Trigonometry equation, odd-function.

Question

So I have the following equation:

$\sin\left(x-\frac{\pi}{6}\right) + \cos\left(x+\frac{\pi}{4}\right)=0$

It should be solved using the fact that Sin is an odd function, I can not really get the gripp of how and what I need to do? Any sugestions?

Answer 1

$$\cos(x+\frac\pi4)=\sin(\frac\pi4-x)$$ and the equation turns to

$$\sin(x-\frac\pi6)=\sin(x-\frac\pi4),$$

i.e.

$$2x-\frac{5\pi}{12}=(2k+1)\pi.$$

Answer 2

So i found the solution using the fact that Sin is odd, here is my solution:

$\sin\left (x-\frac{\pi}{6}\right ) + \cos\left (x+\frac{\pi}{4}\right ) = 0\Leftrightarrow \cos\left (x+\frac{\pi}{4}\right ) = -\sin\left (x-\frac{\pi}{6}\right )$

Since sin is an odd function, $-\sin\left (x-\frac{\pi}{6}\right ) = \sin\left (x-\frac{\pi}{6} + \pi\right )$

This the gives the following: $\cos\left (x+\frac{\pi}{4}\right ) = \sin\left (x-\frac{\pi}{6} + \pi\right ))\Leftrightarrow \cos\left (x+\frac{\pi}{4}\right ) = \cos\left (\frac{\pi}{2} -\frac{5\pi}{6} -x\right )$

This happends when: $x + \frac{\pi}{4} = \frac{-2\pi}{6}-x+n2\pi \Leftrightarrow x=\frac{-7\pi}{24}+n\pi$

Thanks alot for the help, special thanx to @BLAZE, you gave me the hints to find the solution.

Answer 3

Sorry that this approach does not use the oddness of the Sine function, after I finishing writing this it was too late for me to turn back and utilize that property.

Using trigonometric addition formulae as mentioned in the comments:

$$\sin(A \pm B)= \sin A\cos B\pm\cos A\sin B$$ $$\cos(A \pm B)= \cos A\cos B\mp\sin A\sin B$$ Write $$\sin\left(x-\frac{\pi}{6}\right) + \cos\left(x+\frac{\pi}{4}\right)=0$$

as $$\sin(x)\cos\left(\frac{\pi}{6}\right)-\cos(x)\sin\left(\frac{\pi}{6}\right)+\cos(x)\cos\left(\frac{\pi}{4}\right)-\sin(x)\sin\left(\frac{\pi}{4}\right)$$

$$=\frac{\sqrt{3}}{2}\sin(x)-\frac{1}{2}\cos(x)+\frac{1}{\sqrt{2}}\cos(x)-\frac{1}{\sqrt{2}}\sin(x)$$

$$=\cos(x)\left(\frac{1}{\sqrt{2}}-\frac{1}{2}\right)+\sin(x)\left(\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\right)$$

$$\implies\cos(x)\left(\frac{2-\sqrt{2}}{2\sqrt{2}}\right)+\sin(x)\left(\frac{\sqrt{6}-2}{2\sqrt{2}}\right)=0$$

Since $$\cos(x)\ne 0$$

$$\implies\left(\frac{2-\sqrt{2}}{2\sqrt{2}}\right)+\tan(x)\left(\frac{\sqrt{6}-2}{2\sqrt{2}}\right)=0$$

$$\implies\tan(x)\left(\frac{\sqrt{6}-2}{2\sqrt{2}}\right)=\left(\frac{\sqrt{2}-2}{2\sqrt{2}}\right)$$

$$x=\tan^{-1}\left(\frac{\sqrt{2}-2}{\sqrt{6}-2}\right)$$

$$x=\tan^{-1}\left(\frac{(\sqrt{2}-2)(\sqrt{6} + 2)}{(\sqrt{6}-2)(\sqrt{6} + 2)}\right)=\tan^{-1}\left(\frac{(\sqrt{2}-2)(\sqrt{6} + 2)}{2}\right)$$

$$=\tan^{-1}{(\sqrt{2}+\sqrt{3}-\sqrt{6}-2)}+k\pi$$

where $k \in \mathbb{N}$

Answer 4

More unorthodoxly, we first take the midpoint of $x-\frac{\pi}{6}$ and $x + \frac{\pi}{4}$ before applying the angle sum and difference identities:

Let $A = x + \frac{\pi}{24}$ (the midpoint). Then we need

$$\sin\left(A - \frac{5\pi}{24}\right) + \cos \left(A + \frac{5\pi}{24}\right) = 0$$ $$\sin A\cos\frac{5\pi}{24} - \cos A\sin\frac{5\pi}{24} + \cos A\cos\frac{5\pi}{24} - \sin A\sin\frac{5\pi}{24} = 0$$ $$\sin A\left(\cos \frac{5\pi}{24} - \sin \frac{5\pi}{24}\right) + \cos A\left(-\sin\frac{5\pi}{24} + \cos \frac{5\pi}{24}\right) = 0$$

$$\left(-\sin\frac{5\pi}{24} + \cos \frac{5\pi}{24}\right)\left(\sin A + \cos A\right) = 0$$

so we need

$$\sin A + \cos A = 0$$ $$\tan A = -1$$ $$\tan \left( x + \frac{\pi}{24}\right) = -1$$

which can be solved directly as per normal.