prove that 2sinxcosx-cosx/1 -sinx+sin^2x-cos^2x=cotx
12,629 results-
trig
express this in sinx (1/ cscx + cotx )+ (1/cscx- cotx) i got 2sinx is that right?? and B) express in cosx problem: is 1 + cotx/cscx - sin^2x i get to the step of 1 + cos-sin^2x and im stuck..help! (1/cscx + cotx) + (1/cscx - cotx) = (sinx + cosx/sinx) + -
Trigonometry.
( tanx/1-cotx )+ (cotx/1-tanx)= (1+secxcscx) Good one! Generally these are done by changing everything to sines and cosines, unless you see some obvious identities. Also generally, it is best to start with the more complicated side and try to change it to -
math
prove that 2sinxcosx-cosx/1 -sinx+sin^2x-cos^2x=cotx -
Trig.......
I need to prove that the following is true. Thanks (2tanx /1-tan^x)+(1/2cos^2x-1)= (cosx+sinx)/(cosx - sinx) and thanks ........... check your typing. I tried 30º, the two sides are not equal, they differ by 1 oh , thank you Mr Reiny I'll tell my teacher -
Simplifying with Trigonometry Identities
Hi, I am a senior in High School having a really difficult time with two problems. I have to prove using the trigonometric identities that they equal each other but I am having a really hard time trying to get them to equal each other. I've spent over 2 -
Precalculus/Trig
I can't seem to prove these trig identities and would really appreciate help: 1. cosx + 1/sin^3x = cscx/1 - cosx I changed the 1: cosx/sin^3x + sin^3x/sin^3x = cscx/1-cosx Simplified: cosx + sin^3x/sin^3x = cscx/1-cosx I don't know where to go from there. -
Trig........
I need to prove that the following is true. Thanks (cosx / 1-sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds of problems when I was still teaching. it requires a little "trick" L.S. =cosx/(1-sinx) multiply top and bottom by 1+sinx, -
Trigonometry
Prove the following trigonometric identities. please give a detailed answer because I don't understand this at all. a. sin(x)tan(x)=cos(x)/cot^2 (x) b. (1+tanx)^2=sec^2 (x)+2tan(x) c. 1/sin(x) + 1/cos(x) = (cosx+sinx)(secx)(cscx) d. tan^2 (x)(1+1/tan^2 x) -
math;)
The equation 2sinx+sqrt(3)cotx=sinx is partially solved below. 2sinx+sqrt(3)cotx=sinx sinx(2sinx+sqrt(3)cotx)=sinx(sinx) 2sin^2x+sqrt(3)cosx=sin^2x sin^2x+sqrt(3)cosx=0 Which of the following steps could be included in the completed solution? a. -
Math - Trig - Double Angles
Prove: sin2x / 1 - cos2x = cotx My Attempt: LS: = 2sinxcosx / - 1 - (1 - 2sin^2x) = 2sinxcosx / - 1 + 2sin^2x = cosx / sinx - 1 = cosx / sinx - 1/1 = cosx / sinx - sinx / sinx -- Prove: 2sin(x+y)sin(x-y) = cos2y - cos2x My Attempt: RS: = 1 - 2sin^2y - 1 - -
Confused! Pre-Cal
Verify that each equation is an identity.. tan A= sec a/csca I have notes (i wasn't here that day and teacher refuses to reteach) but I don't understand them here is the notes... Problem w/ same directions: Cos x= cotx/csc x = Cosx/Sin x / 1/sinx = cosx I -
Calculus
