Найдите значения: sin(a), cos(a), tg(а), если tg(а
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tgα∗ctgα=1

а) tg \alpha =2tgα=2 ctg \alpha =1:2= 0,5ctgα=1:2=0,5

\frac{tg a+ctg a}{tg a-ctg a}= \frac{2+0,5}{2-0,5}= \frac{2,5}{1,5}= \frac{5}{3}=1 \frac{2}{3}

tga−ctga

tga+ctga

=

2−0,5

2+0,5

=

1,5

2,5

=

3

5

=1

3

2

б) \frac{sin \alpha }{cos \alpha }=2

cosα

sinα

=2 sin \alpha =2*cos \alphasinα=2∗cosα

\frac{sin a -cos a}{sin a+cos a} = \frac{2*cos a-cos a}{2*cos a+cos a}= \frac{cosa}{3cosa} = \frac{1}{3}

sina+cosa

sina−cosa

=

2∗cosa+cosa

2∗cosa−cosa

=

3cosa

cosa

=

3

1

в) \frac{2sin a+3cos a}{3sin a-7cos a} = \frac{4cos a+3cos a}{6cos a-7cos a} = \frac{7cos a}{-cos a}= \frac{7}{-1}=-7

3sina−7cosa

2sina+3cosa

=

6cosa−7cosa

4cosa+3cosa

=

−cosa

7cosa

=

−1

7

=−7

г) \frac{sin^2a+2cos^2 a}{sin^2a-2cos^2 a}= \frac{(2*cos a)^2+2cos^2 a}{(2*cos a)^2-2cos^2 a}= \frac{4cos^2 a+2cos^2 a}{4cos^2 a-2cos^2 a}= \frac{6cos^2 a}{2cos^2 a} = \frac{6}{2}=3

sin

2

a−2cos

2

a

sin

2

a+2cos

2

a

=

(2∗cosa)

2

−2cos

2

a

(2∗cosa)

2

+2cos

2

a

=

4cos

2

a−2cos

2

a

4cos

2

a+2cos

2

a

=

2cos

2

a

6cos

2

a

=

2

6

=3