Тригонометрични тъждества - задачи с решения

Решение:
Отговор: $\frac{2}{\sin x}=\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}$

$\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=\frac{\sin ^{2}x+\left( 1+\cos x\right) ^{2}}{\sin x\left( 1+\cos x\right) }=\frac{\sin ^{2}x+1+2\cos x+\cos ^{2}x}{\sin x\left( 1+\cos x\right) }=\frac{1+2\cos x+\left( \sin ^{2}x+\cos ^{2}x\right) }{\sin x\left( 1+\cos x\right) }$

Така $=\frac{2+2\cos x}{\sin x\left( 1+\cos x\right) }=\frac{2\left( 1+\cos x\right) }{\sin x\left( 1+\cos x\right) }=\frac{2}{\sin x}$

Тогава $\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=\frac{2}{\sin x}$

Решение:
Отговор: $\frac{1-\sin x}{\cos x}=\frac{\cos x}{1+\sin x}$

$\frac{\cos x}{1+\sin x}=\frac{\cos ^{2}x}{\cos x\left( 1+\sin x\right) }=\frac{1-\sin ^{2}x}{\cos x\left( 1+\sin x\right) }=\frac{\left( 1-\sin x\right) \left( 1+\sin x\right) }{\cos x\left( 1+\sin x\right) }=\frac{1-\sin x}{\cos x}$

Доказахме, че $\frac{1-\sin x}{\cos x}=\frac{\cos x}{1+\sin x}$

Решение:
Отговор: $\frac{\sin x-\cos x}{\sin x+\cos x}=\frac{\text{tg} x-1}{\text{tg} x+1}$

$\frac{\sin x-\cos x}{\sin x+\cos x}=\frac{\frac{1}{\cos x}-\frac{1}{\sin x}}{\frac{1}{\cos x}+\frac{1}{\sin x}}=\frac{\frac{\sin x-\cos x}{\sin x\cos x}}{\frac{\sin x+\cos x}{\sin x\cos x}}=\frac{\frac{\sin x-\cos x}{\cos x}}{\frac{\sin x+\cos x}{\cos x}}=\frac{\frac{\sin x}{\cos x}-1}{\frac{\sin x}{\cos x}+1}=\frac{\text{tg} x-1}{\text{tg} x+1}$

Доказахме, че $\frac{\sin x-\cos x}{\sin x+\cos x}=\frac{\text{tg} x-1}{\text{tg}x+1}$

Решение:
Отговор: $\frac{\text{tg} x-\sin x}{\sin^{3}x}=\frac{1}{\cos x+\cos^{2}x}$

$\frac{\text{tg} x-\sin x}{\sin^{3}x}=\frac{\frac{\sin x}{\cos x}-\sin x}{\sin ^{3}x}=\frac{\sin x-\sin x\cos x}{\cos x\sin ^{3}x}=\frac{\sin x\left( 1-\cos x\right) }{\cos x\sin ^{3}x}$
$= \frac{1-\cos x}{\cos x\sin^{2}x}=\frac{1-\cos x}{\cos x\left( 1-\cos ^{2}x\right) }=\frac{1}{\cos x\left( 1+\cos x\right) }=\frac{1}{\cos x+\cos ^{2}x}$

Доказахме, че $\frac{\text{tg} x-\sin x}{\sin^{3}x}=\frac{1}{\cos x+\cos^{2}x}$

Решение:
Отговор: $\frac{\cos ^{3}x-\sin ^{3}x}{\cos x-\sin x}=1+\sin x\cos x$

$\frac{\cos ^{3}x-\sin^{3}x}{\cos x-\sin x} =\frac{\left( \cos x-\sin x\right) \left( \cos^{2}x+\cos x\sin x+\sin ^{2}x\right) }{\cos x-\sin x}=\left( \cos ^{2}x+\sin ^{2}x\right) +\cos x\sin x=$
$=1+\cos x\sin x$

Доказахме, че $\frac{\cos^{3}x-\sin^{3}x}{\cos x-\sin x}=1+\sin x\cos x$

Решение:
Отговор: $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{\sin \theta +1}{\cos \theta }$

$\frac{\sin \theta +1\ }{\cos \theta }=\frac{\left( \sin \theta +1\right) \left( \sin \theta +\cos \theta -1\right) }{\cos \theta \left( \sin \theta +\cos \theta -1\right) }=\frac{\sin ^{2}\theta +\sin \theta \cos \theta +\cos \theta -1}{\cos \theta \left( \sin \theta +\cos \theta -1\right) }= \frac{\left( \sin ^{2}\theta -1\right) +\sin \theta \cos \theta +\cos \theta }{\cos \theta \left( \sin \theta +\cos \theta -1\right) }$