Integrate 1/sinx dx using the identity sinx=2(sin(x/2)cos(x/2)). I rewrote the integral to 1/2 ∫ 1/(sin(x/2)cos(x/2))dx, but I don't know how to continue. Thanks for the help. Calculus - Steve, Tuesday, January 12, 2016 at 12:45am 1/2 ∫ -
trigonometry
can i use factoring to simplify this trig identity? the problem is sinx + cotx * cosx i know the answer is cscx and i know how to get it but i want to know if i can do factoring to get it bc i tried to but it wont give me the answer . this is the step i -
Pre-calc
prove the identity: (cosx)(tanx + sinx cotx)=sinx+cos(squared)x i need steps to show how i got the answer generally, it is a good idea to change all trig ratios to sines and cosines, and start with the more complicated-looking side. so.... LS = -
Pre-Calculus
How would you prove/Verify this Identity. x=theta (cosx+sinx-sin^3x)/(sinx)=cotx+cos^2x -
Trig
prove the identity (sinX)^6 +(cosX)^6= 1 - 3(sinX)^2 (cosX)^2 sinX^6= sinx^2 ^3 = (1-cosX^2)^3 = (1-2CosX^2 + cos^4) (1-cosX^2) then multiply that out 1-2CosX^2 + cos^4 - cosX^2 + 2cos^4 -cos^6 add that on the left to the cos^6, and combine terms.. -
Math Trigonometric Identities
Is this correct? (sinx - cosx)^2 = 1 - 2sinxcosx LS = sin^2x - 2sinxcosx + cos^2x = 1 - cos^2x - 2sinxcosx + cos^2x = 1 - cos^2x + cos^2x - 2sinxcosx = 1 - 2sinxcosx LS = RS -
Math trigonometric Identities
Is this correct? (sinx - cosx)^2 = 1 - 2sinxcosx LS = sin^2x - 2sinxcosx + cos^2x = 1 - cos^2x - 2sinxcosx + cos^2x = 1 - cos^2x + cos^2x - 2sinxcosx = 1 - 2sinxcosx LS = RS -
trig
I need to prove equality. a) (sina + cosa)^2 -1 / ctga - sinacosa = 2tg^2a b) (sin^2x/sinx-cosx) - (sinx+cosx/tg^2x+1) = sinx + cosx c) sin^4a - sin^2a - cos^4a + cos^2a = cosπ/2 How to do these? -
math
i have a few questions cosx + cosx =2secx ----- ----- 1+sinx 1-sinx cos(x-B)-cos(x-B) = 2sinxsinB csc2x= 1 secx cscx --- 2 cotx= cos5x+cos3x ----------- sin 5x-sin 3x -
Pre-Calc
Trigonometric Identities Prove: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx + cosx/cosx)/(sinx/cos x - 1/cosx + cosx/cosx)= tanx + cosx (just working on the left side) ((sinx + 1 - cosx)/cosx)/((sinx - 1 + -
Math
Im really struggling with these proving identities problems can somebody please show me how to do these? I'm only aloud to manipulate one side of the equation and it has to equal the other side of the equation at the end Problem 1. Sinx/(cotx+1) + -
Math (trigonometric identities)
I was given 21 questions for homework and I can't get the last few no matter how hard and how many times I try. 17. Sinx-1/sinx+1 = -cos^2x/(sinx+1)^2 18. Sin^4x + 2sin^2xcos^2x + cos^4x = 1 19. 4/cos^2x - 5 = 4tan^2x - 1 20. Cosx - sinx - cos^3x/Cosx = -
Trig Identities
Prove the following identities: 13. tan(x) + sec(x) = (cos(x)) / (1-sin(x)) *Sorry for any confusing parenthesis.* My work: I simplified the left side to a. ((sinx) / (cosx)) + (1 / cosx) , then b. (sinx + 1) / cosx = (cos(x)) / (1-sin(x)) I don't know how -
TRIG..............
