Number of results: 4,114
Trig
draw the triangles and you will see that tan(s) = 12/5 tan(t) = 4/3 Now just plug in your sum formula: tan(s+t) = (tan s + tan t)/(1 - tan s * tan t)
Tuesday, March 19, 2013 at 3:22pm by Steve
trig
tan(x+y)=(tan(x)+tan(y))/(1-tan(x)tan(y)) substitute x=y to get tan(2x) = 2tan(x)/(1-tan²(x)) Substitute tan(2x)=1 and solve for tan(x). Reject the negative root knowing that tan(x)≥0 for 0<x<π/2.
Tuesday, August 24, 2010 at 3:40am by MathMate
pre calc
tan^5 - 9tan = 0 tan(tan^4-9) = 0 tan(tan^2-3)(tan^2+3) = 0 tan(tan-√3)(tan+√3)(tan^2+9) = 0 so, tan x = 0 or √3 or -√3 now it should be a cinch
Sunday, April 14, 2013 at 5:02pm by Steve
Trigonometry
(1+tan)/(1-tan) multiply top and bottom by 1+tan (1+tan)^2 / (1-tan)(1+tan) (1 + 2tan + tan^2)/(1-tan^2) now, since sec^2 = 1+tan^2, (sec^2 + 2tan)/(1-tan^2)
Friday, August 3, 2012 at 3:31pm by Steve
maths --plse help me..
make a sketch, label the height as h distance from last position to the tower as x tan (alpha) = h/(x+a) h = (x+a)tan alpha in the same way h = xtan beta (x+a)tan alpha = xtan beta x tan alpha - xtan beta = - a tan alpha x(tan alpah - tan beta) = -a tan alpha x = -a tan alpha...
Thursday, December 13, 2012 at 12:22pm by Reiny
Math
tan (pi/4+x) = (tan pi/4 + tan x)/(1 -tan pi/4 tan x) tan (pi/4+x) = (tan pi/4 - tan x)/(1 +tan pi/4 tan x) note tan pi/4 = 1 add and put over common denominator [(1+tan x)(1+tan x)+(1-tan x)(1-tan x)]/(1 - tan^2 x) expand top [1+2 tanx +tan^2 x +1 - 2 tan x +tan^2 x]/(1 - tan...
Sunday, January 4, 2009 at 2:48am by Damon
trig
tan(x) has a period of π, so tan(x)=tan(x+π), in fact, tan(x+kπ)=tan(x), where k∈Z So to evaluate tan(257π/4) =tan((257/4)π) =tan(64π + π/4) =tan(π/4) I will let you figure out which answer to choose.
Wednesday, May 23, 2012 at 9:44pm by MathMate
Math
tan^-1(1/2)=x=.463648 tan^-1(2/3)=y.588003 x+y+pi/2 = -z=2.62245 z=-2.62245 cos (z) = -.868243 and tan a= -1/2 tan b= -2/3 tan (a+b)=(tan a +tan b)/(1-tan a.tan b) so tan (a+b)= -7/4 sec^2 x-tan^2 x=1 so sec (a+b)=√(1+49/16) =√(65/16) (a+b)=1/sec (a+b) so ...
Tuesday, June 9, 2009 at 1:36pm by Sophie
Trig
Did you recoginize that your expression matches the formula for tan 2A ? tan 2A = 2tanA/(1-tan^2 A) so 2tan(5pi/12)/1-tan²(5pi/12) = tan(5pi/6) or tan 150º tan 150 = -tan 30 or - tan(pi/6) = -1/√3
Sunday, April 19, 2009 at 2:51pm by Reiny
pre-calc
1 + 2 tan x + tan^2 x = 2 tan x + 2 tan^2 x - 1 = 0 tan x = 1 or -1 x = 45 or 135 or or 225 or 315 degrees
Tuesday, November 20, 2012 at 3:20pm by Damon
trig
recall tan(A+B) = (tanA + tanB)/(1 - tanAtanB) I believe you meant to type (tan 25° + tan 5°) /( 1- tan 25° tan 5°) now it fits the above pattern and it equals tan(25+5) = tan 30° from my 30-60-90 triangle I know that tan 30° = 1/√3 or &#...
