I'm really confused about this linearization problem... I have particularly difficulty with sin/trig functions so adding that onto the problem itself and I've got a mess!
Here's a screenshot of the problem: http://oi42.tinypic.com/2rrp451.jpg
How do you even find a "nice" point near 25? I know the linearization formula and you need a number a/nice number but, hah, no clue. I tried using 30 degrees as well as (sqrt3)/(2) but I'm just getting waaaay incorrect numbers.
Update:I mean, I get that x-a is the same as dx... but yeah, the way we were taught to use that equation was just derpy and.. okay, I'm not really explaining it well but trust me, it just made it unnecessarily difficult. Like instead of doing 25-30, we were actually doing x-30 in the equation, solving for all that, and then plugging in the 25 so somewhere along the way I did some bad algebra because ajfl;dsfjl;afj mess. MESS.
Answers & Comments
Verified answer
I too would've started with sin(30 deg) = 1/2
and sin(25 deg) = approximately sin(30 deg) + (dx) cos(30 deg)
Now here's the trick: "dx" has to be in radians!
It's -5 degrees, which is -5 pi/180 radians or -pi/36 radians, so
sin(25 deg) = about 0.5 - pi sqrt(3)/(72)
= 0.4244
The four-digit value of sin(25 deg) is 0.4226,
so I wouldn't say the 0.4244 was "waaaay" incorrect.
Maybe you thought "dx" was -5. Nope!