Q.1 Prove the following identities:- (i) tan^3x/1+tan^2x + cot^3x/1+cot^2 = 1-2sin^x cos^x/sinx cosx (ii) (1+cotx+tanx)(sinx-cosx)/sec^3x-cosec^3x = sin^2xcos^2x. -
TRIGONOMETRY *(MATHS)
Q.1 Prove the following identities:- (i) tan^3x/1+tan^2x + cot^3x/1+cot^2 = 1-2sin^x cos^x/sinx cosx (ii) (1+cotx+tanx)(sinx-cosx)/sec^3x-cosec^3x = sin^2xcos^2x. -
Math - Pre- Clac
Prove that each of these equations is an identity. A) (1 + sinx + cos x)/(1 + sinx + cosx)=(1 + cosx)/sinx B) (1 + sinx + cosx)/(1 - sinx + cosx)= (1 + sin x)/cosx Please and thankyou! -
Calculus 2 Trigonometric Substitution
I'm working this problem: ∫ [1-tan^2 (x)] / [sec^2 (x)] dx ∫(1/secx)-[(sin^2x/cos^2x)/(1/cosx) ∫cosx-sinx(sinx/cosx) ∫cosx-∫sin^2(x)/cosx sinx-∫(1-cos^2(x))/cosx sinx-∫(1/cosx)-cosx sinx-∫secx-∫cosx sinx-sinx-∫secx =-ln |secxtanx|+C Can -
maths
hey, i would really appreciate some help solving for x when: sin2x=cosx Use the identity sin 2A = 2sinAcosA so: sin 2x = cos x 2sinxcosx - cosx = 0 cosx(2sinx - 1)=0 cosx = 0 or 2sinx=1, yielding sinx=1/2 from cosx=0 and by looking at the cosine graph, we -
drwls
My previous question: Verify that (secx/sinx)*(cotx/cscx)=cscx is an identity. (secx/sinx)*(cotx/cscx) = (secx/cscx)(cotx/sinx) = (sinx/cosx)*cotx*(1/sinx) "The last steps should be obvious" Not to me. I can convert (sinx/cosx) to tanx if that's even what -
Please help
Write in terms of cos and sin function. cotx*secx I know cotx = cosx/sinx and sec x= 1/cos x. but how do i solve it? -
another precal
i have two questions for you to check please! 6cos^2x+5cosx-4=0 where 0degrees < or equal to 0 which is < or equal to 360. I got that this factors to (3x+4)(2x-1) so the answer is x= -4/3 and x= 1/2 ?? right? and the second question: determine sin2theta, -
Trigonometry Check
Simplify #3: [cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] = [cosx-((1)cosx-(0)sinx)sinx]/[cosx-((-1)cosx+(0)sinx)tanx] = [cosx-cosxsinx]/[cosx+cosxtanx] = [cosx(1-sinx]/[cosx(1+tanx] = -
Math help again
cos(3π/4+x) + sin (3π/4 -x) = 0 = cos(3π/4)cosx + sin(3π/4)sinx + sin(3π/4)cosx - cos(3π/4)sinx = -1/sqrt2cosx + 1/sqrt2sinx + 1/sqrt2cosx - (-1/sqrt2sinx) I canceled out -1/sqrt2cosx and 1/sqrt2cosx Now I have 1/sqrt sinx + 1/sqrt2 sinx And that -
math
Write in terms of cos and sin function. cotx*secx Show work. I know cotx = cosx/sinx and secx = 1/cosx, would that just be the answer or can i solve it? -
Calculus
Q: If y=sinx/(1+tanx), find value of x not greater than pi, corresponding to maxima or minima value of y. I have proceeded thus- Equating dy/dx=0 we get{ (1+tanx)cosx-sinx.sec^2 x}/(1+tanx)^2=0……..(A) Or cosx+sinx=sinx.sec^2 x or cosx= sinx.sec^2-sinx -
Math
1) evaluate without a calculator: a)sin(3.14/4) b) cos(-3(3.14)/4) c) tan(4(3.14)/3) d) arccos(- square root of three/2) e) csctheata=2 2) verify the following identities: a) cotxcosx+sinx=cscx b)[(1+sinx)/ cosx] + [cosx/ (1+sinx)]= 2secx c) (sin^2x + -
Precalculus/Trig
Prove the following identity: 1/tanx + tanx = 1/sinxcosx I can't seem to prove it. This is my work, I must've made a mistake somewhere: Converted 1/tanx: 1/sinx/cosx + sinx/cosx = 1/sinxcosx Simplified 1/sinx/cosx: cosx/sinx + sinx/cosx = 1/sinxcosx Found -
Math