Friday, April 30, 2010 at 10:07pm by Reiny
calculus
can anyone tell me if tan-1x= 1 over tan x? No. They are different. tan^-1 (x) is a frequently used way of writing arctan x, which means the angle whose tangent is x. tan (tan^-1 x) = x
Sunday, February 11, 2007 at 12:16pm by Elizabeth
Precalculus
Hint: use the following trigonometric identity tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b))
Wednesday, February 23, 2011 at 7:17pm by MathMate
precal
LS = tan(3x) = tan(2x + x) = (tan2x + tanx)/(1- tan2x tanx) = [2tanx/(1-tan^2 x) + tanx ] / [(1 - 2tanx(tanx)/(1 - tan^2 x) ] multiply top and bottom by 1 - tan^2 x (2tanx + tanx + tan^3 x)/(1 - tan^2 x - tan^2 x) = (3tanx + tan^3 x)/(1 - 3tan^2 x) = RS
Monday, February 6, 2012 at 1:56pm by Reiny
precalculus
For each of the following determine whether or not it is an identity and prove your result. a. cos(x)sec(x)-sin^2(x)=cos^2(x) b. tan(x+(pi/4))= (tan(x)+1)/(1-tan(x)) c. (cos(x+y))/(cos(x-y))= (1-tan(x)tan(y))/(1+tan(x)tan(y)) d. (tan(x)+sin(x))/(1+cos(x))=tan(x) e. (sin(x-y...
Sunday, April 14, 2013 at 3:27pm by anonymous
Trigonometry
Find the exact value of tan(a-b) sin a = 4/5, -3pi/2<a<-pi; tan b = -sqrt2, pi/2<b<pi identity used is: tan(a-b)=(tan a-tan b)/1+tan a tan b simplify answer using radicals. (a is alpha, b is beta)
Monday, November 8, 2010 at 11:40pm by Dee
Math
Unfortunately it doesn't look right, ad... from the very start. Could you rechceck your source from where you got the expression? [ 1 - (tan α)(tan β) ] / [ 1 + (tan α)(tan β)] represents, according to me, cos(α+β)/cos...
Tuesday, June 9, 2009 at 1:36pm by MathMate
trig
Both cot x and tan x must either be 1 or -1, since tan x = 1/cot x tan^2 x = 1 tan x = +/-1 tan 1 = 1 at 45 and 225 degrees. tan x = -1 at 135 and 315 degrees.
Monday, February 18, 2008 at 11:05pm by drwls
math
In most cases I suggest changing everything to sines and cosines, but sometimes the variations of the Pythagorean identities come in handy that is, by dividing sin^2 A + cos^2 A = 1 by cos^2 A we get tan^2 A + 1 = sec^2 A so the denominator of the first one is -tan^2 e and the...
Sunday, February 22, 2009 at 12:41pm by Reiny
Trig
Using tan(A-B) = (tanA - tanB)/(1+tanAtanB) RS = tan(x-6π) = (tanx - tan 6π)/(1 + tanxtan 6π) but 6π is coterminal with 2π (6π is 3 rotations, and 2π is one rotation so tan 6π = tan 2π = 0 = (tanx...
Wednesday, January 23, 2013 at 8:14pm by Reiny
trig
tan(345)? Tan(2*345)=tan(690)=tan(720-30)=tan(-30) tan(2*345)= 2*tan(345)/(1-tan^2(345) lets call tan(345) Y tan(-30)=2y/(1-y^2) Call Tan(-30) X x-XY^2=2Y XY^2+2Y-x=0 use the quadratic to solve for y in terms of X (tan(-30)=-1/sqrt3)
Thursday, November 19, 2009 at 7:38pm by bobpursley
math
if you need tan 7.5 degrees you would get .1316525 if you want tan 7.5 radians that would be 2.706 if the question asked for the "exact" value of tan 7.5º you would first find tan 15º by using tan 15º = tan(45º - 30º) = (tan45 - tan30)/(1 + ...
Saturday, August 15, 2009 at 10:07pm by Reiny
calculus
Actually, since tan(pi/2-x) = cot(x) and cot = 1/tan it falls right out. Or, using the sum of tangents formula, tan(arctanx + arctan(1/x)) = [tan(arctan(x)) + tan(arctan(1/x))][1 - tan(arctan(x))*tan(arctan(1/x))] = [x + 1/x]/[1 - x*1/x] = (x + 1/x)/0 = oo tan pi/2 = oo
Wednesday, December 7, 2011 at 3:59pm by Anonymous
Pre-Calculus
LS: sec/(sec-tan) multiply by (sec+tan) top and bottom sec(sec+tan) / (sec^2-tan^2) now recall that sec^2 = 1+tan^2, so we have (sec^2 + sec tan) / 1 sec^2 + sec tan = RS
Monday, November 12, 2012 at 1:28pm by Steve
Pre-Calculus
your text should have tan(a-b) = (tan a - tan b)/(1 + tanatanb) so you will need tan a and tan b cos a = -3/5 and a is in the second quadrant, so by sketching a diagram it is easy to see that sin a = 4/5 and tan a = - 4/3 similarly sin b = 5/13 and b is in quadrant II, so cos ...
