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# find sin 2x, cos 2x, tan 2x given that cos x = 4/5, 3pi/2<x<2pi?

i have to show work too!!!

### 3 Answers

- la consoleLv 71 month ago
cos²(x) + sin²(x) = 1

sin²(x) = 1 - cos²(x) → given that: cos(x) = 4/5

sin²(x) = 1 - (4/5)²

sin²(x) = (25/25) - (16/25)

sin²(x) = 9/25

sin(x) = ± 3/5 → given that: 3π/2 < x < 2π → sin(x) < 0

sin(x) = - 3/5

Do you know this identity?

sin(a + b) = sin(a).cos(b) + cos(a).sin(b) → suppose that: a = b = x

sin(x + x) = sin(x).cos(x) + cos(x).sin(x)

sin(2x) = 2.sin(x).cos(x)

sin(2x) = 2.(- 3/5).(4/5)

sin(2x) = - 24/25

Do you know this identity?

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) → suppose that: a = b = x

cos(x + x) = cos(x).cos(x) - sin(x).sin(x)

cos(2x) = cos²(x) - sin²(x)

cos(2x) = (4/5)² - (9/25)

cos(2x) = (16/25) - (9/25)

cos(2x) = 7/25

tan(2x) = sin(2x)/cos(2x)

tan(2x) = (- 24/25)/(7/25)

tan(2x) = - 24/7

- AshLv 71 month ago
sin²x + cos²x = 1

sin²x + (⅘)² = 1

sin²x + 16/25 = 1

sin²x = 1 - 16/25

sin²x = 9/25

sinx = ±3/5

Given 3𝜋/2 < x < 2𝜋 means x is in 4th quadrant

In that quadrant sin is negative.

so we get

sinx = -⅗

sin2x = 2 sinx cosx

sin2x = 2(-⅗)(⅘)

sin2x = -24/25

cos2x = 2cos²x - 1

cos2x = 2(⅘)² - 1

cos2x = 7/25

tan2x = sin2x / cos2x

tan2x = (-24/25)/(7/25)

tan2x = -24/7

- ?Lv 71 month ago
Hint: use your trig identities. Do you know what these are?

Hopefully no one will spoil you the answers. That would be very irresponsible of them.