Тригонометриялық теңбе-теңдіктерді пайдаланып өрнектерді түрлендіру
Кез-келген бұрыштың бір тригонометриялық функциясы берілген жағдайда, негізгі теңбе-теңдіктердің көмегімен басқа функцияларын тауып алуға болады.
Төменде негізгі тригонометриялық теңбе-теңдіктер көрсетілген:
\[\boxed{{{\sin }^2}\alpha + {{\cos }^2}\alpha = 1}\] \[\boxed{\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }}}\quad ,\quad \left( {\alpha \ne \frac{\pi }{2}(2n + 1),\,n \in \mathbb{Z}} \right)\] \[\boxed{\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }}}\quad ,\quad (\alpha \ne \pi n,\,n \in \mathbb{Z})\] \[\boxed{\tan \alpha \cdot \cot \alpha = 1}\quad ,\quad \left( {\alpha \ne \frac{{\pi n}}{2},\,n \in \mathbb{Z}} \right)\] \[\boxed{1 + {{\tan }^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}}\quad ,\quad \left( {\alpha \ne \frac{\pi }{2}(2n + 1),n \in \mathbb{Z}} \right)\] \[\boxed{1 + {{\cot }^2}\alpha = \frac{1}{{{{\sin }^2}\alpha }}}\quad ,\quad (\alpha \ne \pi n,n \in \mathbb{Z})\]
Жақшаның ішінде аргументтің мүмкін мәндер жиыны көрсетілген.
Тригонометриялық функциялардың кейбір бұрыштардағы мәндерін табуға мысалдар қарастырайық.
1.1. Есептеңіз.
1) $\dfrac{{1 + {{\tan }^6}\frac{\pi }{3}}}{{1 – {{\tan }^2}\frac{\pi }{3} + {{\tan }^4}\frac{\pi }{3}}}$.
Шешуі
$$\frac{{1 + {{\tan }^6}\frac{\pi }{3}}}{{1 – {{\tan }^2}\frac{\pi }{3} + {{\tan }^4}\frac{\pi }{3}}} = \frac{{1 + {{\left( {{{\tan }^2}\frac{\pi }{3}} \right)}^3}}}{{1 – {{\tan }^2}\frac{\pi }{3} + {{\tan }^4}\frac{\pi }{3}}} = $$ $$\frac{{\left( {1 + {{\tan }^2}\frac{\pi }{3}} \right)\left( {1 – {{\tan }^2}\frac{\pi }{3} + {{\tan }^4}\frac{\pi }{3}} \right)}}{{1 – {{\tan }^2}\frac{\pi }{3} + {{\tan }^4}\frac{\pi }{3}}} = 1 + {\tan ^2}\frac{\pi }{3} = 1 + {(\sqrt 3 )^2} = 4$$
Жауабы: 4.
2) $2\tan 180^\circ + \cos 180^\circ – {\cos ^2}15^\circ – {\sin ^2}15^\circ $.
Шешуі
$$2\tan 180^\circ + \cos 180^\circ – {\cos ^2}15^\circ – {\sin ^2}15^\circ = $$ $$ = 2\tan 180^\circ + \cos 180^\circ – \left( {{{\cos }^2}15^\circ + {{\sin }^2}15^\circ } \right) = 2 \cdot 0 – 1 – 1 = – 2$$
Жауабы: -2.
3) $\left| {\sin \left( { – \frac{1}{4}\pi } \right)} \right| + \cos \left( { – \frac{\pi }{4}} \right) + 1,5{\tan ^2}\left( { – \frac{\pi }{6}} \right) – \frac{1}{2}\left| {\cot \left( { – \frac{\pi }{4}} \right)} \right|$.
Шешуі
$$\left| {\sin \left( { – \frac{1}{4}\pi } \right)} \right| + \cos \left( { – \frac{\pi }{4}} \right) + 1,5{\tan ^2}\left( { – \frac{\pi }{6}} \right) – \frac{1}{2}\left| {\cot \left( { – \frac{\pi }{4}} \right)} \right| = $$ $$ = \sin \frac{\pi }{4} + \cos \frac{\pi }{4} + 1,5{\tan ^2}\frac{\pi }{6} – \frac{1}{2}\cot \frac{\pi }{4} = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} + 1,5 \cdot \frac{1}{3} – \frac{1}{2} = $$ $$ = \sqrt 2 + 0,5 – 0,5 = \sqrt 2 $$
Жауабы: $\sqrt 2 $.
