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It is given that $0 < x < 180^\circ$ and $\sin x+\cos x=\frac12$, Find $\tan x $.

I tried all identities I know but I have no idea how to proceed. Any help would be appreciated.

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  • $\begingroup$ What identities have you tried? Can you relate $\tan$ with $\sin$ and $\cos$? Given the information you have, can you at least determine which quadrant $x$ is in ($0 < x < 90$ or $90 < x < 180$)? $\endgroup$ – user61527 Jul 25 '14 at 5:19
  • $\begingroup$ Or you can just solve $\sin x+\cos x=1/2$ for $x$ and use that to find $\tan x$. $\endgroup$ – Quang Hoang Jul 25 '14 at 5:25
  • $\begingroup$ If you knew $\cos x$, say, you would probably be fine. So you could write $\sin x=1/2 -\cos x$ and square both sides. $\endgroup$ – André Nicolas Jul 25 '14 at 5:26
  • $\begingroup$ Yes it's in the 2 quadrant 90 < x < 180. $\endgroup$ – Kanishk Jul 25 '14 at 5:34
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Use the identity $$\sin{x}+\cos{x}=\sqrt{2}\sin(x+\pi/4)$$ Then $$\sin(x+\pi/4)=\frac{1}{2\sqrt{2}}$$ \begin{align} \tan(x+\pi/4) &=\frac{1}{\sqrt{7}}\\ &=\frac{1+\tan x}{1-\tan x}\\ \end{align} The rest is easy.

In general we have $$a\sin{x}+b\cos{x}=\sqrt{a^2+b^2}\sin(x+\arctan(b/a))$$ This can be shown using the compound angle formula.

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  • $\begingroup$ Is this answer correct? (1-root7)/(1+root7) $\endgroup$ – Kanishk Jul 25 '14 at 6:09
  • $\begingroup$ @Kanishk Yes it is. I hope the answers were useful to you. $\endgroup$ – SuperAbound Jul 25 '14 at 6:16
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Hint : \begin{align} \sin x+\cos x&=\frac12\\ \sin x&=\frac12-\cos x\\ \sin^2 x&=\left(\frac12-\cos x\right)^2\\ 1-\cos^2 x&=\frac14-\cos x+\cos^2 x\\ 2\cos^2 x-\cos x-\frac34&=0\\ 8\cos^2 x-4\cos x-3&=0 \end{align}

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$\sin x+\cos x=\frac12$ $\Rightarrow$
$(\sin x+\cos x)^2=\frac14$ $\Rightarrow$
$\sin^2 x+2\sin x\cos x+\cos^2 x=\frac14$ $\Rightarrow$
$1+2\sin x\cos x=\frac14$ $\Rightarrow$
$2\sin x\cos x = -\frac34$ $\Rightarrow$
$\sin x\cos x=-\frac38$.

If $a=\sin x$, $b=\cos x$, we have the system of equations $$ \begin{align} a+b&=\frac12\\ ab&=-\frac38 \end{align} $$ which leads to $$a-\frac3{8a}=\frac12\\ a^2-\frac12a-\frac38=0$$ with the solutions $a=\frac{1\pm\sqrt7}4$.

So we get $$ \sin x=\frac{1+\sqrt7}4, \qquad \cos x= \frac{1-\sqrt7}4 \qquad \Rightarrow \qquad \tan x = \frac{1+\sqrt7}{1-\sqrt7}\\ \sin x=\frac{1-\sqrt7}4, \qquad \cos x= \frac{1+\sqrt7}4 \qquad \Rightarrow \qquad \tan x = \frac{1-\sqrt7}{1+\sqrt7} $$

You could also simplify the result: $$\frac{1+\sqrt7}{\frac1-\sqrt7} = \frac{(1+\sqrt7)^2}{1-7} = -\frac{8+2\sqrt7}6= -\frac{4+\sqrt7}3$$

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HINT:

Avoid squaring whenever possible as it immediately introduces extraneous root which needs to be discarded

Use Weierstrass substitution to form a Quadratic Equation in $\tan\dfrac x2$

Now use $\displaystyle\tan2A=\frac{2\tan A}{1-\tan^2A}$

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Divide through by $\cos x$ to get \begin{align} \tan x + 1 & = \frac 12 \sec x \tag{1}\\ \text{Square both sides: } \qquad\qquad\tan^2 x + 2 \tan x + 1 & = \frac 14 \sec^2 x = \frac{1}{4} (\tan^2 x + 1) \\ 3 \tan^2 x + 8 \tan x + 3 & = 0 \\ \tan x & = \frac{-8 \pm \sqrt{28}}{6} = \frac{-4 \pm \sqrt{7}}{3}. \end{align} Check the two solutions with the original equation before squaring (equation $(1)$). Because $x \in [0, \pi]$ and $\tan x < 0$ in both cases, we know that $\sin x \ge 0$ and $\cos x \le 0$, so $\sec x \le 0$.

If $\tan x = \frac{-4 + \sqrt{7}}3$, then $\tan x + 1 = \frac{-1 + \sqrt{7}}3 > 0$, contradicting $(1)$ and $\sec x \le 0$.

If $\tan x = \frac{-4 - \sqrt{7}}3$, then $\tan x + 1 = \frac{-1 - \sqrt{7}}3 < 0$. This looks good.


