# Angle measures in radians & degrees

• Sep 3rd 2009, 08:14 AM
fezz349
Angle measures in radians & degrees
just a couple problems i need some help with.

1. Find the measure of the angle in radians and degrees.
sin^-1 (0.6)

2. Find the six trigonometric values of theta= cos^-1 (3/7). Give exact answers. I remember doing this during the school year but it's been a long summer and i've competely lost touch with this topic unfortunately (Worried)

3. Solve for x: e^-0.2x = 4
• Sep 3rd 2009, 11:06 AM
Jameson
Quote:

Originally Posted by fezz349
just a couple problems i need some help with.

1. Find the measure of the angle in radians and degrees.
sin^-1 (0.6)

2. Find the six trigonometric values of theta= cos^-1 (3/7). Give exact answers. I remember doing this during the school year but it's been a long summer and i've competely lost touch with this topic unfortunately (Worried)

3. Solve for x: e^-0.2x = 4

Take the natural log of both sides.

$\displaystyle \ln(e^{-0.2x})=\ln(4)$

Also remember the rule that $\displaystyle \ln(a^b)=b\ln(a)$
• Sep 3rd 2009, 01:59 PM
fezz349
right, so i took what u said and got 0.2xlne = ln4

i feel like this is wrong but i divide ln4 by ln of e in my calculator. it shows as ln(e^( . so i'm assuming the is being raised to the first power right? I got 1.386294361. Then divide by 0.2 on each side and my final answer was x=6.931471806. Just doesn't sound right to me for some reason.
• Sep 3rd 2009, 02:04 PM
Jameson
$\displaystyle \ln(e)=1$

That's a very important identity. I wouldn't a decimal answer. Just keep the natural logs as is.
• Sep 3rd 2009, 02:13 PM
fezz349
Gotcha, it's starting to come back to me now from precalc last year. Just another quick question, from the six trig functions questions, i thoguht about it for a bit and cos^-1 = (3/7) is equivalent to secant=3/7 right? Meaning cosine would be 7/3. I draw my triangle, and solve for the last side by using pythagorean theorem... i think
• Sep 3rd 2009, 02:38 PM
Jameson
Quote:

Originally Posted by fezz349
Gotcha, it's starting to come back to me now from precalc last year. Just another quick question, from the six trig functions questions, i thoguht about it for a bit and cos^-1 = (3/7) is equivalent to secant=3/7 right? Meaning cosine would be 7/3. I draw my triangle, and solve for the last side by using pythagorean theorem... i think

No, that's not true. It's a common mistake. cos^(-1) means inverse cosine, not 1 over cosine. When you take the cosine of an angle you get a ratio. When you are given a ratio you use the inverse function to find the angle.