Saturday, August 30, 2008 at 10:12pm by Reiny
Trig.
sec^2xcotx-cotx=tanx (1/cos)^2 times (1/tan)-(1/tan)=tan (1/cos^2) times (-2/tan)=tan (-2/cos^2tan)times tan=tan(tan) sq. root of (-2/cos^2)= sq. root of (tan^2) sq. root of (2i)/cos=tan I'm not sure if I did this right. If I didn't, can you show me the correct steps? ...
Saturday, January 9, 2010 at 10:27am by Trig
trig
h t t p : / / i m g 4 0 . i m a g e s h a c k . u s / c o n t e n t . p h p ? p a g e = d o n e& l = i m g 4 0 / 1 4 8 / a s d a s d y e . j p g & v i a = m u p l o a d (def of tan theta = a^-1 o)a = o opposite = adjacent tan theta written with respect to the first ...
Sunday, September 6, 2009 at 5:06pm by trig
math
Using reference angles, (take 225 and subtract 180), you get 45. Since 225° is in quadrant 3, it is positive. Now you have 1 + tan^2 (45°)… Tan^2 is the same as tan x tan, so you have 1 + tan(45) x tan (45). If you use a 45-45-90 triangle, you get tan(45) =1. So, ...
Sunday, September 23, 2012 at 9:05am by Monica
calculus
The angle joining P to the origin is A = tan^-1 (y/x) You can invert that to read y = x tan A In addition to that, along the parabola, y = x^2 Therefore x = tan A and y = tan^2 A
Wednesday, September 8, 2010 at 10:14pm by drwls
Pre calc
1. 2 x 67.5 = 135 2. tan 135 = - tan 45 = -1 3. tan 2A = [2tan A]/[1-(tan A)^2] 4. Put A = 67.5 in (3) and get tan135 = [2tan67.5]/[1-(tan67.5)^2] -1 = [2tan 67.5]/[1-(tan 67.5)^2] 5. Let x = tan 67.5. Then, -1 = 2x/[1 - x^2] x^2 - 2x - 1 = 0 By quadratic formula, x = 1 + sqrt...
Sunday, January 22, 2012 at 10:42pm by Mick from cannotta
math
Volume = Integral pi [y(x)]^2 dx x = 0 to pi/4 The integral of tan^4(x), which you will need, can be found in a table of integrals. It uses a recursion formula. INT tan^4(x) = (1/3)tan^3x - (1/2)tan^2x + tan x - x
Friday, January 30, 2009 at 9:13pm by drwls
Calculus
use tan^2x = sec^2x - 1 tan^6 = tan^4 (sec^2 - 1) = tan^4 sec^2 - tan^4 = tan^4 sec^2 - tan^2 sec^2 + tan^2 = tan^4 sec^2 - tan^2 sec^2 + sec^2 - 1 now d(tanx) = sec^2x dx, so what you have is u^4 du - u^2 du + du - dx = 1/5 u^5 - 1/3 u^3 + u - x = 1/5 tan^5x - 1/3 tan^3x + ...
Sunday, January 8, 2012 at 5:56pm by Steve
trig
tangent has a period of π so tan(x) = tan(x+π) = tan(x + 2π) ... Also, tan(x)=-tan(π - x) so tan(23π/6) =tan(5π/6 + 3π) =tan(5π/6) =-tan(π/6) =-√3/3 A reference angle is the acute angle between the ...
Sunday, October 4, 2009 at 8:17pm by MathMate
Precalculus troubles
It will be the same as -tan 15. They probably want you to use the formula tan(x+y) = [tanx + tany]/[1 - tanx*tany] tan 300 = -tan 60 = -sqrt3 tan 45 = 1 tan 345 = (1 - sqrt3)/[1 + sqrt3) = (1 - sqrt3)^2/(1-3) = (-1/2)(sqrt3 -1)^2 = -0.26795
Tuesday, January 12, 2010 at 11:15pm by drwls
Math
cot(4c-π/4) = 1/tan(4c-π/4) = 1/[ (tan4c) - tanπ/4)/(1 + tan4c(tanπ/4))] = (1+tan4ctanπ/4)/(tan4c - tanπ/4) = (1 + tan4c)/(tan4c - 1) , since tan π/4 = 1 also from tan 2A = 2tanA/(1 - tan^2 A) the above is = [ 1 + ...
