# P1 Trigonometry

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#1
Solve the following equation in the interval:
0 degrees (< or equal to) theta < 360 degrees

cos^2 (theta) - 1 = 0
cos^2 (theta) = 1
cos (theta) = sq. root (1)

therefore: (theta) = inverse cos (sq. root of 1)
therefore: (theta) = 0 degrees (1 solution)

this is the only solution as the interval does not include 360 degrees
for its solutions to the equation.

ok, is this answer correct then.
pleae could someone check. thanx.
0
16 years ago
#2
(Original post by guess_who)
Solve the following equation in the interval:
0 degrees (< or equal to) theta < 360 degrees

cos^2 (theta) - 1 = 0
cos^2 (theta) = 1
cos (theta) = sq. root (1)

therefore: (theta) = inverse cos (sq. root of 1)
therefore: (theta) = 0 degrees (1 solution)

this is the only solution as the interval does not include 360 degrees
for its solutions to the equation.

ok, is this answer correct then.
pleae could someone check. thanx.
cos(theta) can also be square root of -1.
0
16 years ago
#3
(Original post by Ralfskini)
cos(theta) can also be square root of -1.
You mean the negative square root of 1, ie. -(1^1/2) and not (-1)^1/2. The square root of -1 is not a real number.
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#4
i dont get that. i got one solution of (theta) = 0. this gives a value of 1 when i look on the graph. you are saying that:

(theta) = inverse cosine -(sqr. root of 1) is also another solution.

then this gives: (theta) = 180 as the other solution.

yes/ but i dont understand that. when i get (theta) = 0 as the first solution, on teh graph it gives a y value of 1. i then look across to see if tehre are any other values of (theta) thst give 1. 180 degrees gives -1 on the graph of y = sinx.

help!!!!!!!!!
0
16 years ago
#5
haven't you guessed who this is yet?
0
16 years ago
#6
I think the name says it all...
0
16 years ago
#7
(Original post by elpaw)
haven't you guessed who this is yet?
You're joking right? If not, nooooooo.
0
16 years ago
#8
no go on, who is it
0
16 years ago
#9
(Original post by kikzen)
no go on, who is it
0
16 years ago
#10
(Original post by Nylex)
really? 0
16 years ago
#11
(Original post by elpaw)
really? Who then?
0
16 years ago
#12
(Original post by Nylex)
but he's being so... normal... *touches wood + crosses fingers*

lou xxx
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#13
ahhhh.......i get it...when ever its "square root of something...." the solution can always be plus or minus the square root of a number...i remember. its like with quadratic equations. when u use that formula, its always blahhh blahh plus or minus the root of whatever.

they do that so you ahve to find more solutions and it takes longer.

thanks for amking me remeber guys.. AND YES IT IS MASKALL, BUT HONEST I JUST MADE A SILLY MISTAKE.
0
16 years ago
#14
(Original post by guess_who)
ahhhh.......i get it...when ever its "square root of something...." the solution can always be plus or minus the square root of a number...i remember. its like with quadratic equations. when u use that formula, its always blahhh blahh plus or minus the root of whatever.

they do that so you ahve to find more solutions and it takes longer.

thanks for amking me remeber guys.. AND YES IT IS MASKALL, BUT HONEST I JUST MADE A SILLY MISTAKE.
gosh i was bad at maths a few months ago!!

of course u will have 2 solutions for a square root of a value.

because, for instance, u have +/- sq.root of 25 as ur solution.

1.) +sq.root25 = 5 -------> 5^2 = 25
2.) - sq.root25 = -5 ----> -5 ^ 2 = 25

hence, u can have 2 solutions e.g.) as evident when u r finding the root of a quadratic equation.

we can have both +/- sq. root of the value, as this value squared will give the exact same figure either way as shown above.

clueless................
0
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