Verify the identity . (cscX-cotX)^2=1-cosX/1+cosX _______ sorry i cant help you (cscX-cotX)=1/sinX - cosX/sinX = (1-cosX)/sinX If you square this you have (1-cosX)^2/(sinX)^2 Now use (sinX)^2 = 1 - (cosX)^2 to get (1-cosX)^2 / 1 - (cosX)^2 = -
Trigonometry
Simplify #1: cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer cscxtanx/sinx+cosx? Simplify #2: sin2x/1+cos2X = ??? I'm stuck on this one. I don't know what I should do. Simplify #3: -
Math 12
Simplify #1: cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer cscxtanx/sinx+cosx? Simplify #2: sin2x/1+cos2X = ??? I'm stuck on this one. I don't know what I should do. Simplify #3: -
trigonometric identities
Prove the following trigonometric identities 1. tanx = sinx + sin^2x/cosx(1+sinx) 2. cos^3+(cosx)(sin^2x) = 1/secx -
Trigonometry
How do you simplify: (1/(sin^2x-cos^2x))-(2/cosx-sinx)? I tried factoring and creating a LCD of (sinx+cosx)(sinx cosx) (cosx-sinx), but cannot come up with the right answer. The answer is (1+2sinx+2cosx)/sin^2x+cos^2x,but I don't know how the book arrived -
pre-calc
Solve: cos(2x-180) - sin(x-90)=0 my work: cos2xcos180 + sin2xsin180= sinxcos90 - sin90cosx -cos2x - sin2x= cosx -cos^2x + sin^2x -2sinxcosx=cosx I'm stuck here. I tried subtracting cosx from both sides and making sin^2x into 1- cos^2x, but I still can't -
maths - trigonometry
I've asked about this same question before, and someone gave me the way to finish, which I understand to some extent. I need help figuring out what they did in the second step though. How they got to the third step from the second. [sinx + tanx]/[cosx + 1] -
Trigonometic
Solving Trigonometic Equations solve for x and give the answers as a equations : ( by radian) 1)cos(sinx)=1 We know sin 2x = 2(sinx)(cosx) so (sinx)(cos)=1/2(sin 2x) So we can change your equation from (sinx)(cosx)=1 to 1/2(sin 2x) = 1 (sin 2x) = 2 But the -
Trig
Prove: (cotx sinx)(secx-cosx)=sin^2(X) -
Math
I'm only aloud to manipulate one side of the problem and the end result has to match the other side of the equation Problem 1. sinx + cosx + sinx + tanx + cosxcotx = secx + cscx Problem 2. ((sinx + cosx)/(1 + tanx))^2 + ((sinx - cos^2x)/(1 - cotx))^2 = 1 -
trig
Verify that each equation is an identity. 16. 1+tanx/sinx+cosx =secx ok i have a clue on how to do it. i multiplyed the denominator by sinx-cosx and i also did the top but when i do i get this weird fraction with all these cos and sin and then i get -
Mathematics - Trigonometric Identities - Reiny
Mathematics - Trigonometric Identities - Reiny, Friday, November 9, 2007 at 10:30pm (sinx - 1 -cos^2x) (sinx + 1 - cos^2x) should have been (sinx - 1 + cos^2x) (sinx + 1 - cos^2x) and then the next line should be sin^2x + sinx - cos^2xsinx - sinx - 1 + -
trig
Solve for [0, 360) 2sinxcosx + cosx =0 2sinxcosx = -cosx 2sinx = -cosx/cosx sinx = -1/2 {210, 330) Is this correct? -
Trig (Last URGENT)
sin2x+cosx=0 , [-180,180) = 2sinxcosx+cosx=0 = cosx(2sinx+1)=0 cosx=0 x1=cos^-1(0) x1=90 x2=360-90 x2=270 270 doesn't fit in [-180,180) what do I do? Or maybe I did something wrong. sinx=1/2 x1=sin^-1(1/2) x1=30 x2=180-30 x2=150 is this correct? Please and -
Pre-calculus
