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$$3\sin x + 4\cos x = 2$$

To solve an equation like the one above, we were taught to use the double angle identity formula to get two equations in the form of $R\cos\alpha = y$ where $R$ is a coefficient and $\alpha$ is the second angle being added to $x$ when using the double angle identity.

Why can't we use the identity $\sin(x) = \cos(x-90)$ to get $3\cos(x-90) + 4\cos x = 2$? Is this equation difficult to simplify further?

Additionally, why was the relationship between $sinx$ and $cosx$ in the pythagorean theorem (modified for the unit circle) not put to use? I did the following:

$$\sin^2x = 1 - \cos^2x$$

$$\therefore \sin x = ±\sqrt{1-\cos^2x}$$

If:

$$\sin x = y$$

Then,

$$y = ±\sqrt{1-\cos^2x}$$

Meaning that,

$$\cos x = ±\sqrt{1-y^2}$$

Inputting this into the original equation,

$$3y + 4\sqrt{1-y^2} = 2$$

We see, $$3y-2=-4\sqrt{1-y^2}$$ So, $$(3y-2)^2=16-16y^2$$ Therefore, $$9y^2-12y+4=16-16y^2$$ Rearranging gives, $$25y^2-12y-12=0$$ And so the solutions are, $$y_0,y_1=\frac{12}{50}\pm\frac{1}{50}\sqrt{144+1200}$$ And simplifying yields, $$y_0,y_1=\frac{6\pm4\sqrt{21}}{25}$$ Checking these solutions will give us the unique solution: $$y_0=\frac{6-4\sqrt{21}}{25}$$

$$q.e.d.$$

Can the above method be generalised? Has it been generalised?

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  • $\begingroup$ Well you found $\sin(x)$ now, but you still don't have $x$. Also, your final solution are not all solutions since you actually needed to solve $3y-4\sqrt{1-y^2}=2$ as well $\endgroup$ – vrugtehagel Feb 4 '16 at 11:32
  • $\begingroup$ The method described at the very top involving the use of the double angle identity formula to get two equations in the form of Rcosα=y where R is a coefficient and α is the second angle being added to x when using the double angle identity would also get us at a point where we have Rsin(x+y) = [some fraction], which would subsequently require a calculator after that step... $\endgroup$ – StopReadingThisUsername Feb 4 '16 at 11:34
  • $\begingroup$ Related : math.stackexchange.com/questions/757497/… $\endgroup$ – lab bhattacharjee Feb 4 '16 at 11:36
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$$\begin{align}3\sin x + 4\cos x &= \sqrt{3^2 + 4^2 }\sin\left(x+\arctan\frac{4}{3}\right) \\ &= 5\sin(x+\arctan\frac{4}{3})\end{align}$$


In general, if $$a\sin x + b\cos x = c$$

$$\frac{a\sin x}{\sqrt{a^2 + b^2}}+\frac{b \cos x}{\sqrt{a^2 + b^2}} = \frac{c}{\sqrt{a^2 + b^2}}$$

Let $\cos\phi = \dfrac{a}{\sqrt{a^2 + b^2}} \ \ \therefore \sin\phi = \dfrac{b}{\sqrt{a^2 + b^2}}$

Hence $\tan\phi = \dfrac{b}{a}$. Substituting in above eqn,

$$\sin x \cos \phi + \cos x \sin\phi = \frac{c}{\sqrt{a^2 + b^2}}$$

$$\sin (x + \phi)\sqrt{a^2 + b^2} = c$$ $$\implies c = \sqrt{a^2 + b^2}\ \sin\left(x + \arctan\frac{b}{a}\right)$$

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  • $\begingroup$ Yes, that is exactly the method I was referring to in the beginning. Can this approach described above not be a perfectly valid method to get to the solution? Furthermore, can this method be generalised? $\endgroup$ – StopReadingThisUsername Feb 4 '16 at 11:38
  • $\begingroup$ @Arjun I have added a possible way to show that it works for any a,b $\endgroup$ – Max Payne Feb 4 '16 at 13:06
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A different approach:

You can trade the trigonometric functions for a rational expression with

$$3\sin x+4\cos x=3\frac{2t}{1+t^2}+4\frac{1-t^2}{1+t^2}=2.$$

Thus, the quadratic equation

$$6t^2-6t-2=0,$$ which you can readily solve.

Then

$$\tan x=\frac{2t}{1-t^2}$$ and $$x=\arctan\frac{-12\pm2\sqrt{21}}5+k\pi.$$

To determine $k$, you must ensure that the angle lies in the quadrant of $(1-t^2,2t)$.


To avoid the computation of $\dfrac{2t}{1-t^2}$, you can also use

$$\tan x=\tan2\frac x2=\frac{2\tan\frac x2}{1-\tan^2\frac x2}=\frac{2t}{1-t^2}$$ or

$$\tan\frac x2=t,$$

$$x=2\arctan t+2k\pi.$$

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  • $\begingroup$ Nice use of the fact that $\sin \theta$ $\cos \theta$ and $\tan \theta$ can be written in terms of $\tan\frac{\theta}{2}$ $\endgroup$ – Max Payne Feb 4 '16 at 14:10
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    $\begingroup$ I'm sorry if this is a ridiculous question, but where did the first equation come from? $\endgroup$ – StopReadingThisUsername Feb 4 '16 at 14:28
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    $\begingroup$ @Arjun He used the formulas for $\sin\theta, \cos\theta, \tan\theta$ in terms of $\tan\frac{\theta}{2}$ ($t = \tan\frac{\theta}{2}$) $\endgroup$ – Max Payne Feb 4 '16 at 14:34
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By complex numbers:

By the complex definition of the trigonometric functions, setting $z=e^{ix}$,

$$3\frac{z+z^{-1}}{2i}+4\frac{z+z^{-1}}2=2.$$

Multiplying by $z$ and rearranging,

$$\left(2+i\frac32\right)z^2-2z+\left(2-i\frac32\right)=0.$$

Then solving the quadratic equation $$z=\frac{(8+6i)\pm\sqrt{21}(3-4i)}{25}.$$

The real and imaginary parts give $\cos x$ and $\sin x$ and $x$ is the imaginary part of the logarithm of $z$.

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