Rahisisha mlinganyo uliotolewa.

Mlinganyo uliotolewa ni:

$$\sin (30^\circ - 45^\circ) = \cos (\theta + 15^\circ)$$

Rahisisha upande wa kushoto:

$$\sin (-15^\circ) = \cos (\theta + 15^\circ)$$

Tunajua kwamba $\sin x = \cos (90^\circ - x)$.
Hivyo, $\sin (-15^\circ)$ inaweza kuandikwa kama:

$$\sin (-15^\circ) = \cos (90^\circ - (-15^\circ)) = \cos (90^\circ + 15^\circ) = \cos (105^\circ)$$

Sasa mlinganyo unakuwa:

$$\cos (105^\circ) = \cos (\theta + 15^\circ)$$

Ikiwa $\cos A = \cos B$, basi $A = \pm B + 360^\circ n$, ambapo $n$ ni nambari kamili.
Hivyo, tuna visa viwili:

• Kesi ya 1: $105^\circ = \theta + 15^\circ + 360^\circ n$ Chukua $n=0$ kwa suluhisho rahisi: $105^\circ = \theta + 15^\circ$ $\theta = 105^\circ - 15^\circ$ $\theta = 90^\circ$ Ikiwa $\theta = 90^\circ$, basi $\tan \theta = \tan(90^\circ)$, ambayo haijafafanuliwa. Kwa kawaida, hii si jibu linalotarajiwa kwa swali la "tafuta $\tan \theta$".

• Kesi ya 2: $105^\circ = -(\theta + 15^\circ) + 360^\circ n$ Chukua $n=0$ kwa suluhisho rahisi: $105^\circ = -\theta - 15^\circ$ $\theta = -15^\circ - 105^\circ$ $\theta = -120^\circ$

Tunatumia $\theta = -120^\circ$.

$$\tan (-120^\circ)$$

Tunajua kwamba $\tan (-x) = -\tan x$.
Hivyo, $\tan (-120^\circ) = -\tan (120^\circ)$.
Pia, $\tan (120^\circ) = \tan (180^\circ - 60^\circ)$. Kwa kuwa $120^\circ$ iko katika roboduara ya pili, $\tan$ ni hasi.
$\tan (180^\circ - 60^\circ) = -\tan (60^\circ)$.
Tunajua $\tan (60^\circ) = \sqrt{3}$.
Hivyo, $\tan (120^\circ) = -\sqrt{3}$.
Kwa hiyo, $\tan (-120^\circ) = -(-\sqrt{3}) = \sqrt{3}$.

Jibu la mwisho ni $\sqrt{3}$.