$=\frac{-\cos^{2}\theta +\sin \theta \cos \theta +\cos \theta }{\cos \theta \left( \sin \theta +\cos \theta -1\right) }=\frac{\cos \theta \left( \sin \theta -\cos \theta +1\right) }{\cos \theta \left( \sin \theta +\cos \theta -1\right) }=\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}$
Доказахме, че $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{\sin \theta +1\ }{\cos \theta }$

Решение:
Отговор: $\text{tg} \left( \alpha +\beta \right) =\frac{\text{tg} \alpha +\text{tg} \beta }{1-\text{tg} \alpha \text{tg} \beta }$

$\text{tg} \left( \alpha +\beta \right) =\frac{\sin \left( \alpha +\beta \right) }{\cos \left( \alpha +\beta \right) }=\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta -\sin \alpha \sin \beta }$

Разделяме числителя и знаменателя на $\cos \alpha \cos \beta $

So $\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta -\sin \alpha \sin \beta }=\frac{\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta }}{\frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}=\frac{\frac{\sin \alpha \cos \beta }{\cos \alpha \cos \beta }+\frac{\cos \alpha \sin \beta }{\cos \alpha \cos \beta }}{\frac{\cos \alpha \cos \beta }{\cos \alpha \cos \beta }-\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}=\frac{\text{tg} \alpha +\text{tg} \beta }{1-\text{tg} \alpha \text{tg} \beta }$

We proved that $\text{tg} \left( \alpha +\beta \right) =\frac{\text{tg} \alpha +\text{tg} \beta }{1-\text{tg} \alpha \text{tg} \beta }$

Решение:
Отговор: $\sin 15^{\circ }=\frac{\sqrt{2}}{4}\left( \sqrt{3}-1\right) \quad \cos 15^{\circ }=\frac{\sqrt{2}}{4}\left( \sqrt{3}+1\right) \quad \text{tg} 15^{\circ }=2-\sqrt{3}$

$\sin 15^{\circ }=\sin \left( 45^{\circ }-30^{\circ }\right) =\sin 45^{\circ }\cos 30^{\circ }-\cos 45^{\circ }\sin 30^{\circ }=\tfrac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\cdot \frac{1}{2}=\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}\left( \sqrt{3}-1\right) $

$\cos 15^{\circ }=\cos \left( 45^{\circ }-30^{\circ }\right) =\cos 45^{\circ }\cos 30^{\circ }+\sin 45^{\circ }\sin 30^{\circ }=\tfrac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\cdot \frac{1}{2}=\frac{\sqrt{2}}{4}\left( \sqrt{3}+1\right) $

$\text{tg} 15^{\circ }=\text{tg} \left( 45^{\circ }-30^{\circ }\right) =\frac{\text{tg} 45^{\circ }-\text{tg} 30^{\circ }}{1+\text{tg} 45^{\circ }\text{tg} 30^{\circ }}=\frac{1-\frac{1}{\sqrt{3}}}{1+1\left( \frac{1}{\sqrt{3}}\right) }=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$

Решение:
Отговор: $\sin 75^{\circ }=\frac{\sqrt{2}}{4}\left( \sqrt{3}+1\right) \quad \cos 75^{\circ }=\frac{\sqrt{2}}{4}\left( \sqrt{3}-1\right) \quad \text{tg} 75^{\circ }=\frac{\left( \sqrt{3}+1\right) ^{2}}{2}$

$\sin 75^{\circ }=\sin \left( 90^{\circ }-15^{\circ }\right) =\sin 90^{\circ }\cos 15^{\circ }-\cos 90^{\circ }\sin 15^{\circ }=$

$1\cdot \frac{\sqrt{2}}{4}\left( \sqrt{3}+1\right) -0\cdot \frac{\sqrt{2}}{4}% \left( \sqrt{3}-1\right) =\frac{\sqrt{2}}{4}\left( \sqrt{3}+1\right) $

$\cos 75^{\circ }=\cos \left( 90^{\circ }-15^{\circ }\right) =\cos 90^{\circ }\cos 15^{\circ }+\sin 90^{\circ }\sin 15^{\circ }=$

$0\cdot \frac{\sqrt{2}}{4}\left( \sqrt{3}+1\right) +1\cdot \frac{\sqrt{2}}{4}\left( \sqrt{3}-1\right) =\frac{\sqrt{2}}{4}\left( \sqrt{3}-1\right) $

$\text{tg} 75^{\circ }=\frac{\sin 75^{\circ }}{\cos 75^{\circ }}=\frac{\frac{\sqrt{2}}{4}\left( \sqrt{3}+1\right) }{\frac{\sqrt{2}}{4}\left( \sqrt{3}-1\right) }=\frac{\left( \sqrt{3}+1\right)^{2}}{\left( \sqrt{3}-1\right) \left( \sqrt{3}+1\right) }=\frac{\left( \sqrt{3}+1\right)^{2}}{2}$

