# Find sin2x, cos2x, tan2x

+1 vote
sec=-4 find sin(2x), cos(2x), tan(2x)

These problems seem straight forward but I keep getting the wrong answer. If you could please in detail show me how to solve this I would really, really, really appreciate it so I don't throw my Trig book out the window.

sec(x) = -4

Reciprocal identities: sec(x) = 1/cos(x)

1/cos(x) = -4 then cos(x) = -1/4

Pythagorean theorem: a2 + b2 = c2

So, a2 + (-1)2 = (4)2 ⇒ a2 = 15 ⇒ a = √(15)

sin(x) = √15/4

Double Angle Formulas: sin2θ=2sinθcosθ, cos2θ=cos2θ-sin2θ and tan2θ = sin2θ/cos2θ.

sin(2x) = 2sin(x)cos(x) = 2(√15/4)(-1/4) = -2√15/16 = - 2√15/8

cos(2x) = cos2x - sin2x = (-1/4)2- (√15/4)2 = 1/16 - 15/16 = -14/16 = -7/8.

tan(2x) = sin(2x)/cos(2x) = (- 2√15/8) / (-7/8.) = 2√15 / 7.

The values of sin(2x) = -√15/8, cos(2x) = - 7/8, and tan(2x) = √15/7.

Sec x = - 4.

Using reciprocal identity : sec x = 1/cos x.

Sec x = - 4 ⇒ 1/cos x = - 4 ⇒ cos x = - 1/4.

Cos x = adjacent / hypotenuse = - 1/4.

Pythagorean theorem : hypotenuse2 = opp2 + adj2 .

42 = opp2 + (- 1) 2

16 = opp2 + 1

opp2 = 16 - 1 = 15

opp = √15.

Sin x = opposite / hypotenuse = √15/4.

Double Angle identities : sin2θ = 2sinθcosθ,

cos2θ =cos2θ - sin2θ, and

tan2θ = sin2θ/cos2θ.

• Sin 2x = 2 sin x cos x

= 2 (√15/4)(- 1/4)

= -√15/8.

• cos2x=cos2x - sin2x.

= (- 1/4)2 - (√15/4)2

= 1/16 - 15/16

= - 14/16

= - 7/8.

• tan(2x) = sin(2x)/cos(2x)

= (- √15/8) / (-7/8)

= √15/7.