4) $\sqrt {{{\left( {1 – 2\sin 45^\circ } \right)}^2}} – \sqrt {{{\left( {1 – 2\cos 30^\circ } \right)}^2}} $.
Шешуі
$$\sqrt {{{\left( {1 – 2\sin 45^\circ } \right)}^2}} – \sqrt {{{\left( {1 – 2\cos 30^\circ } \right)}^2}} = \left| {1 – 2 \cdot \frac{{\sqrt 2 }}{2}} \right| – \left| {1 – 2 \cdot \frac{{\sqrt 3 }}{2}} \right| = $$ $$ = \left| {1 – \sqrt 2 } \right| – \left| {1 – \sqrt 3 } \right| = – \left( {1 – \sqrt 2 } \right) + \left( {1 – \sqrt 3 } \right) = $$ $$ = – 1 + \sqrt 2 + 1 – \sqrt 3 = \sqrt 2 – \sqrt 3 $$
Жауабы: $\sqrt 2 – \sqrt 3 $.
5) $\sqrt {1 + {{\tan }^2}2} $.
Шешуі
$$\sqrt {1 + {{\tan }^2}2} = \sqrt {\frac{1}{{{{\cos }^2}2}}} = \frac{1}{{\left| {\cos 2} \right|}} = \left\| \begin{array}{l}\text{2 радиан бұрышы } \\ \text{ІІ ширекке тиесілі,} \\ \text{сондықтан } \cos 2 \lt 0 \end{array} \right\| = – \frac{1}{{\cos 2}}$$
Жауабы: $ – \dfrac{1}{{\cos 2}}$.
6) $\sqrt{1-\cos ^2 4}$.
Шешуі
$$\sqrt {1 – {{\cos }^2}4} = \sqrt {{{\sin }^2}4} = \left| {\sin 4} \right| = \left\| \begin{array}{l}\text{4 радиан бұрышы } \\ \text{ІІІ ширекке тиесілі,} \\ \text{сондықтан } \sin 4 \lt 0 \end{array} \right\| = – \sin 4$$
Жауабы: $ – \sin 4$.
7) $1+\sin ^4 10^{\circ}+\cos ^2 10^{\circ}+\sin ^2 10^{\circ} \cdot \cos ^2 10^{\circ}$.
Шешуі
$$1 + {\sin ^4}10^\circ + {\cos ^2}10^\circ + {\sin ^2}10^\circ \cdot {\cos ^2}10^\circ = $$ $$ = 1 + {\cos ^2}10^\circ + {\sin ^2}10^\circ \cdot \left( {{{\sin }^2}10^\circ + {{\cos }^2}10^\circ } \right) = $$ $$ = 1 + {\cos ^2}10^\circ + {\sin ^2}10^\circ = 1 + 1 = 2$$
Жауабы: 2.
1.2. Есептеңіз.
1) $2 \sin ^2 \alpha+\sqrt{2} \cos \alpha+\operatorname{tg} \alpha$ , егер $\operatorname{ctg} \alpha=1, \quad 0 \lt \alpha \lt \frac{\pi}{2}$.
Шешуі
$\operatorname{ctg} \alpha=1$ және $0 \lt \alpha \lt \frac{\pi}{2}$ шарттарынан $\alpha=\frac{\pi}{4}$ екенін білеміз.
Онда $$2 \sin ^2 \alpha+\sqrt{2} \cos \alpha+\operatorname{tg} \alpha=2 \sin ^2 \frac{\pi}{4}+\sqrt{2} \cos \frac{\pi}{4}+\operatorname{tg} \frac{\pi}{4}=$$ $$=2 \cdot\left(\frac{\sqrt{2}}{2}\right)^2+\sqrt{2} \cdot \frac{\sqrt{2}}{2}+1=1+1+1=3$$
Жауабы: 3.
2) $\cos \alpha$, егер $\operatorname{ctg} \alpha=-\frac{1}{2}, \quad \frac{\pi}{2} \lt \alpha \lt \pi$.
