Wednesday

September 17, 2014

September 17, 2014

Posted by **Chris** on Monday, April 26, 2010 at 9:43pm.

Cos(2 Arccos(5/13))

2. Solve the following equation for

0° ≤ Θ < 360°

Sec(Θ)= tan(Θ) + cos(Θ)

- Trigonometry -
**Reiny**, Monday, April 26, 2010 at 11:26pmCos(2 Arccos(5/13))

let's look at the arccos 5/13

we would have a right-angled triangle with angle Ø so that cos Ø = 5/13

so then we want cos(2Ø)

but cos 2Ø = 2cos^2 Ø - 1

= 2(25/169) - 1 = -119/169

- Trigonometry -
**Reiny**, Monday, April 26, 2010 at 11:31pmSec(Θ)= tan(Θ) + cos(Θ)

1/cosØ = sinØ/cosØ + cosØ

1 = sinØ + cos^2Ø

1 = sinØ + 1 - sin^2Ø

sin^2Ø - sinØ = 0

sinØ(sinØ - 1) = 0

sinØ = 0 ----> Ø = 0,180,360

or sinØ = 1 ---> Ø = 90

so Ø = 0,90,180, and 360°

**Answer this Question**

**Related Questions**

please - Given tan Θ = - 8/5 and sin Θ < 0, find sin Θ, cos &#...

math-Trig - 1. Write the algebraic expression which shows Cos((ArcSin(4/X)), 2...

Math-Trig - Trig Questions- 1. Write the algebraic expression which shows Cos((...

Trig - Identies/equation, please help! - 2Sin(Θ+47°)=1 ΘЄ[0°, ...

Trigonometry - Solve the equation cotΘ-abtanΘ = a-b Answer-tanΘ=1...

Trigonometry - Find all the angles between 0° and 90° which satisfy the equation...

math - Solve the following equations giving any roots in terms of pi in the...

TRIGONOMETRY HELP - Prove that: cos²Θ - sin²Θ = 2cos²Θ - 1

Trig - check my answers plz! - 1. (P -15/17, -8/17) is found on the unit circle...

Trigonometry - Solve the equations secΘ - 1 = (√2 - 1)tanΘ