A}{2} = 180^{\circ} n \Leftrightarrow α = A + 360^{\circ} n

where n is any integer
Else

\tan \frac{α + A}{2} = 1 = {\tan 45}^{\circ}

\Rightarrow \frac{α + A}{2} = m 180^{\circ} + 45^{\circ} \Leftrightarrow α = 360^{\circ} m + 90^{\circ} - A

where m is any integer

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