Saturday, June 18, 2011 at 10:50pm by Reiny
Trig
tan(a-b) = [ tan a - tan b)/(1 + tan a tan b) You better learn those trig identities.
Tuesday, December 27, 2011 at 6:42pm by Damon
math
because there is an exact relationship between π and the circle, thus its angles e.g. tan 45° = tan π/4 or tan .7854 ( I had to round it off) set your calculator to radians tan π.4 = 1 tan .7854 = 1.000003673
Monday, March 25, 2013 at 9:08pm by Reiny
Trig - tan 15° using composite argument?
tan 15° tan (45°-30°) (tan 45° - tan 30°)/1+ tan 45°tan30° (1-√3/3)/(1+1√3/3) then i donnt what to do/ chancel out. Can someone finish it if i didnt get it wrong. thanks in advance
Wednesday, November 12, 2008 at 7:58pm by Anonymous
Integration
Intergrate ¡ì sec^3(x) dx could anybody please check this answer. are the steps correct? thanks. = ¡ì sec x d tan x = sec x tan x - ¡ì tan x d sec x = sec x tan x - ¡ì sec x tan^2(x) dx = sec x tan x + ¡ì sec x ...
Sunday, October 15, 2006 at 2:06am by Vidal
Solving Trig equation
you should recognize that they are using the formula tan(A+B) = (tanA + tanB)/(1 - tanAtanB) so (tanx + tan π/3)/(1 - tanxtan(π/3) = √3 becomes tan(x+π/3) = √3 , I know tan 60° or tan 240° = √3 OR tan π/...
Friday, April 27, 2012 at 11:41pm by Reiny
trig
tan x = 13 cot x tan x = 13/tan x tan^2 x = 13 tan x = ±√13 x = 74.5º,105.5º, 254.5º, 285.5º
Thursday, May 1, 2008 at 12:12am by Reiny
maths
(1+tan x)/(1-tan x) + (1-tan x)/(1+tan x) = [(1+tan x)^2 + (1-tan x)^2]/(1-tan^2 x) = (2 + 2tan^2 x)/(1-tan^2 x) = 2(cos^2 x + sin^2 x)/(cos^2 x - sin^2 x) = 2/cos 2x = 2sec 2x
Friday, August 10, 2012 at 5:09am by Steve
Pre calculus
Let the building height be Y Let the horizontal distance to the building be h. h tan 42 = a h tan 8 = b Y = a + b = h (tan42,+ tan 8) For the value of h, use h tan 8 = 10 m
Sunday, July 15, 2012 at 8:26pm by drwls
Maths
Since 540 = 3*180 and the period of tan(x) is 180 tan(540-x) = tan(-x) = -tan(x) = -1/5 However, to answer your question, yo need to go clockwise, since you have tan(-x).
Tuesday, May 22, 2012 at 5:28am by Steve
trig
looks like you mean y = tan(2x + pi/4) = (tan 2x + tan pi/4)/(1 - (tan 2x)(tan pi/4) = (tan 2x + 1)/(1 - tan 2x), since tan pi/4 = 1
Friday, March 5, 2010 at 4:11pm by Reiny
math
(tan x -1) / (1/tan x - 1) (tan^2 x - tan x) / (1 - tan x) tan x (tan x - 1)/(1 - tan x) -tan x
Sunday, December 7, 2008 at 5:20pm by Damon
calculus
additional note: Since tan(x)=tan(x±π), you can calculate tan(x-π) instead of tan(x). This way, the cosine series converges more rapidly.
Saturday, February 6, 2010 at 8:01pm by MathMate
Math - help really needed
I'm really sorry. I don't want to sound stupid, but when I add tan^2x/tan^2x to 1/tan^2x I get 1+tan^2x/tan^2x and then the tans cancel out and I'm left with just one. Obviously, that's not correct, but I can't find my mistake.
