2 zadania z trygonometrii są one w załączniku daje dużo pkt.... Question from @szustollo94

October 2018 2 14 Report

2 zadania z trygonometrii są one w załączniku daje dużo pkt

platon1984

Zadanie 1.

$\frac{2\tan\alpha}{\sin\alpha+1/\tan\beta}=\frac{2\cdot1}{0.5\sqrt{2}+\sqrt{3}}=\frac{2(0.5\sqrt{2}-\sqrt{3})}{0.5-3}=\frac{2(\sqrt{3}-0.5\sqrt{2})}{2.5}=\\=\frac{4(\sqrt{3}-0.5\sqrt{2})}{5}$

Zadanie 2.

$\frac{1-\tan^2\alpha}{1+\tan^2\alpha}=\frac{1-\frac{\sin^2\alpha}{\cos^2\alpha}}{1+\frac{\sin^2\alpha}{\cos^2\alpha}}=\\=\frac{cos^2\alpha-\sin^2\alpha}{1}=1-\sin^2\alpha-\sin^2\alpha=1-2\sin^2\alpha$

co było do udowodnienia.

pozdrawiam

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"non enim possumus quae vidimus et audivimus non loqui"

4 votes Thanks 1

wik8947201

$\\\frac{2}{{\frac{\sqrt2}{2}+\sqrt3}}=\frac{4}{\sqrt2+2\sqrt3}= \\\frac{4*(\sqrt2-2\sqrt3)}{2-12}=-0,4(\sqrt2-2\sqrt3)=0,8\sqrt3-0,4\sqrt2 \\\frac{1-tg^2\alpha}{1+tg^2\alpha}= \\\frac{cos^2\alpha-sin^2\alpha}{cos^2\alpha+sin^2\alpha}= \\1-sin^2\alpha-sin^2\alpha=1-2sin^2\alpha$

0 votes Thanks 0

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