If you want to be more sure, compute $\sin x$ and $\cos x$ from the chosen $\tan x$: \begin{align} \sin x & = \frac{4 + \sqrt 7}{\sqrt{3^2 + (-4 - \sqrt{7})^2}} \\ & = \frac{4 + \sqrt 7}{\sqrt{32 + 8\sqrt 7}} = \frac{1}{4}\sqrt{8 + 2\sqrt{7}} = \frac{1}{4}(1 + \sqrt{7}) \\ \cos x & = -\sqrt{1 - \sin^2 x} = -\frac{1}{4}\sqrt{8 - 2\sqrt 7} = \frac{1}{4}(1 - \sqrt{7}). \end{align} It's obvious that $\sin x + \cos x = \frac 12$, $\sin x \ge 0$ and $\cos x \le 0$.

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HINT: $$\sin⁡\Big( x+\dfrac{\pi}{4}\Big)=\dfrac{1}{\sqrt{2}}(\sin ⁡x+\cos ⁡x )=\dfrac{1}{2\sqrt{2}}$$
$$\tan⁡\Big(x+\dfrac{\pi}{4}\Big)=±\dfrac{\sin\Big( x+\dfrac{π}{4}\Big)}{\sqrt{1-\sin^2⁡\Big(x+\dfrac{π}{4}\Big) }}=±\dfrac{1}{\sqrt{7}}$$
$$\dfrac{1+\tan ⁡x}{1-\tan ⁡x}=±\dfrac{1 }{\sqrt{7}}$$
Then you have to solve above two equations separately.

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Here is a geometric way of looking at this problem. I will change the angle variable to $t.$ We need to find $\tan t$ if $\sin t + \cos t = 1/2.$ I will use $x = \cos t, y = \sin t$ so we need to solve $$x^2 + y^2 = 1, x + y = 1/2$$ for $x, y.$ that is, to find the two points where the line cuts the unit circle. By eliminating one variable, say $x,$ gives $$1 = x^2 +(1/2-x)^2= 2x^2 - 1/2x + 1/4$$ which is a quadratic equation $$8x^2 - 4x -3 = 0 $$ in $x.$ Now use the quadratic formulae to find $x,$ then $y = 1/2 -x$ and $\tan t = y/x ={1\over 2x } - 1$ can then be computed. Solving the quadratic equation for $${1 \over x} = {-4 \pm 2\sqrt{28} \over 6} = {2(-1 \pm \sqrt 7)/3},\ {1 \over 2x} -1= {-4 \pm 2\sqrt{28} \over 6} = {(-4 \pm \sqrt 7) \over 3}. $$

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Or: $1 + 2\sin{x }\cos{x} = \frac{1}{4},$ so $\sin{x}\cos{x} = \frac{-3}{8},$ leading to $\sin{x} - \cos{x} = \pm \sqrt{ \frac{1}{4} + \frac{3}{2}} = \pm \frac{\sqrt{7}}{2},$ allowing you to solve for $\sin{x}$ and $\cos{x}$ (making use of the fact that $0 < x < \pi$ (radians) so that $\sin{x} >0$).

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With no quadratic steps, only using trig identities:

Multiplying through by $\frac1{\sqrt{2}}$:

$$\begin{align} \sin(x)\frac1{\sqrt{2}}+\cos(x)\frac1{\sqrt{2}}&=\frac{1}{\sqrt{8}}\\ \sin(x)\cos(\pi/4)+\cos(x)\sin(\pi/4)&=\frac{1}{\sqrt{8}}\\ \sin(x+\pi/4)&=\frac{1}{\sqrt{8}} \end{align}$$

Given that $x$ is in $[0,\pi]$, $x$ must be in $(\pi/2,3\pi/4)$ because

  • if $x$ is in $[0,\pi/2]$ then both $\sin(x)$ and $\cos(x)$ are nonnegative, and $\max(\sin(x),\cos(x))\geq\frac{1}{\sqrt{2}}>\frac12$
  • if $x$ is in $[3\pi/4,\pi]$ then $\cos(x)$ is negative and outweighs $\sin(x)$

So $x+\pi/4$ is in $(3\pi/4,\pi)$, and: $$\begin{align} \sin(x+\pi/4)&=\frac{1}{\sqrt{8}}\\ x+\pi/4&=\pi-\arcsin\left(\frac{1}{\sqrt{8}}\right)\\ x&=\pi-\arcsin\left(\frac{1}{\sqrt{8}}\right)-\pi/4\\ \tan(x)&=\tan\left(\pi-\arcsin\left(\frac{1}{\sqrt{8}}\right)-\pi/4\right)\\ \tan(x)&=-\tan\left(\arcsin\left(\frac{1}{\sqrt{8}}\right)+\pi/4\right)\\ \tan(x)&=-\frac{\tan\left(\arcsin\left(\frac{1}{\sqrt{8}}\right)\right)+1}{1-\tan\left(\arcsin\left(\frac{1}{\sqrt{8}}\right)\right)}\\ \tan(x)&=-\frac{\frac{\frac{1}{\sqrt{8}}}{\sqrt{1-\frac{1}{8}}}+1}{1-\frac{\frac{1}{\sqrt{8}}}{\sqrt{1-\frac{1}{8}}}}\\ \tan(x)&=-\frac{\frac{1}{\sqrt{8}}+\sqrt{\frac78}}{\sqrt{\frac78}-\frac{1}{\sqrt{8}}}\\ \tan(x)&=\frac{1+\sqrt{7}}{1-\sqrt{7}}\\\end{align}$$

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