Tuesday, December 2, 2008 at 9:33pm by Rani
Maths Calculus Derivatives
Find the first derivative for the following functions 1) f(x) = sin(cos^2x) cos(sin^3x) 2) f(x) = ( tan 2x - tan x ) / ( 1 + tan x tan 2x ) 3) f(x) = sin { tan ( sqrt x^3 + 6 ) } 4) f(x) = {sec^2(100x) - tan^2(100x)} / x
Saturday, October 22, 2011 at 7:31am by Yousef
trigo math
tan*sin+cos = sin^2/cos + cos = (sin^2+cos^2)/cos = 1/cos = sec (tan*cos^2 + sin^2)/sin = (sin*cos + sin^2)/sin = sin(cos+sin)/sin = cos+sin I think you mean (1+tan)/(1-tan) = (1+tan)^2/(1-tan^2) = (1+2tan+tan^2)/(1-tan^2) = (sec^2+2tan)/(1-tan^2) the last one needs some ...
Thursday, December 13, 2012 at 10:58am by Steve
Maths
cos^2(360-A) = cos^2(A) tan^2(180+A) = tan^2(A) sec^2(180-A) = sec^2(A) sin^2(540+A) = sin^2(A) csc^2(-A) = csc^2(A) cot^2(90+A) = tan^2(A) so, using those simplifications, we have cos^2 * tan^2 * sec^2 = tan^2 sin^2 * csc^2 * tan^2 = tan^2 and the fraction is just 1 How far ...
Thursday, May 2, 2013 at 11:18pm by Steve
calculus
Let's define I_{n} = Integral of tan^n(x)dx You can express I_{n} in terms of I_{n-2} as follows. We write the integrand as: tan^n(x) = sin^n(x)/cos^n(x) Take out a factor sin^2(x): tan^n(x) = sin^2(x)sin^(n-2)/cos^n(x) And then substitute: sin^2(x) = 1 - cos^2(x) tan^n(x...
Thursday, September 25, 2008 at 6:50pm by Count Iblis
trig
H = Hight of lighthouse L = Distance between a ship and a lighthouse tan( theta ) = H / L If 19^ mean 19° then: tan( theta ) = H / L tan( 19° ) = 120 / L L * tan( 19° ) = 120 Divide both sides with tan( 19° ) L = 120 / tan ( 19° ) L = 120 / 0.34433 L = 348....
Saturday, October 1, 2011 at 11:19am by Anonymous
Calc
Make a table where the first column is x, the second column is tan(3x), the third is tan(5x), and the last is tan(3x)/tan(5x). Set your calculator to radians, and calculate the values of tan(3x) and tan(5x) for x=0.1, 0.01, 0.001... From the values suggested by the last column...
Sunday, September 26, 2010 at 3:58pm by MathMate
Trig
sec^4(x) - tan^4(x) = = (1 + tan^2(x))^2 - tan^4(x) = 1 + 2tan^2(x) + tan^4(x) - tan^4(x) = 1 + 2tan^2(x) QED
Tuesday, March 1, 2011 at 5:23pm by agrin04
Calculus
sin^3 x tan x sin^3 x (d tan x /dx ) + tan x ( 3 sin^2x) sin^3 x (sec^2 x) + 3 tan x sin^2 x sin^3 x (1/cos^2 x) + 3 tan x sin^2 x sin x tan^2 x + 3 tan x sin^2x sin x tan x ( tan x + 3 sin x)
Monday, April 19, 2010 at 9:30pm by Damon
more trig.... how fun!!!!
if you can't help me with my first question hopw you can help me with this one. sec(-x)/csc(-x)=tan(x) thanx to anyone who can help From the definition of the sec and csc functions, and the tan function, sec(-x)/csc(-x) = sin(-x)/cos(-x) = tan(-x) However, tan (x) does not...
Monday, November 6, 2006 at 10:23pm by berhana
Pre-Cal
from you second line tan^2(x)/tan(x) - sec^2x/tanx = (tan^2 x - sec^2 x)/tanx = (tan^2 x - (1 + tan^2 x))/tanx = -1/tanx or -cotx or -sinx/cosx
Friday, March 5, 2010 at 6:31pm by Reiny
Calculus
f'(x) = (1/2)(tan(1-x))^(-1/2)*sec^2 (1-x)(-1) = -(1/2) [tan(1-x)]^(-1/2) [sec(1-x)]^2 now we need the product rule, what a mess ... f''(x) = -(1/2) [tan(1-x)]^(-1/2) (2)(sec(1-x)) (sec(1-x))(tan(1-x))(-1) + (-1/2)[sec(1-x)]^2 (-1/2)tan(1-x))^(-3/2)(sec^2(1-x))(-1...