Prove the following identities. 1. 1+cosx/1-cosx = secx + 1/secx -1 2. (tanx + cotx)^2=sec^2x csc^2x 3. cos(x+y) cos(x-y)= cos^2x - sin^2y -
math
1. Prove the identity:cosx/1-tanx+sinx/1-cotx=sinx+cosx 2.Prove the identity: cosx/1+sinx+1+sinx/cosx=2secx -
trig
(cotxsec2x - cotx)/ (sinxtanx + cosx) = sinx prove. i keep getting (if i start from the left side) = cosx instead of sinx please help! -
trigonometry (please double check this)
Solve the following trig equations. give all the positive values of the angle between 0 degrees and 360 degrees that will satisfy each. give any approximate value to the nearest minute only. 1. sin2Į = (sqrt 3)/2 2. sin^2Į = cos^2Į + 1/2 3. sin 2x -
math
this is for proving identies and its fustrating i can't solve this one question! lol x=feta (btw the first part is supposed to be divided by the bottom) 1 + 2sinxcosx + sinxcosx sinx + cosx (1 + 2sinxcosx / sinx + cosx) + sinxcosx -
Calculus
Find the dervatives: 1. f(x)=(3x+1)e^x^2 2. y = e^(sin x) ln(x) 3. f(x)=(x^(2)+x)^23 4. f(x)=sin(2x)/cosx 5. square root of x/(3x+1) 6. f(x)=sin^4(3x=1)-sin(3x+1) 7. x+y=cos(xy) The answers I got: 1. 6x^(2)e^(x^2)+2xe^(x^2)+3e^(x^2) 2. e^sinx/x + -
Mathematics-Integration
Question: For any positive integer n,show that integrate.[(sin x)^2n dx ] from o - π/2 = [(2n)!*π]/[(2)^(2n+1)*(n!)^2 ] What I thought: Let I =int.[(sinx)^2n dx] And again I= int.[ (sin(π/2-x))^2n dx] = int.[ (cos)^2n dx] 2I= int.[(sin x)^2n + (cos -
pre-cal
Simplify the given expression........? (2sin2x)(cos6x) sin 2x and cos 6x can be expressed as a series of terms that involve sin x or cos x only, but the end result is not a simplification. sin 2x = 2 sinx cosx cos 6x = 32 cos^6 x -48 cos^4 x + 18 cos^2 x - -
Trig
Prove the following functions: (sinx+sin2x)/(1+cosx+cos2x)=tan x (cos3x/sinx)+(sin3x/cosx)=2cot2x tan2x=(2/cotx-tanx) theses are due in the am. Fastness would be great? -
trig
prove that (sin^4x-cos^4x)/(sinx-cosx) = sinx+cosx -
calculus
Could someone check my reasoning? thanx Find the derivative of the function. sin(sin[sinx]) I need to use the chain rule to solve. So I take the derivative sin(sin[sinx) first. Then multiply that by the inside which is the derivative of sin[sinx]. And -
Trig Help
Prove the following: [1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)] =[sinx+sin^2x]/[sinx+1] =[sinx+(1-cos^2x)]/[sinx+1] =??? This is where I'm stuck. Can someone help me. Please check what I got is -
Pre-Calculus
How can this identity be proved/verified? cscx-sinx=1/(secxtanx) I have tried starting on both sides of the equation to get to the other. By starting on one side of the equation and manipulating it to make it look like the other side, the identity will be -
calc
Where do I start to prove this identity: sinx/cosx= 1-cos2x/sin2x please help!! Hint: Fractions are evil. Get rid of them. Well, cos2x = cos2x - sin2x, so 1-coscx = 1 - cos2x - sin2x = 1 - cos2x + sin2x You should be able to simplify this to 2*something -
Pre-Calc : Verify the Identity
Verify the Identity: sin(x+π)/cos(x+3π/2) =tan^2x-sec^2x I've done: sinxcosπ+cosxsinπ / cosxcos(3π/2) - sinxsin(3π/2) sinx(-1) + cosx(0) / cosx(0)- sinx(-1) -sinx/sinx What do I do from here? Or what did I do wrong? -
Trig