Решение:
$\sin \left( \alpha +\beta \right) +\sin \left( \alpha -\beta \right) =\sin \alpha \cos \beta +\sin \beta \cos \alpha +\sin \alpha \cos \beta -\sin \beta \cos \alpha$

Тогава $\sin \left( \alpha +\beta \right) +\sin \left( \alpha -\beta \right) =2\sin \alpha \cos \beta$

Решение:
$\sin \left( \alpha +\beta \right) -\sin \left( \alpha -\beta \right) =\sin \alpha \cos \beta +\sin \beta \cos \alpha -\left( \sin \alpha \cos \beta-\sin \beta \cos \alpha \right) $
$=2\cos \alpha \sin \beta $

Решение:
$\cos \left( \alpha +\beta \right) +\cos \left( \alpha -\beta \right) =\left( \cos \alpha \cos \beta -\sin \alpha \sin \beta \right) +\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right) =2\cos \alpha \cos \beta$

Решение:
$\cos \left( \alpha +\beta \right) -\cos \left( \alpha -\beta \right) =\left( \cos \alpha \cos \beta -\sin \alpha \sin \beta \right) -\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right) =-2\sin \alpha \sin \beta $

Решение:
Използвайки тъждеството $\text{tg} (A-B)=\frac{\text{tg} A-\text{tg} B}{1+\text{tg} A\text{tg} B}$ получаваме
$\frac{\text{tg} \left( \alpha +\beta \right) -\text{tg} \alpha }{1+\text{tg} \left( \alpha +\beta \right) \text{tg} \alpha }=\text{tg} \left[ \left( \alpha +\beta \right) -\alpha \right] =\text{tg} \beta $

Доказахме, че $\frac{\text{tg} \left( \alpha +\beta \right) -\text{tg} \alpha }{1+\text{tg} \left( \alpha +\beta \right) \text{tg} \alpha }=\text{tg} \beta $

Решение:
Тъй като $\sin \left( \alpha -\beta \right) =\sin \alpha \cos \beta -\cos \alpha \sin \beta $ и $\cos \left( \alpha -\beta \right) =\cos \alpha \cos \beta +\sin \alpha \sin \beta $
Тогава
$\left( \sin \alpha \cos \beta -\cos \alpha \sin \beta \right) ^{2}+\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right) ^{2}=\sin ^{2}\left( \alpha -\beta \right) +\cos ^{2}\left( \alpha -\beta \right) =1$

Решение:
$\text{cotg} \left( \alpha +\beta \right) =\frac{1}{\text{tg} \left( \alpha +\beta \right) }=\frac{1-\text{tg} \alpha \text{tg} \beta }{\text{tg} \alpha +\text{tg} \beta }=\frac{1-\frac{1}{\text{cotg} \alpha \text{cotg} \beta }}{\frac{1}{\text{cotg} \alpha }+\frac{1}{\text{cotg} \beta }}=\frac{\frac{\text{cotg} \alpha \text{cotg} \beta -1}{\text{cotg} \alpha \text{cotg} \beta }}{\frac{\text{cotg} \beta +\text{cotg} \alpha }{\text{cotg} \alpha \text{cotg} \beta }}=\frac{\text{cotg} \alpha \text{cotg} \beta -1}{\text{cotg} \beta +\text{cotg} \alpha }$

Доказахме, че $\text{cotg} \left( \alpha +\beta \right ) =\frac{\text{cotg} \alpha \text{cotg} \beta -1}{\text{cotg} \beta +\text{cotg} \alpha }$

Решение:
Отговор: $\text{cotg }\left( \alpha -\beta \right) =\frac{\text{cotg }\alpha \text{cotg }\beta +1}{\text{cotg }\beta -\text{cotg }\alpha }$

Тъй като $\text{cotg }\left( -\beta \right) =-\text{cotg }\left( \beta \right) $
и ако използваме, че $\text{cotg }\left( \alpha +\beta \right) =\frac{\text{cotg }\alpha \text{cotg }\beta -1}{\text{cotg }\beta +\text{cotg }\alpha }$

Then $\text{cotg }\left( \alpha -\beta \right) =\text{cotg }\left[ \alpha +\left( -\beta \right) \right] =\frac{\text{cotg }\alpha \text{cotg }\left( -\beta \right) -1}{\text{cotg }\left( -\beta \right) +\text{cotg }\alpha }=\frac{-\text{cotg }\alpha \text{cotg }\beta -1}{-\text{cotg }\beta +\text{cotg }\alpha }=\frac{\text{cotg }\alpha \text{cotg }\beta +1}{\text{cotg }\beta -\text{cotg }\alpha }$

Доказахме, че $\text{cotg } \left( \alpha +\beta \right) =\frac{\text{cotg }\alpha \text{cotg }\beta +1}{\text{cotg }\beta -\text{cotg }\alpha }$

