# Inconsitent general solutions of trigonometric equations

General solutions of trigonometric equations are given by:

If $\sin(x) = \sin(y)$, then $x = n \pi + (-1)^ny$
If $\cos(x) = \cos(y)$, then $x = 2n \pi \pm y$

If we consider an example,

$$\sin(x) = \sin(30^\circ) \\ \implies x = n \pi + (-1)^n30^\circ \\ [where \pi=180^\circ]$$ which can be worked out for the values of n from -1 to 10, which are

$$\begin{array}{cc} n&x (\text{degrees})\\ -1&-210\\ 0&30\\ 1&150\\ 2&390\\ 3&510\\ 4&750\\ 5&870\\ 6&1110\\ 7&1230\\ 8&1470\\ 9&1590\\ 10&1830\\ \end{array}$$

Now $$\sin(x)=\sin(30^\circ)=\cos(60^\circ) \\ \implies \sqrt{1-\cos^2x} = cos(60^\circ) \\ \implies 1-\cos^2x = (\frac{1}{2})^2 \\ \implies \cos^2x = 1 - (\frac{1}{2})^2 = \frac{3}{4} \\ \implies \cos(x) = \frac{\sqrt{3}}{2} = \cos(30^\circ)$$

So for $\cos(x) = \cos(30^\circ)$, we have $$x = 2n\pi \pm 30^\circ$$

Solving for values yields

$$\begin{array}{cc} n&x (\text{degrees})\\ -1&-390,330\\ 0&-30,30\\ 1&330,390\\ 2&690,750\\ 3&1050,1110\\ 4&1410, 1470\\ 5&1770, 1830\\ 6&2130,2190\\ 7&2490,2550\\ 8&2850, 2910\\ 9&3210,3270\\ 10&3570,3630\\ \end{array}$$

If you compare the two, you'll find that only some of the values exist in both solutions, and they both have values that don't exist in the other value set.

I was expecting both of these to be exactly equal. What is the cause of this inconsistency?

One "practical" answer that I was able to come up with is :

If you're provided with the information that $\sin(x) = \sin(30^\circ)$, AND $\cos(x) = \cos(30^\circ)$, then both those conditions would invalidate the extra results in both value sets (since those values will satisfy only one equation),

Which leads me to the conclusion that both statements $\sin(x) = \sin(30^\circ)$ and $\cos(x) = \cos(30^\circ)$ are not equivalent.
But does that mean there's a fault in the conversion of equation from sin terms to cos terms?

If so, how could I start with $\sin(x)=\sin(y)$, correctly convert these to cosine terms and proceed to get the the same results as I would with sin terms?

• It is because you take square root of $cos^2(x)$, assuming $cos(x)$ is positive. – crbah Mar 11 '16 at 13:30
• Angles should be in radians: nπ+30° is inconsistent. – Bernard Mar 11 '16 at 13:48
• @Bernard fixed that technicality – Peeyush Kushwaha Mar 21 '16 at 6:41
• @corbah You are correct. Upon considering the negative values, they give the solutions for x that positive values didn't, e.g. $x = 210^\circ$. However, all the extra values, they'll just simply have to be rejected because they didn't fit the original equation of sin x = sin 30deg, correct? – Peeyush Kushwaha Mar 21 '16 at 6:47
• @Bernard - Actually, it's no more inconsistent than writing 3ft + 4 in is inconsistent. – steven gregory Jan 1 at 16:49

Its best in situations like this to think in terms of the unit circle.

$\sin x=\sin y,x=n\pi+(-1)^ny$ where $n$ is any integer

For even $n=2m$(say) $x=2m\pi+y\ \ \ \ (1)$

For odd $n=2m+1$(say) $x=(2m+1)\pi-y$

Now let $X=\dfrac\pi2-x, Y=\dfrac\pi2-y\implies\cos x=\cos y$

$(1)\implies\dfrac\pi2-X=2m\pi+\dfrac\pi2-Y\iff X=-2m\pi+Y$

$(2)\implies\dfrac\pi2-X=(2m+1)\pi-\left(\dfrac\pi2-Y\right)\iff X=-2m\pi-Y$

which is same as $$\cos X=\cos Y\implies X=2r\pi\pm Y$$ where $r$ is any integer

A) If $\sin(x) = \sin(y)$, then $x = n \pi + (-1)^ny$
B) If $\cos(x) = \cos(y)$, then $x = 2n \pi \pm y$

A) and B) are given different independent problems and so they need not be consistent.