Fun with precalc (need help with trig identities and such)

[sydney_bristow87]

Well, our teacher managed to go through 3 lessons in about 20 minutes... and lets just say I'm clueless... I kind of get what I'm supposed to be doing, but anyway, here are the problems and what i've got so far:

- Find the exact value of sin pi/12 ... so far, i've got nothing.
- Find the exact value of tan 7pi/12... what I got so far was:
tan7pi/12= 3pi/12 and 4pi/12, which can be simplified into pi/4 and pi/3

i used pi/4 for tanx and pi/3 as tany in the addition identity
tan(x + y)=
tan x + tan y
___________
1 - tanxtany

so then, i got

tan(x + y) =
pi/4 +pi/3
_________
1 - pi/4(pi/3)

which is actually
1 + sq.rt. 3
_________
1-1(sq.rt.3)
which, after i simplified, turned into 1.... but i get the idea that that's wrong, since in the back of the book the answer is something totally different.... any explainations on how to solve it?

-Simplify the given expression: cos (x+y) cos y + sin (x+y) sin y
first i distributed to get cosx + cosy(cosy) + sin x + siny(siny) which i simplified into cosx+cos^2 y + sin x + sin^2 y .... and again, i get the idea that i did something totally abnormal there, as the answer in the back of the book was cos x.

- If cos x = -1/5 and pi (less than) x (less than) 3pi/2, then sin(pi/3 - x)= ?
I didn't even know where to start on that one.... and those problems above were just the ones I didn't get from the first lesson =/ I've got tons more that I'm gonna see if I can figure out... any help would seriously be appreciated... thanks

 

[galactus]

Re: Fun with precalc (need help with trig identities and suc

Well, our teacher managed to go through 3 lessons in about 20 minutes... and lets just say I'm clueless... I kind of get what I'm supposed to be doing, but anyway, here are the problems and what i've got so far:

- Find the exact value of sin pi/12 ... so far, i've got nothing.
- Find the exact value of tan 7pi/12... what I got so far was:
tan7pi/12= 3pi/12 and 4pi/12, which can be simplified into pi/4 and pi/3

i used pi/4 for tanx and pi/3 as tany in the addition identity
tan(x + y)=
tan x + tan y
___________
1 - tanxtany

so then, i got

tan(x + y) =
pi/4 +pi/3
_________
1 - pi/4(pi/3)

which is actually
1 + sq.rt. 3
_________
1-1(sq.rt.3)
which, after i simplified, turned into 1.... but i get the idea that that's wrong, since in the back of the book the answer is something totally different.... any explainations on how to solve it?

-Simplify the given expression: cos (x+y) cos y + sin (x+y) sin y
first i distributed to get cosx + cosy(cosy) + sin x + siny(siny) which i simplified into cosx+cos^2 y + sin x + sin^2 y .... and again, i get the idea that i did something totally abnormal there, as the answer in the back of the book was cos x.
Use the trig addition forumulas.

cos(x+y)=cos(x)cos(y)-sin(x)sin(y) and sin(x+y)=sin(x)cos(y)+cos(x)sin(y).

It'll come together. Remember, sin²y+cos²y=1

- If cos x = -1/5 and pi (less than) x (less than) 3pi/2, then sin(pi/3 - x)= ?
I didn't even know where to start on that one.... and those problems above were just the ones I didn't get from the first lesson =/ I've got tons more that I'm gonna see if I can figure out... any help would seriously be appreciated... thanks

 

[sydney_bristow87]

Heh... I shoulda thought of that... the lession is ON addition and subtraction identities....
Well... now that I understand where I went wrong ( i needed to plug in the identities instead of distributing), I'm still a little confused as to what to do with these: cos (x + y) cos y +sin (x + y) sin y
.... now I've got:

cos x cos y - sin x sin y (cos y) + sin x cos y + cos x sin y (sin y)

which i GUESS (and my guesses usually are pretty off) could be:

cos x cos y - sin x sin y (cos y) + sin x cos y + cos x sin^2 y

which still seems like i'm missing something =/

 

[Unco]

- Find the exact value of sin pi/12 ... so far, i've got nothing.

Hint: 1/12 = 1/3 - 1/4

:wink:

 

[Daniel_Feldman]

1+root3 1+root3
______ x ________

1-root3 1+root3





=1+2root3+3

__________

1-3



=4+2root3
_________

-2


=-2-root3


Is that the book's answer?

 

[soroban]

Re: Fun with precalc (need help with trig identities and suc

Hello, sydney!

Find the exact value of: sin (π/12)

Use the same technique you used on the second problem.

Since .π/12 .= .4π/12 - 3π/12 .= .π/3 - π/4

you can use: .sin(π/3 - π/4) .= .sin(π/3)cos(π/4) - sin(π/4)cos(π/3)

Simplify the given expression: cos (x+y) cos y + sin (x+y) sin y

Do you recognize the expression as the right side of the formula:
. . . . . cos(A - B) .= .cos(A)cos(B) + sin(A)sin(B) . . . . (Ha! Who would??)

Reading the formula from right to left, we have: A = x + y, B = y

So the left side is: .cos([x + y] - y) .= .cos(x)

 

[Unco]

Edit2: Am I losing my sight?

 

[galactus]

Heh... I shoulda thought of that... the lession is ON addition and subtraction identities....
Well... now that I understand where I went wrong ( i needed to plug in the identities instead of distributing), I'm still a little confused as to what to do with these: cos (x + y) cos y +sin (x + y) sin y
.... now I've got:

cos x cos y - sin x sin y (cos y) + sin x cos y + cos x sin y (sin y)

cos(x)cos²(y)-sin(x)sin(y)cos(y)+cos(x)sin²(y)+sin(x)sin(y)cos(y).

Now, try again. It should fall into place now.

which i GUESS (and my guesses usually are pretty off) could be:

cos x cos y - sin x sin y (cos y) + sin x cos y + cos x sin^2 y

which still seems like i'm missing something =/

 

[sydney_bristow87]

ARgh I still don't get any of it.... and being my stupid self, I accidentally erased the problems I had already done and gotten right.... now they look wierd....
for the cos (x + y) cos y + sin (x + y) sin y one, I've somehow mutated it into: cos x cos^2y + sin x cos y + cos x sin^2y

aaanddd i have no idea what i'm doing. at all.

thanks for all of your help so far, though... i understood most of what everyone said, but i dunno, i just get stuck on wierd parts and will have no idea how to get them.