# math

posted by isaiah

if cosØ=0.6 and 0°<Ø<90°, find the exact value of 2Ø

1. Reiny

remember cos^2 Ø + sin^2 Ø = 1
.36 + sin^2 Ø = 1
sin^2 Ø = .64
sinØ = .8

sin 2Ø = 2sinØcosØ
= 2(.8)(.6)
= .96

2Ø = sin^-1 (.96) or arcsin(.96)

## Similar Questions

1. ### trig

how do you solve this problem? 1+sinØ/cosØ +cosØ/1+sinØ = 2secØ and cosß- cosß/1-tanß = sinßcosß/sinß-cosß
2. ### MATH

1.)Find the exact solution algebriacally, if possible: (PLEASE SHOW ALL STEPS) sin 2x - sin x = 0 2.) GIVEN: sin u = 3/5, 0 < u < ï/2 Find the exact values of: sin 2u, cos 2u and tan 2u using the double-angle formulas. 3.)Use …
3. ### Math

How can I prove this identity? (1 + sinØ + cosØ):(1 - sinØ + cosØ) = (1 + sinØ) : cosØ
4. ### maths

if sinØ=cosØ then find the value of Ø
5. ### maths

Find the Cartesian form of the equation, r3 = 3r cosØ
6. ### Topology

5. Find the Cartesian form of the equation, r3 = 3r cosØ.
7. ### Topology

Find the Cartesian form of the equation, r3 = 3r cosØ.
8. ### math

if Ø is an angle in the first quadrant and cosØ= 1/2, find the exact value of cos2Ø
9. ### maths

if cosØ=x/y, find cosecØ
10. ### Math

sinØ (1+ tanØ) + cosØ(1+ cotØ) = (secØ + cosecØ)

More Similar Questions