Tuesday, April 13, 2010 at 4:01pm by Reiny
trignonmetry
PLEASE tell your classmates to use parentheses to clarify the problems! I think you mean (tan*cos^2 + sin^2)/sin = (sin*cos + sin^2)/sin = sin(cos+sin)/sin = cos+sin I think you mean (1+tan)/(1-tan) = (1+tan)^2/(1-tan^2) = (1+2tan+tan^2)/(1-tan^2) = (sec^2+2tan)/(1-tan^2) I ...
Sunday, December 16, 2012 at 1:27pm by Steve
trig
Draw a diagram. If the nearer point is a distance x from the base of the mountain, then the height h can be figured from h/x = tan 43° h/(x+235) = tan 30° x = h/tan 43° x = h/tan 30° - 235 h/tan 43° = h/tan 30° - 235 h = 235/(cot30° - cot43°) h...
Wednesday, March 13, 2013 at 10:45pm by Steve
Prove trig identity
(sec^2 (x) - 1)(csc^2 (x) - 1) = 1 sec^2 (x) - 1 = tan^2 (x) csc^2 (x) - 1 = cot^2 (x) (tan^2 (x) )(cot^2 (x)) = 1 cot^2 (x) = 1/tan^2 (x) (tan^2 (x) )(1/tan^2 (x))= 1 tan^2 (x)/tan^2 (x)= 1 1 = 1
Sunday, January 9, 2011 at 3:01pm by helper
trig
cos 2t = 2cos^2(t) - 1 = 2/sec^2(t) - 1 sec^2 = 1 + tan^2 2/(1 + tan^2(t)) - 1 = [2 - (1 + tan^2 t)]/(1 + tan^2 t) x = 4tan t tan t = x/4 [2 - ( 1 + x^2 / 16)]/(1 + x^2 / 16) (32 - 1 - x^2)/(16 + x^2) (31 - x^2)/(16 + x^2)
Monday, October 17, 2011 at 10:21pm by Steve
calculus
Yo can find a reduction formula for the integral of tan^n(x) as follows. We have that tan^n(x) = sin^n(x)/cos^n(x) = sin^(n-2)(x)/cos^n(x) sin^2(x) = sin^(n-2)(x)/cos^n(x) [1-cos^2(x)] = sin^(n-2)(x)/cos^n(x) - sin^(n-2)(x)/cos^(n-2)(x) = sin^(n-2)(x)/cos^(n-2)(x) 1/cos^2(x...
Thursday, June 3, 2010 at 11:22pm by Count Iblis
Trig
s is in Q1 (sin>0, cos>0) so cos(s)=1/3, sin(s)=+√(1-(1/3)²) =√(8/9) tan(s)=√(8/9) / (1/3) =√8 t is in Q4 (sin<0, cos>0) sin(t)=-1/2 cos(t)=+√(1-(1/2)²) =√(3/4) tan(t)=(-1...
Saturday, February 19, 2011 at 11:21pm by MathMate
trig
2 csc 2θ cos 2θ = 2cot 2θ = 2/tan 2θ = 2(1-tan^2 θ)/(2tanθ) = 1/tanθ - tan^2θ/tanθ = cotθ - tanθ
Wednesday, March 13, 2013 at 5:14pm by Steve
Math - help really needed
I'll give you a hint: Try replacing the 1 with tan^2x/tan^2x. After add that to 1/tan^2x, replace the denominator (the tan^2x) with sin^2/cos^2.
Tuesday, December 2, 2008 at 9:33pm by Bob
Trigonometry
Verify the identity algebraically. This problem is very intriguing and awesome at the same time. It's wonderfully amazing! 1.) TAN³α-1/TAN α-1= TAN²α + TAN α + 1
Sunday, February 17, 2013 at 1:57am by AwesomeGuy
trig
That's not what I get. (1-cosa)/sina = tan(a/2) so, you have 1/tan(a/2) + tan(a/2) = (1 + tan^2(a/2))/tan(a/2) = sec^2(a/2)/tan(a/2) = 1/cos^2(a/2) * cos(a/2)/sin(a/2) = 1/sin(a/2)cos(a/2) = 2/sina
Saturday, November 12, 2011 at 10:45am by Steve
calculus
tan^n(x) = sin^2(x)/cos^2(x)tan^(n-2)(x) sin^2(x)/cos^2(x) = [1-cos^2(x)]/cos^2(x) = 1/cos^2(x) - 1 So: tan^n(x) = tan^(n-2)(x)/cos^2(x) - tan^(n-2)(x) Integral of tan^(n-2)(x)/cos^2(x)dx = tan^(n-1)(x)/(n-1)
Monday, November 5, 2007 at 4:06pm by Count Iblis
Math
Determine each tangent ratio to four decimal places using a calculator. A) tan 74 degrees B) tan 45 degrees C) tan 60 degrees D) tan 89 degrees How do I figure this out !?
