27. Формулы двойного и половинного аргумента

Теорема. Если cosalphane0, то

    [sin2alpha={2{rm tg},alphaover 1+{rm tg}^2alpha}, cos2alpha={1-{rm tg}^2alphaover 1+{rm tg}^2alpha} .]

Доказательство.

    [sin2alpha=2sinalphacosalpha=2{rm tg}, alphacdotcos^2alpha={2{rm tg}, alphaover 1+{rm tg}^2alpha},]

    [cos2alpha=2cos^2alpha-1={2over 1+{rm tg}^2alpha}-1={1-{rm tg}^2alphaover 1+{rm tg}^2alpha} .]

Теорема.

    [begin{array}{ll} {bf 1.}&displaystyle cos^2{alphaover 2}={1+cosalphaover 2}\[3mm] {bf 2.}&displaystyle sin^2{alphaover 2}={1-cosalphaover 2}\[3mm] {bf 3.}&displaystyle {rm tg}, {alphaover 2}={sinalphaover 1+cosalpha}quadleft(cos{alphaover 2}ne0right) end{array}]

Доказательство.

    [begin{array}{ll} displaystyle cosalpha=cos{2alphaover 2}=2cos^2{alphaover 2}-1Rightarrow{bf 1.}\[3mm] displaystyle =1-2sin^2{alphaover 2}Rightarrow{bf 2.}\[3mm] displaystyle {rm tg}, {alphaover 2}={sin{alphaover 2}over cos{alphaover 2}}={2sin{alphaover 2}cos{alphaover 2}over 2cos^2{alphaover 2}}={sinalphaover 1+cosalpha} . end{array}]

Задачи.

1) Упростите

1. displaystyle frac{sin25^{circ}sin65^{circ}}{cos40^{circ}}.

2. (cos^210^{circ}-cos^280^{circ})^2+cos^270^{circ}.

3. sin20^{circ}+2sin^235^{circ}.

4. displaystyle frac{1}{2sin10^{circ}}-2sin70^{circ}.

5. cos20^{circ}cos40^{circ}cos80^{circ}.

6. cosfrac{pi}{5}.

7. displaystyle frac{cos2alpha}{sinalpha+cosalpha}.

Ссылка на основную публикацию