#1 (1/sinx)(sin^2x + cos^2x(sinx/cosx) )/(sinx + cosx) = (sinx + cosx)/(sinx+cosx) = 1 Reiny helped with this question but I don't how Reiny did it. Please explain -
Calculus
If y=3/(sinx+cosx) , find dy/dx A. 3sinx-3cosx B. 3/(sinx+cosx)^2 C. -3/(sinx+cosx)^2 D. 3(cosx-sinx)/(sinx+cosx)^2 E. 3(sinx-cosx)/(1+2sinxcosx) -
Math(Please help)
1)tan Q = -3/4 Find cosQ -3^2 + 4^2 = x^2 9+16 = sqrt 25 = 5 cos = ad/hy = -4/5 Am I correct? 2) Use the sum and difference identites sin[x + pi/4] + sin[x-pi/4] = -1 sinx cospi/4 + cosxsin pi/4 + sinx cos pi/4 - cosx sin pi/4 = -1 2 sin x cos pi/4 =-1 cos -
Math(Please help)
2) Use the sum and difference identites sin[x + pi/4] + sin[x-pi/4] = -1 sinx cospi/4 + cosxsin pi/4 + sinx cos pi/4 - cosx sin pi/4 = -1 2 sin x cos pi/4 =-1 cos pi/4 = sqr2/2 2sin^x(sqrt2/2) = -1 sin x = -sqrt2 x = 7pi/4 and 5pi/4 Am I correct? -
math, calculus
Prove that if y = cotx then dy/dx = - (cscx)^2. Hint: cotx = cosx/sinx -
Pre-Calc
How do I solve this? My work has led me to a dead end. tan(45-x) + cot(45-x) =4 my work: (tan45 - tanx)/(1+ tan45tanx) + (cot45 - cotx)/(1 + cot45cotx) = 4 (1-tanx)/(1+tanx) + (1-cotx)/(1+cotx) = 4 Then I found a common denominator, giving me this: -
Math - help really needed
Prove each idenity. 1+1/tan^2x=1/sin^2x 1/cosx-cosx=sinxtanx 1/sin^2x+1/cos^2x=1/sin^2xcos^2x 1/1-cos^2x+/1+cosx=2/sin^2x and (1-cos^2x)(1+1/tan^2x)= 1 I haven't even gotten 'round to sny of the quedtions because the first one is just so hard. I'm not -
Maths
Solve this equation fo rx in the interval 0 -
maths
if sinX+cosX=m prove that sin^6X+cos^6X=4-3(m^2-1)^2/4 where m^2= -
Math - help really needed
I'm sorry to double post; I don't want to seem impatient, but I really need help with this. Prove each idenity. 1+1/tan^2x=1/sin^2x 1/cosx-cosx=sinxtanx 1/sin^2x+1/cos^2x=1/sin^2xcos^2x 1/1-cos^2x+/1+cosx=2/sin^2x and (1-cos^2x)(1+1/tan^2x)= 1 I haven't -
Math
Prove each of the following identities. a)tan^2x = 1 – cos^2x/cos^2x b)sinx/1 – sin^2x = tanx/cosx -
math, calculus
Calculate the derivatives of the following: (also please could you show the steps, i'm so confused) a) y = sinx - cosx b) f(x) = x^2 + cosx c) y = sin(x^2) d) f(x) = x^5 sin (2x^3) e) y = cos (sinx) f) f(x) = sin^2 x -
Pre-Calculus
Hi, I need a bit of help with verifying identities. The problems are as follows: sinx+cosx/cotx+1=sinx sin2x/sin - cos2x/cosx=secx 1-tan^2θ/1+tan^2θ=cos2θ -
Mathematics-Integration
Question: Prove that [integrate {x*sin2x*sin[π/2*cos x]} dx] /(2x-π) } from (0-π) = [ integrate {sin x*cos x*sin[π/2*cos x} dx ] from (0-π). My thoughts on the question: We know that integrate f(x) dx from (0-a) = integrate f(a-x) dx from (0-a) From -
Mathematics - Trigonometric Identities
Prove: sin^2x - sin^4x = cos^2x - cos^4x What I have, LS = (sinx - sin^2x) (sinx + sin^2x) = (sinx - 1 -cos^2x) (sinx + 1 - cos^2x) = sin^2x + sinx - sinx - cos^2xsinx - cos^2xsinx - 1 - 1 + cos^4x = sin^2x - 2cos^2xsinx - 2 + cos^4x Where did I go wrong? -
Math, please help
Which of the following are trigonometric identities? (Can be more then one answer) tanx cosx cscx = 1 secx-cosx/secs=sin^2x 1-tanxtany=cos(x+y)/cosxcosy 4cosx sinx = 2cosx + 1 - 2sinx Find all solutions to the equation cosx cos(3x) - sinx sin(3x) = 0 on -