Решение:
Знаем, че $\cos 2\alpha =\cos^{2}\alpha -\sin ^{2}\alpha =\left( \cos ^{2}\alpha +\sin ^{2}\alpha \right) -2\sin ^{2}\alpha =1-2\sin ^{2}\alpha $,

Нека $\alpha =\frac{1}{2}\theta $.
Тогава, $\cos \theta =1-2\sin ^{2}\frac{1}{2}\theta $

$\sin ^{2}\frac{1}{2}\theta =\frac{1-\cos \theta }{2}$
Така

$\sin \frac{1}{2}\theta =\pm \sqrt{\frac{1-\cos \theta }{2}}$

Решение:
$\cos 2\alpha =\cos^{2}\alpha -\sin^{2}\alpha =\left( \cos^{2}\alpha +\cos^{2}\alpha \right) -\cos^{2}\alpha -\sin ^{2}\alpha =\left( \cos^{2}\alpha +\cos ^{2}\alpha \right) -\left( \cos ^{2}\alpha +\sin^{2}\alpha \right) $
$=2\cos ^{2}\alpha -1$
Нека $\alpha =\frac{1}{2}\theta $.

$\cos \theta =2\cos ^{2}\frac{1}{2}\theta -1$,then $\cos ^{2}\frac{1}{2}\theta =\frac{1+\cos \theta }{2}$

Така $\cos \frac{1}{2}\theta =\pm \sqrt{\frac{1+\cos \theta }{2}}$

Решение:
$\text{tg} \frac{1}{2}\theta =\frac{\sin \frac{1}{2}\theta }{\cos \frac{1}{2}\theta }=\frac{\pm \sqrt{\frac{1-\cos \theta }{2}}}{\pm \sqrt{\frac{1+\cos \theta }{2}}}=\pm \sqrt{\frac{1-\cos \theta }{1+\cos \theta }}=\pm \sqrt{\frac{\left( 1-\cos \theta \right) \left(1+\cos \theta \right) }{\left( 1+\cos \theta \right) \left( 1+\cos \theta \right) }}=\pm \sqrt{\frac{1-\cos ^{2}\theta }{\left( 1+\cos \theta \right) ^{2}}}=\frac{\sin \theta }{1+\cos \theta }$

Доказахме, че $\text{tg} \frac{1}{2}\theta =\frac{\sin \theta }{1+\cos \theta }$

Решение:
Отговор: $1-\frac{1}{2}\sin 2x=\frac{\sin^{3}x+\cos^{3}x}{\sin x+\cos x}$

Когато разложим на множители числителя получаваме:

$\frac{\sin^{3}x+\cos^{3}x}{\sin x+\cos x}=\frac{\left( \sin x+\cos x\right) \left( \sin^{2}x-\sin x\cos x+\cos ^{2}x\right) }{\sin x+\cos x}=\sin ^{2}x-\sin x\cos x+\cos ^{2}x=$
$=1-\sin x\cos x=1-\frac{1}{2}\left( 2\sin x\cos x\right) =1-\frac{1}{2} \sin 2x$

Доказахме, че $1-\frac{1}{2}\sin 2x=\frac{\sin^{3}x+\cos^{3}x}{\sin x+\cos x}$

Решение:
Since $\sin \left( \theta +30^{\circ }\right) +\cos \left( \theta +60^{\circ }\right) =\left( \sin \theta \cos 30^{\circ }+\cos \theta \sin 30^{\circ}\right) +\left( \cos \theta \cos 60^{\circ }-\sin \theta \sin 60^{\circ}\right) $

Сега заместваме $\cos 30^{\circ },\sin 30^{\circ },\cos 60^{\circ },\sin 60^{\circ }$

$\sin \left( \theta +30^{\circ }\right) +\cos \left( \theta +60^{\circ }\right) =\frac{\sqrt{3}}{2}\sin \theta +\frac{1}{2}\cos \theta +\frac{1}{2}\cos \theta -\frac{\sqrt{3}}{2}\sin \theta =\cos \theta $

Решение:
$\frac{1-\text{tg}^{2}\frac{1}{2}x}{1+\text{tg}^{2}\frac{1}{2}x}=\frac{1-\frac{\sin ^{2}\frac{1}{2}x}{\cos^{2}\frac{1}{2}x}}{\sec^{2}\frac{1}{2}x}=\frac{\left( 1-\frac{\sin^{2}\frac{1}{2}x}{\cos ^{2}\frac{1}{2}x}\right) \cos^{2}\frac{1}{2}x}{\sec^{2}\frac{1}{2}x\cos ^{2}\frac{1}{2}x}=\cos ^{2}\frac{1}{2}x-\sin^{2}\frac{1}{2}x=\cos x$