Thursday, January 19, 2012 at 10:19am by Courtney
Trigonometry
drwls thats wrong :O tan 7pie/12 = tan (3pie/12 + 4pie/12) *break it down* = tan (pie/4 + pie/3) *find tan pie/4 and tan pie/3 on unit circle* = 1 + square root of 3 *answer* thats the exact value =] -Cyborg03
Thursday, September 13, 2007 at 12:40am by William (Cyborg03)
calculus
So I am suppose to evaulate this problem y=tan^4(2x) and I am confused. my friend did this : 3 tan ^4 (2x) d sec^ 2x (2x)= 6 tan ^4 (2x) d sec^2 (2x) She says it's right but what confuses me is she deriving the 4 and made it a three? I did the problem like this: tan^4 (2x...
Saturday, July 28, 2012 at 9:23pm by Anonymous
Math - help really needed
You actually get (1/tan^2x)+1, which is cot^2 + (csc^2x-cot^2x), which is csc^2x, which is 1/sin^2x. And that's your answer. (You could only cancel out the tan^2 if it were (Some #)(tan^2x)/tan^2x, rather than (1)+(tan^2x)/tan^2x. Dividing is the inverse of multiplying, so...
Tuesday, December 2, 2008 at 9:33pm by Bob
Pre-Calculus
Ok i understand until you say that tan(theta) +tan(beta) needs to be divided by 1-tan(theta)tan(beta) where does that come from?
Sunday, November 14, 2010 at 10:00pm by Riley
calculus
Just use the product rule f(θ) = θtanθ secθ = u*v f'(θ) = u'v + uv' = 1tanθ secθ + x(tanθ secθ)' = tanθ secθ + x[sec^2(θ)sec(θ) + tan(θ...
Sunday, October 9, 2011 at 5:58pm by Steve
Trigonometry
1.Solve tan^2x + tan x – 1 = 0 for the principal value(s) to two decimal places. 6.Prove that tan y cos^2 y + sin^2y/sin y = cos y + sin y 10.Prove that 1+tanθ/1-tanθ = sec^2θ+2tanθ/1-tan^2θ 17.Prove that sin^2w-cos^2w/tan ...
Tuesday, January 1, 2013 at 12:51pm by Alonso
Maths- complex numbers
Find tan(3 theta) in terms of tan theta Use the formula tan (a + b) = (tan a + tan b)/[1 - tan a tan b) in two steps. First, let a = b = theta and get a formula for tan (2 theta). tan (2 theta) = 2 tan theta/[(1 - tan theta)^2] Then write down the equation for tan (2 theta + ...
Tuesday, February 13, 2007 at 2:45am by Jake
math
tan^-1a+tan^-1b+tan^-1c=pi,yhen prove that a+b+c=abc
Wednesday, April 3, 2013 at 12:12pm by hasee
calculus
Differentiate e^(2x^3+x) X tan(x^2) tan e^x / ln(x^6/2-3x+3e) cos(tan(e^4x^2))
Sunday, March 30, 2008 at 1:43am by daria
Trigonometry Help
divide by 4 and multiply by tan x tan^2 x - 3 = 0 tan x = ±√3 x = kpi ± pi/3 for integer k
Sunday, November 18, 2012 at 8:06pm by Steve
Calculus : Derivative
If tan^-1(x^2-y^2/x^2+y^2) = a, prove that dy/dx = x(1 - tan a)/y(1 + tan a) Please solve!!!
Wednesday, June 15, 2011 at 3:06pm by Akash
precalculus
multiply by tan(x) tan^2(x) - 3 = 0 tan^2(x) = 3 tan(x) = ±√3 reference angle is thus x = ±pi/3 x = k*pi ± pi/3
Wednesday, November 28, 2012 at 10:42pm by Steve
trig
Given that tan 45=1, use tan(x+y) to show that tan 22.5= /2 - 1. (/is square root sign
Tuesday, August 24, 2010 at 3:40am by Mimi
PreCalculus
Tan(theta+n*PI)=tan(theta) tangent is periodic every pi radians. Tan(theta+n*180)=tan theta -tan380=-tan(380-180)=-tan200 Learn quickly those periodic formulas, or as I do, memorize the sine and cosine curves.