Calc II
For the following question, we need to find the length of the polar curve: r= 2/(1-cosx) from pi/2 -
Math
Choose the three true statements from the following. Options A) The period of y =cosx is 90◦. B) The period of y =sinx is ð radians. C) The period of y =sin(2x) is four times the period of y =sinx. D) The period of y =sin(2x) is half of the period of y -
Math
Choose the three true statements from the following. Options A) The period of y =cosx is 90◦. B )The period of y =sinx is ð radians. C) The period of y =sin(2x) is four times the period of y =sinx. D) The period of y =sin(2x) is half of the period of y -
math trigonometry
Prove the identity Sin^3X sinx*cos^2X=tanx/cosX -
TC Maths
Prove sin^2(x)-cos^2(y)/Sinx cosx-Sinycosy=TAN(x+y) -
math trigonometry
Prove the identity Sin^3X+sinx*cos^2X=tanx/cosX -
Trig
Simplify sin x cos^2x-sinx Here's my book's explanation which I don't totally follow sin x cos^2x-sinx=sinx(cos^2x-1) =-sinx(1-cos^2x) =-sinx(sin^2x) (Where does sine come from and what happend to cosine?) =-sin^3x -
Calculus
Can someone help me find the derivatives of these questions? For these can you only need to tell me the answer so i can see if i have the same thing. y=cos(-4x) y = sinx^2 y = xcosx I need help with these 6 please. y = xcos(1/x) y = (1+sinx)/(1-sin^(2)x) y -
Math
How do I solve this? tan^2x= 2tanxsinx My work so far: tan^2x - 2tanxsinx=0 tanx(tanx - 2sinx)=0 Then the solutions are: TanX=0 and sinX/cosX = 2 sin X Divide through by sinX: we have to check this later to see if allowed (ie sinX is not zero) cosX=1/2 -
mathematics
Knowing that derivative of sinx is cosx and derivative of cosx is -senx, prove by induction, that the 2n derivative of sinx = ((- 1) ^ n ) * sinx and the 2n derivative of cosx = ((- 1) ^ n)* cosx. -
Maths
Prove that 1+cosx+sinx/1-cosx+sinx=cotx -
Math
integrate (sinx)^(5)(cosx)^(12)from 0 to (pi/2) I'm having problems just even integrating it... I tried breaking it up (sin^(2)x)^(2)(cos^(2)x)^(6)(sinx)dx then doing u-sub with u=cosx and du= -sinx but then I get stuck... the answer is 8/3315 just don't -
Trig math
PLEASEE HELPPMEEE!!!! cot2x+sec2x=tan2x+csc2x and 1+2sinxcosx/sinx+cosx=sinx + cosx -
calculus
express in sinx 1 1 ---------- + -------- cscx + cotx cscx - cotx and express in cosx 1 + cot x ------- - sin^2x cscx = 1/sinx so what do i do w. that extra one on the top!?? im so confused cscx Your fractions aren't lining up. This question is also not -
Math (Calculus AB)
If y=cos^2x-sin^2x, then y'= a) -1 b) 0 c) -2(cosx+sinx) d) 2(cosx+sinx) e) -4(cosx)(sinx) I thought the answer was C but the answer key says it is E. Please help. Thanks in advance. -
math (trig)
i have some problems doing trig the first one is: Show that cos(x/2) sin(3x/2) = ½(sinx + sin2x) i know that you are supposed to substitute all those trig function things in it but i kind of forgot how to the only that i can see substituting in is the -
Mathematics - Trigonometric Identities
Prove: (tanx)(sinx) / (tanx) + (sinx) = (tanx) - (sinx) / (tanx)(sinx) What I have so far: L.S. = (sinx / cosx) sinx / (sinx / cosx) + sinx = (sin^2x / cosx) / (sinx + (sinx) (cosx) / cosx) = (sin^2x / cosx) / (cosx / sinx + sinxcosx)