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Find equation of a tangent on $y= \sin2x$ in intersections with $y=\frac{1}{2}$

What I calculated:

Intersections: $$\sin2x= \frac{1}{2}$$ ... $$0=tg^2x-4tgx-1$$

$$tgx_{1}=2+\sqrt{3}$$

$$x_{1}=75+k\pi$$

$$tgx_{2}=2-\sqrt{3}$$

$$x_{2}=15+ k\pi$$

$$T_{1}(75+k\pi,\frac{1}{2})$$ and $$T_{2}(15+k\pi,\frac{1}{2})$$

Derivative of $y= sin2x$ is: $y'= 2cos2x$ $$ y'(75)= -\sqrt {3} =k_{t_{1}} $$ $$y'(15)= \sqrt {3} =k_{t_{2}}$$ formula for calculating tangent that I used: $y-y_{1}=k_{t}(x-x_{1})$

1.)$T_{1}(75+k\pi,\frac{1}{2})$

$k_{t_{1}}=-\sqrt {3}$

$y_{1}-\frac{1}{2}=-\sqrt {3}(x-75-k\pi)$

$y_{1}=-\sqrt {3}x+75\sqrt {3}+\sqrt {3}k\pi+\frac{1}{2}$

2.)$T_{2}(15+k\pi,\frac{1}{2})$

$k_{t_{2}}=\sqrt {3}$

$y_{2}-\frac{1}{2}=\sqrt {3}(x-15 -k\pi)$

$y_{2}=\sqrt {3}x-15\sqrt {3}-\sqrt {3}k\pi+\frac{1}{2}$

Both solutions are wrong; what I should actually get: $y_{1}=\sqrt{3}x-\frac{\sqrt{3}pi-6+12kpi\sqrt{3}}{12} $ and $y_{2}=-\sqrt{3}x-\frac{5\sqrt{3}pi+6+12kpi\sqrt{3}}{12};k=Z $

How could I possibly get that ?!

Ahaan S. Rungta
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user114141
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  • http://math.stackexchange.com/questions/466299/why-is-the-derivative-of-sine-the-cosine-in-radians-but-not-in-degrees. Here $15$ is in radian should be replaced with $\frac\pi{180}\cdot15$ as the value of $x$. http://answers.yahoo.com/question/index?qid=20060930195532AAIus3w and http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/archv014.cgi?read=63525 – lab bhattacharjee Dec 17 '13 at 16:54

3 Answers3

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You calculated your equations using the given $x$ (angles) in terms of degrees, not in radians: specifically, you need to multiply each angle in your equations of the lines $\,(75^\circ, 15^\circ\,)$ by the conversion factor $\frac \pi{180^\circ}$ to obtain those values in radians, which is how the solution equations are given.

Note that you've mixed degrees and radians when expressing the solutions to $x_1, x_2$:

$$x_{1}=75^\circ +k\pi$$ $$x_{2}=15^\circ+ k\pi$$

where you should express them as $$x_1 = \dfrac {5\pi}{12} + k\pi$$ $$x_2 = \dfrac{\pi}{12} + k\pi$$

amWhy
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2

$$\sin2x= \frac{1}{2}$$ $$2x=n\pi+(-1)^n\pi/6$$ Therefore

$$x=\frac{n \pi}{2} + (-1)^n \frac{ \pi}{12} \text{ , } n=Z$$

$f(x)=\sin2x$

$f'(x)=2\cos2x$ where $m=f'(x)$ and c can be find by putting x and $y=\frac{1}{2}$

$y_{1}=\sqrt{3}x-\frac{\sqrt{3}\pi-6+12k\pi\sqrt{3}}{12} $ and $y_{2}=-\sqrt{3}x-\frac{5\sqrt{3}\pi+6+12k \pi\sqrt{3}}{12};k=Z $

amWhy
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john
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1

\begin{align*} \ \sin{2x} &= \frac{1}{2} \\ \ \arcsin{\big(\sin{2x}}\big) &= \arcsin{\frac{1}{2}} \\ \ 2x &= \frac{\pi}{6} \\ \ x &= \frac{\pi}{12} \qquad\text{to be more precise you can say:}\\ \ x &= \frac{\pi}{12} + \pi k\quad ,\quad k\in\mathbb Z\\ \end{align*}

So we know $x=\dfrac{\pi}{12}$ when they intersect.

$y=a(x-x_0) + y_0$ is the formula for the tangent, where $a=f'(x_0)$.

So, $x_0 = \dfrac{\pi}{12} \land y_0=\dfrac{1}{2}$.

$f(x)=\sin{2x} \implies f'(x)=2\cos{2x}$

$f'(x_0) = f'(\dfrac{\pi}{12}) = 2\cos{\big(\dfrac{\pi}{6}\big)} = \sqrt{3} $

Thus, the tangent is $$y=\underline{\underline{\sqrt{3}\big(x-\frac{\pi}{12}\big) + \dfrac{1}{2}}}$$

Erlend
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