Тест по предмету «Алгебра» на тему «Доказательство тригонометрических тождеств»

1. Докажите тождество: \(\sin^4 \alpha + \cos^4 \alpha = 1 - 2\sin^2 \alpha \cos^2 \alpha\).

2. Докажите тождество при \(\sin \alpha \neq 0\): \(\dfrac{1-\cos \alpha}{\sin \alpha} = \dfrac{\sin \alpha}{1+\cos \alpha}\).

3. Докажите тождество при \(\sin \alpha \cos \alpha \neq 0\): \(\operatorname{tg}\alpha + \operatorname{ctg}\alpha = \dfrac{1}{\sin \alpha \cos \alpha}\).

4. Докажите тождество: \((\sin \alpha + \cos \alpha)^2 = 1 + 2\sin \alpha \cos \alpha\).

5. Докажите тождество при \(\cos \alpha \neq 0\): \(\cos^2 \alpha - \sin^2 \alpha = \dfrac{1-\operatorname{tg}^2\alpha}{1+\operatorname{tg}^2\alpha}\).

6. Докажите тождество при \(\cos \alpha \neq 0\): \(\dfrac{1}{1+\operatorname{tg}^2\alpha} = \cos^2 \alpha\).

7. Докажите тождество при \(\sin 2\alpha \neq 0\): \(\dfrac{1-\cos 2\alpha}{\sin 2\alpha} = \operatorname{tg}\alpha\).

8. Докажите тождество при \(\cos \alpha \neq 0\), \(1+\operatorname{tg}\alpha \neq 0\), \(1+\sin 2\alpha \neq 0\): \(\dfrac{\cos 2\alpha}{1+\sin 2\alpha} = \dfrac{1-\operatorname{tg}\alpha}{1+\operatorname{tg}\alpha}\).

Ответы

1. \(\sin^4\alpha+\cos^4\alpha=(\sin^2\alpha+\cos^2\alpha)^2-2\sin^2\alpha\cos^2\alpha=1-2\sin^2\alpha\cos^2\alpha\).

2. \(\dfrac{1-\cos\alpha}{\sin\alpha}=\dfrac{(1-\cos\alpha)(1+\cos\alpha)}{\sin\alpha(1+\cos\alpha)}=\dfrac{\sin^2\alpha}{\sin\alpha(1+\cos\alpha)}=\dfrac{\sin\alpha}{1+\cos\alpha}\).

3. \(\operatorname{tg}\alpha+\operatorname{ctg}\alpha=\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha\cos\alpha}=\dfrac{1}{\sin\alpha\cos\alpha}\).

4. \((\sin\alpha+\cos\alpha)^2=\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha=1+2\sin\alpha\cos\alpha\).

5. \(\dfrac{1-\operatorname{tg}^2\alpha}{1+\operatorname{tg}^2\alpha}=\dfrac{1-\frac{\sin^2\alpha}{\cos^2\alpha}}{1+\frac{\sin^2\alpha}{\cos^2\alpha}}=\dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha+\sin^2\alpha}=\cos^2\alpha-\sin^2\alpha\).

6. \(\dfrac{1}{1+\operatorname{tg}^2\alpha}=\dfrac{1}{1+\frac{\sin^2\alpha}{\cos^2\alpha}}=\dfrac{\cos^2\alpha}{\cos^2\alpha+\sin^2\alpha}=\cos^2\alpha\).

7. \(\dfrac{1-\cos2\alpha}{\sin2\alpha}=\dfrac{2\sin^2\alpha}{2\sin\alpha\cos\alpha}=\dfrac{\sin\alpha}{\cos\alpha}=\operatorname{tg}\alpha\).

8. \(\dfrac{\cos2\alpha}{1+\sin2\alpha}=\dfrac{\cos^2\alpha-\sin^2\alpha}{(\sin\alpha+\cos\alpha)^2}=\dfrac{\cos\alpha-\sin\alpha}{\cos\alpha+\sin\alpha}=\dfrac{1-\operatorname{tg}\alpha}{1+\operatorname{tg}\alpha}\).