Question

Trigonometry question

Answer 1

\(\large \color{black}{\begin{align}
& \sin x+\cos x=\dfrac{1}{5} \hspace{.33em}\\~\\
& 0\leq x \leq \pi \hspace{.33em}\\~\\
& \normalsize \text{Find}\ \large \tan x
\end{align}}\)

Answer 2

\[\sin x + \cos x =\frac{ 1 }{ 5 }\]
From the fundamental equality of trigonometry we can deduce that:
\[\sin x + (\sqrt{1-\sin^2x})=\frac{ 1 }{ 5 }\]
Now that everything is written under one trigonometric operation, we can limit ourselves into solving for it, just as you did back in HS.
Squaring both sides now in order to get rid of the square root will be a problem, so we sustract "sin x " to bothe sides:
\[\sqrt{1- \sin^2x}=\frac{ 1 }{ 5 }-\sin x \iff (\sqrt{1-\sin^2x})^2=(\frac{ 1 }{ 5 }-\sin x)^2\]
\[\iff 1-\sin^2x=(\frac{ 1 }{ 5 })^2-2(\frac{ 1 }{ 5 })(\sin x)+\sin^2x \]
ordering it up:
\[(-\sin^2x-\sin^2x)+2(\frac{ 1 }{ 5 })(\sin x ) +3=0\]
\[\iff -2\sin^2 x+2(\frac{ 1 }{ 5 })\sin x +3 =0\]
And that's a complete quadratic equation, which by now you should be able to solve.
After you find the possible values for sin x, ditch the one that would not belong in the desired interval and calculate cosx from there.
Remember that \(\tan x = \frac{ \sin x }{ \cos x }\).

Answer 3

this is working but quite long

Answer 4

Wait, I forgot a value there:
\[1-\sin^2 x= (\frac{ 1 }{ 5 })^2-2(\frac{ 1 }{ 5 })\sin x + \sin^2x\]
\[\iff -2\sin^2x+2\frac{ 1 }{ 5 }\sin x+(1-\frac{ 1 }{ 25 })=0\]

Answer 5

it will be long no matter what you do
even if you used the addition formula
http://www.cut-the-knot.org/triangle/SinCosFormula.shtml

Answer 6

multiply by \(\frac{\sqrt{2}}{2}\) to get\[\sin \left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{10}\]how is this?

Answer 7

and note that\[\tan(\arcsin(x))=\frac{x}{\sqrt{1-x^2}}\]

Answer 8

I'm very sure I wrote a reply here earlier. On squaring both sides,\[1 + \sin 2x = \frac{1}{25}\]\[\Rightarrow \sin 2x = \frac{2\tan x}{1-\tan^2 x} = \frac{-24}{25}\]This is a quadratic in \(\tan x\) and you can solve it.