trigonometry and tangent problem, please help.

lqq889 · 2012-03-14 05:58:28

1. how to slove

for , i know how to do, but can not get the answer right. so please help.

my working:

2y +4 = cos^-1(-2/3)
= 0.841
= pi +0.841, pi - 0.841, 2pi + (pi +0.841), 2pi - (pi +0.841)
= 3.98 , 2.30,...

y= (3.98 -4)/2

is my working correct?

answer: y = 2.29, 3.13, 5.43, 6.27

2. (a) find the equation of the tangent to the curve

at the point A (1,1). (b) and if the tangent meets the curve at another point B, find the coordinates of B.

i found the answer for part (a)

. how to find the part (b), thanks!

bobbym · 2012-03-14 08:46:01

Hi lqq889;

For 1)

Those answers are what I am getting for 2 places after the decimal point.

( 2.29, 5.43 ) and ( 3.13 ,6.27 )

For 2b)

You have both equations

Set them equal to each other to form the cubic

You know that (1,1) is a root becuse that is the tangent. So you can factor out (x-1)

Factor that to

you now have two x coordinates to check, x = 1 and x = - 2. Ypu already know that x = 1 is solution to both equations. Just plug in x = -2 into the top equations to get the other intersection. It will be at (-2,-8).

hammana · 2012-03-14 09:30:18

if the cos(x) is negative then x>pi/2
I think that 2y+4=pi-0.841

Bob · 2012-03-14 10:20:10

hi lqq889

Welcome to the forum.

Where did 0.841 come from?

Cosine is negative in the second quadrant (ie from pi/2 to pi) so you shouldn't be getting this value.

Check you are working in radians and have the bracket in your calculation.

I'm getting ARCCOS(-2/3) = 2.3005....

That leads to the expected answers.

Bob

lqq889 · 2012-03-14 13:54:53

hi bobbym,

the answer for 2nd question is very detail, i know how to do already, thanks for your help, i am very appreciate. thanks!

lqq889 · 2012-03-14 14:04:38

hi bob bundy,

thanks for you help too.

sorry i forgot the bracket in my calculation.

ARCCOS(-2/3) = 2.3005

so, 2y + 4 = pi +2.3005, pi - 2.3005, 2pi + (pi +2.3005), 2pi - (pi -2.3005)
= 5.442, 0.841, 11.73, 5.442
2y = 5.442-4, 0.841-4, 11.73-4, 5.442 - 4

y = 0.721, -1.57, 3.865, 0.721

bob bundy, can help me check my working above, i still can not get the answer correct.
can explain to me in more details for this question, thanks very much.

Bob · 2012-03-14 20:51:59

hi llq889

The restriction "between 0 and 2 pi" applies to y, not to ARCCOS(-2/3)

So you need to find all the angles in one period of COSINE that could give -2/3 and then consider adding (or subtracting) multiples of 2 PI from these to find more angles giving -2/3

Then calculate 'y' from these and reject those out of range.

Have a look at my diagram. COSINE is negative in the 2nd and 3rd quadrants. 2.30052398... is the angle in the 2nd quadrant. We need to find the angle in the 3rd quadrant.

2 PI - 2.30052398 = 3.98266132...

so angles to consider are 2.30052398..... and 3.98266132....

So calculate further angles by adding multiples of 2 PI.

acos(-2/3)

2.300523983 3.982661324 8.58370929 10.26584663 14.8668946 16.54903194 ........

Now calculate y from each of these:

-0.849738008 -0.008669338 2.291854645 3.132923316 5.433447299 6.274515969 ........

You can stop now because values beyond these are out of range. > 2 PI

And cross off the first two as we want 0 < y

That leaves the 4 values you had as the expected answers.

Bob

Math Is Fun Forum

#1 2012-03-14 05:58:28

trigonometry and tangent problem, please help.

#2 2012-03-14 08:46:01

Re: trigonometry and tangent problem, please help.

#3 2012-03-14 09:30:18

Re: trigonometry and tangent problem, please help.

#4 2012-03-14 10:20:10

Re: trigonometry and tangent problem, please help.

#5 2012-03-14 13:54:53

Re: trigonometry and tangent problem, please help.

#6 2012-03-14 14:04:38

Re: trigonometry and tangent problem, please help.

#7 2012-03-14 20:51:59

Re: trigonometry and tangent problem, please help.

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