Find sin2x, cos2x, tan2x

Question

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.

Answer 1

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.

Comments

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

Answer 2

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.