Шешуі
$\alpha$ — ІІ ширектің бұрышы екенін ескеріп, бұрыштың синусы мен косинусын тауып аламыз.
$$1 + {{\mathop{\rm ctg}\nolimits} ^2}\alpha = \frac{1}{{{{\sin }^2}\alpha }},$$ $$1\frac{1}{4} = \frac{1}{{{{\sin }^2}\alpha }},$$ $${\sin ^2}\alpha = \frac{4}{5} \Rightarrow \sin \alpha = \frac{{2\sqrt 5 }}{5},$$ $$\cos \alpha = \pm \sqrt {1 – {{\sin }^2}\alpha } \Rightarrow \cos \alpha = – \sqrt {1 – \frac{4}{5}} = – \frac{{\sqrt 5 }}{5}.$$
Жауабы: $ – \frac{{\sqrt 5 }}{5}$.
3) $\dfrac{\operatorname{tg} \alpha}{\operatorname{tg} \alpha+\operatorname{ctg} \alpha}$, егер $ \cos \alpha=-0,4$.
Шешуі
$\cos \alpha=-0,4$ $$\frac{{\tan \alpha }}{{\tan \alpha + \cot \alpha }} = \frac{{\tan \alpha }}{{\tan \alpha + \frac{1}{{\tan \alpha }}}} = \frac{{{{\tan }^2}\alpha }}{{{{\tan }^2}\alpha + 1}} = \frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }} = $$ $$ = {\sin ^2}\alpha = 1 – {\cos ^2}\alpha = 1 – {( – 0,4)^2} = 0,84$$
Жауабы: 0,84.
4) $\dfrac{3 \cos \alpha+5 \sin \alpha}{2 \cos \alpha-\sin \alpha}$, егер $\operatorname{tg} \alpha=1$.
Шешуі
$$\frac{3 \cos \alpha+5 \sin \alpha}{2 \cos \alpha-\sin \alpha}=\frac{\frac{3 \cos \alpha}{\cos \alpha}+\frac{5 \sin \alpha}{\cos \alpha}}{\frac{2 \cos \alpha}{\cos \alpha}-\frac{\sin \alpha}{\cos \alpha}}=\frac{3+5 \operatorname{tg} \alpha}{2-\operatorname{tg} \alpha}=\frac{3+5 \cdot 1}{2-1}=8$$
Жауабы: .
5) $\dfrac{\sin \alpha \cdot \cos \alpha}{\sin ^2 \alpha-\cos ^2 \alpha}$, егер $\operatorname{ctg} \alpha=\frac{3}{4}$.
Шешуі
$$\frac{{\sin \alpha \cdot \cos \alpha }}{{{{\sin }^2}\alpha – {{\cos }^2}\alpha }} = \frac{{\frac{{\sin \alpha \cdot \cos \alpha }}{{{{\sin }^2}\alpha }}}}{{\frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} – \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }}}} = \frac{{\cot \alpha }}{{1 – {{\cot }^2}\alpha }} = \frac{{\frac{3}{4}}}{{1 – \frac{9}{{16}}}} = \frac{{12}}{7}$$
Жауабы: $\dfrac{12}{7}$.
6) $\dfrac{\sin ^2 \alpha-3 \cos ^2 \alpha}{2 \sin ^2 \alpha+\cos ^2 \alpha}$, егер $\operatorname{tg} \alpha=3$ .
Шешуі
$$\frac{{{{\sin }^2}\alpha – 3{{\cos }^2}\alpha }}{{2{{\sin }^2}\alpha + {{\cos }^2}\alpha }} = \frac{{\frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} – \frac{{3{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }}}}{{\frac{{2{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }}}} = \frac{{{{\tan }^2}\alpha – 3}}{{2{{\tan }^2}\alpha + 1}} = \frac{{9 – 3}}{{2 \cdot 9 + 1}} = \frac{6}{{19}}$$
Жауабы: $\dfrac{6}{19}$.
7) $\sin \alpha \cdot \cos \alpha $, егер $ \sin \alpha+\cos \alpha=a$.
Шешуі
$$\sin \alpha + \cos \alpha = a$$ $${(\sin \alpha + \cos \alpha )^2} = {a^2}$$ $${\sin ^2}\alpha + {\cos ^2}\alpha + 2\sin \alpha \cdot \cos \alpha = {a^2}$$ $$2\sin \alpha \cdot \cos \alpha = {a^2} – 1 \Rightarrow \sin \alpha \cdot \cos \alpha = \frac{{{a^2} – 1}}{2}$$
Жауабы: $\dfrac{{a^2} – 1}{2}$.