Sunday, November 15, 2009 at 11:26pm by bobpursley
trigonometry
recalling that sec^2 = 1+tan^2, rearrange things to have 3tan^2 y + 5 tan y - 2 = 0 (3tan y - 1)(tan y + 2) = 0 tan y = 1/3 or -2 tan is positive in QI and QIII, so tany = 1/3 means that y = 18.43° or 198.43° tany = -2 means y = 116.57° or 296.57°
Wednesday, May 2, 2012 at 4:15pm by Steve
Math
Could someone explain the steps for this question. Simplify the expression tan(5pi/8)-tan(3pi/8)/1+tan(5pi/8)tan(3pi/8)
Tuesday, January 24, 2012 at 9:49pm by Jake
math
3,4,5 right triangle in quadrant 2 cos t = -4/5 tan t = sin t/cos t = -3/4 tan 2 t = 2 tan t / (1-tan^2 t) = 2 (-3/4) / (1-9/16) = (-3/2)/(7/16) = -3 /(7/8) = -24/7 =-3.43
Friday, April 10, 2009 at 1:38am by Damon
angle sum formula
Use the sum formula for tangents: tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b))
Tuesday, January 12, 2010 at 11:15pm by MathMate
physics
tan^-1(Vz/Vx) = tan^-1(az/ax) = tan^-1(5.9/3.2) = 61.5 degrees The angle will be independent of time.
Thursday, June 23, 2011 at 12:10am by drwls
Math (trigonometry)
*Sorry, should be: I'm just wondering...why did you write (1 + tan x) / x on the *right* side? Instead of (1 + tan x) / tan x?
Sunday, April 13, 2008 at 9:53pm by Lucy
calculus (check my work please)
Not sure if it is right, I have check with the answer in the book and a few integral calculators but they seem to get a different answer ∫ sec^3(x)tan^3(x) dx ∫ sec^3(x)tan(x)(sec^2(x)-1) dx ∫ tan(x)sec(x)[sec^4(x)-sec^2(x)] dx ∫ tan(x)...
Thursday, June 7, 2012 at 12:15am by Jay
ambiguity
Ambiguous question do not get answered as fast as well-defined ones. Here's another case. Do you mean tan(3x)+1=sec(3x), or tan³x+1=sec³x, or even tan(3x+1)=sec(3x) ? In textbooks, the argument of tan(x) or exponents are indicated in a different type. ...
Tuesday, March 16, 2010 at 12:56pm by MathMate
Math
tan 110 is negative (-2.75) tan 70 is positive ( +2.75) to have negatve tangent the angle must be in quadrants 2 or 4 tan 290 = -2.75 same as tan 110
Saturday, November 24, 2012 at 8:03pm by Damon
math
tan(pi - pi/12), Factor out pi: tan(pi(1 - 1/12)), tan(pi(11/12), tan(11pi/12), tan(11pi/12)*180/pi), tan(165)deg. Now, I'm not sure what else you want done. The trig ratios(functions) cannot be written in radians or degrees, because they have no units. The angles ONLY can...
Thursday, January 13, 2011 at 12:06am by Henry
Precalculus/Trig 5
Find the exact value of thet following trigonometric functions: tan(5pi/6) tan(7pi/6) tan(11pi/6)
Monday, October 8, 2012 at 9:03am by Greg
trig
please help me. use trig. identities to find the exact value. tan 25° + tan 5° / 1- tan 25° tan 5°
Friday, April 30, 2010 at 10:14pm by ann
trig
please help me. use trig. identities to find the exact value. tan 25° + tan 5° / 1- tan 25° tan 5°
Friday, April 30, 2010 at 10:07pm by ann
maths-trignometry
Prove that tan-1 (1/7) + tan-1 (1/13) = tan-1(2/9)
Saturday, March 16, 2013 at 2:41pm by anoynomous
math
105 = 45 +60 tan 45 = 1 tan 60 = sqrt 3 tan(a+b) = (tan a+tan b)/(1-tan a tanb) =(1+sqrt 3)/(1-sqrt 3) = (1+sqrt3)(1+sqrt 3)/[(1-sqrt 3)(1+sqrt 3)] = (4+2sqrt3)/(1-3) = -(2 + sqrt 3)
Friday, April 10, 2009 at 1:34am by Damon
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