– yırım trigonometriyalıq ańlatpalardı logarifmlew ushın qolaylı túrge keltiriw maqsetinde túrlendiriwlerge kirgiziletuǵın múyesh. Máselen, $A\pm B$ túrindegi trigonometriyalıq ańlatpalar járdemshi $\varphi$ múyeshin kirgiziw arqalı tómendegishe túrlendiriledi: $$A\pm B=A\left(1\pm \dfrac{B}{A}\right)=A(\text{tg}45^{\circ}\pm\text{tg}\varphi)=\dfrac{A\sin(45^{\circ}\pm\varphi)}{\cos45^{\circ}\cdot\cos\varphi}$$ yamasa $$A\pm B=A\left(1\pm \dfrac{B}{A}\right)=A(\text{ctg}45^{\circ}\pm\text{сtg}\varphi)=\dfrac{A\sin(45^{\circ}\pm\varphi)}{\sin45^{\circ}\cdot\sin\varphi}, $$ bunda $\varphi$ járdemshi múyeshi $\text{tg}\varphi=\dfrac{B}{A}$ yamasa $\text{ctg}\varphi=\dfrac{B}{A}$ teńliklerinen tablica boyınsha anıqlanadı. Bul usıl menen tómendegi túrdegi ańlatpalar kóbeymege ańsat túrlendiriledi: $a+b\sin\alpha$; $a+b\cos\alpha$; $a+b\text{tg}\alpha$; $a+b\text{ctg}\alpha$; $a-b\text{tg}^2\alpha$; $a\sin\alpha+b\cos\alpha$ hám basqa da ańlatpalar. Al, $a\sin\alpha+b\cos\alpha$ túrindegi ańlatpalardı kóbeymege túrlendiriw ushın $\sqrt{a^2+b^2}$ ańlatpasın qawsırmanıń sırtına shıǵarıw múmkin. Sonda $$ a\sin\alpha+b\cos\alpha=\sqrt{a^2+b^2}\left(\dfrac{a}{\sqrt{a^2+b^2}}\sin\alpha+\dfrac{b}{\sqrt{a^2+b^2}}\cos\alpha\right).$$ Endi $$\left(\dfrac{a}{\sqrt{a^2+b^2}}\right)^2+\left(\dfrac{b}{\sqrt{a^2+b^2}}\right)^2=1$$ teńligin esapqa alıp, $$ \dfrac{a}{\sqrt{a^2+b^2}}=\cos\varphi, \quad \dfrac{b}{\sqrt{a^2+b^2}}=\sin\varphi $$ belgilewlerin kirgiziwge boladı (bunda $\varphi$ - járdemshi múyesh). Sonda $a\sin\alpha+b\cos\alpha=A\sin(\alpha+\varphi),$ bunda $A=\sqrt{a^2+b^2}$. Járdemshi $\varphi$ múyeshin kirgiziw usılı menen $$a\sin x+b\cos x=c, \quad \text{bunda} \quad a\neq0, \ b\neq0, \ c\neq 0$$ túrindegi trigonometriyalıq teńlemeler sheshiledi. Haqıyqatında da, joqarıdaǵıday túrlendiriw jolı menen bul berilgen teńlemeni $$\sin x \cos\varphi+\cos x\sin\varphi=\dfrac{c}{\sqrt{a^2+b^2}}, \quad \sin(x+\varphi)=\dfrac{c}{\sqrt{a^2+b^2}} $$ túrine keltiriwge boladı. Bunnan, eger $\left|\dfrac{c}{\sqrt{a^2+b^2}}\right|\leq 1$ bolsa, onda $$ x=\pi k-\varphi+(-1)^k \text{arcsin}\dfrac{c}{\sqrt{a^2+b^2}}. $$ Bundaǵı $\varphi$ múyeshi $$\cos\varphi=\dfrac{a}{\sqrt{a^2+b^2}} \ \text{hám} \ \sin\varphi=\dfrac{b}{\sqrt{a^2+b^2}}$$ teńliklerinen anıqlanadı.