1.3. Ықшамдаңыз.
1) $\dfrac{1}{(\operatorname{tg} \alpha+\operatorname{ctg} \alpha) \cdot \sin ^2 \alpha}$.
Шешуі
$$\frac{1}{{(\tan \alpha + \cot \alpha ) \cdot {{\sin }^2}\alpha }} = \frac{1}{{\left( {\frac{{\sin \alpha }}{{\cos \alpha }} + \frac{{\cos \alpha }}{{\sin \alpha }}} \right) \cdot {{\sin }^2}\alpha }} = $$ $$ = \frac{{\sin \alpha \cdot \cos \alpha }}{{\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) \cdot {{\sin }^2}\alpha }} = \frac{{\cos \alpha }}{{\sin \alpha }} = \cot \alpha $$
2) $\sin ^4 \alpha+\sin ^2 \alpha \cdot \cos ^2 \alpha+\cos ^2 \alpha$.
Шешуі
$${\sin ^4}\alpha + {\sin ^2}\alpha \cdot {\cos ^2}\alpha + {\cos ^2}\alpha = {\sin ^2}\alpha \cdot \left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + {\cos ^2}\alpha = $$ $$ = {\sin ^2}\alpha + {\cos ^2}\alpha = 1$$
3) $\dfrac{\sin \alpha+\cos \alpha}{1+2 \sin \alpha \cdot \cos \alpha}$.
Шешуі
$$\frac{{\sin \alpha + \cos \alpha }}{{1 + 2\sin \alpha \cdot \cos \alpha }} = \frac{{\sin \alpha + \cos \alpha }}{{{{\sin }^2}\alpha + {{\cos }^2}\alpha + 2\sin \alpha \cdot \cos \alpha }} = $$ $$ = \frac{{\sin \alpha + \cos \alpha }}{{{{(\sin \alpha + \cos \alpha )}^2}}} = \frac{1}{{\sin \alpha + \cos \alpha }}$$
4) $\dfrac{\sin \alpha}{1-\cos \alpha}-\frac{\sin \alpha}{1+\cos \alpha}$.
Шешуі
$$\frac{{\sin \alpha }}{{1 – \cos \alpha }} – \frac{{\sin \alpha }}{{1 + \cos \alpha }} = \frac{{\sin \alpha + \sin \alpha \cdot \cos \alpha – \sin \alpha + \sin \alpha \cdot \cos \alpha }}{{(1 – \cos \alpha )(1 + \cos \alpha )}} = $$ $$ = \frac{{2\sin \alpha \cdot \cos \alpha }}{{1 – {{\cos }^2}\alpha }} = \frac{{2\sin \alpha \cdot \cos \alpha }}{{{{\sin }^2}\alpha }} = \frac{{2\cos \alpha }}{{\sin \alpha }} = 2\cot \alpha $$
5) $1+\dfrac{1-\cos ^2 \alpha+\operatorname{tg}^2 \alpha \cdot \cos ^2 \alpha}{\sin ^2 \alpha}$.
Шешуі
$$1 + \frac{{1 – {{\cos }^2}\alpha + {{\tan }^2}\alpha \cdot {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = 1 + \frac{{{{\sin }^2}\alpha + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} \cdot {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = 1 + \frac{{2{{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} = 3$$
6) $\sin ^6 \alpha+\cos ^6 \alpha+3 \sin ^2 \alpha \cdot \cos ^2 \alpha$.
Шешуі
$${\sin ^6}\alpha + {\cos ^6}\alpha + 3{\sin ^2}\alpha \cdot {\cos ^2}\alpha = {\left( {{{\sin }^2}\alpha } \right)^3} + {\left( {{{\cos }^2}\alpha } \right)^3} + 3{\sin ^2}\alpha \cdot {\cos ^2}\alpha = $$ $$ = \left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)\left( {{{\sin }^4}\alpha – {{\sin }^2}\alpha \cdot {{\cos }^2}\alpha + {{\cos }^4}\alpha } \right) + 3{\sin ^2}\alpha \cdot {\cos ^2}\alpha = $$ $$ = {\sin ^4}\alpha + 2{\sin ^2}\alpha \cdot {\cos ^2}\alpha + {\cos ^4}\alpha = {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^2} = 1$$