Question

вычислить sin^2 $\pi /6$ + cos^2$\pi /3$ + tg^3 $3\pi /4$

Answer 1

№ 1
2 sin 60 + cos 90 - tg 45

[ sin 60 = √3/2
tg 45 = 1 ]

2* √3/2 - 1 = √3 - 1

Ответ 2)

№ 2
sin 120 = sin( 90+ 30) = cos 30 > 0
cos 195 = cos(180 + 15) = -cos 15 < 0
cos 359 = cos (-1) = cos 1 > 0
Ответ: 4)

№ 3
6 cos^2 Pi/4 + tg^2 (-Pi/3) - ctg( - Pi/2)

[ cos Pi/4 = cos 45 = √2/2
cos^2 Pi/4 = 1/2
6 cos ^2 Pi/4 = 3
tg -Pi/3 = tg -60 = -tg 60 = -√3
tg^2 Pi/3 = 3
ctg( -Pi/2) = 0 ]

3+3-0=6
Ответ: 3)

№ 4
(sin ( PI + a) * cos (Pi - a) ) / ctg(3*Pi/2 - a)

[ sin (Pi+a) = sin a
cos (Pi-a) = -cos a
ctg (3Pi/2 - a) = tg a = sin a/cos a ]

- sin a * cos a/ sin a /cos a = -cos^2 a
Ответ: 1)

№ 5

sin a * cos a * ctg a - 1 = sin a * cos a * cos a/ sin a - 1 = cos^2 a -1 = - (1-cos^2 a) = -sin^2 a
Ответ: 3)

№ 6

(sin^2 a - cos^2 a)/(sin a * cos a) = -cos 2 a / sin 2a/2 = -ctg 2a/2 = -0.5 ctg 2a
Ответ: видимо, 4)

№ 7
2 sin 15 * cos 15 = sin 2*15 = sin 30 = 1/2
Ответ: 4)

№ 8
cos 7Pi/4 = cos (-Pi/4) = cos Pi/4 = cos 45 = √2/2
Ответ: 1)

№ 9
sin 105 = sin (60 + 45) = sin 60 * cos 45 + cos 60 *sin 45 = √3/2 * √2/2 + 1/2 * √2/2 = √6/4 + √2/4 = (√6+√2)/4
Ответ: 2)

№ 10
[ sin = -3/5
tg 2 a = ? ]

cos^2 a = 1 - sin^2 a = 1 - 9/25 = 16/25 => cos a = 4/5. Но Pi < a < 3PI/2 => cos a = -4/5

tg 2a = sin 2a/cos2a

[ sin 2 a = 2sin a * cos a = 2 *12/25 = 24/25
cos 2 a = cos^2 a - sin^2 a = 16/25 - 9/25 = 7/25 ]

tg 2a = 24/25 / 7/25 = 24/7 = 3(3/7)
Ответ: 4)