Which of the following expressions is equivalent to (cos(3x))/sin(x)cos(x))? csc(x) cos(2x) - sec(x) sin(2x) sec(x) cos(2x) - csc(x) sin(2x) sec(x) cos(x) - csc(x) sin(x) csc(x) cos(x) - sec(x) sin(x) This is my last question and I've tried solving it repeatedly
17,058 results-
Math
Which of the following expressions is equivalent to (cos(3x))/sin(x)cos(x))? csc(x) cos(2x) - sec(x) sin(2x) sec(x) cos(2x) - csc(x) sin(2x) sec(x) cos(x) - csc(x) sin(x) csc(x) cos(x) - sec(x) sin(x) This is my last question and I've tried solving it -
TRIG!
Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 x = (sin^2x)^3 + -
Pre-Calculus
I don't understand,please be clear! Prove that each equation is an identity. I tried to do the problems, but I am stuck. 1. cos^4 t-sin^4 t=1-2sin^2 t 2. 1/cos s= csc^2 s - csc s cot s 3. (cos x/ sec x -1)- (cos x/ tan^2x)=cot^2 x 4. sin^3 z cos^2 z= sin^3 -
math
Proving Trigonometric Identities 1. sec^2x + csc^2x= (sec^2 x)(csc^2 x) 2. sin ^3 x / sin x - cos 3x / cos x = 2 3. 1- cos x/ sin x= sin x/ 1+ cos x 4. 2 sin x cos ^2 (x/2)- 1/x sin (2x) = sinx 5. cos 2 x + sin x/ 1- sin x= 1+ 2 sin x -
math
How would you establish this identity: (1+sec(beta))/(sec(beta))=(sin^2(beta))/(1-cos(beta)) on the right, sin^2 = 1-cos^2, that factor to 1-cos * `1+cos, then the denominator makes the entire right side 1+cosB which is 1+1/sec which is 1/sec (sec+1) qed -
MATH
Hi, I really need help with these questions. I did some of them halfway, but then I got stuck. Would you please help me? Thank you so much. Prove the identity.... 1. sec x + tan x(1-sin x/cos x)=1 1/cos x + sin x/cos x(cos^2 x/cos x)=1 1+sin x/cos -
Pre-Calculus
Prove that each equation is an identity. I tried to do the problems, but I am stuck. 1. cos^4 t-sin^4 t=1-2sin^2 t 2. 1/cos s= csc^2 s - csc s cot s 3. (cos x/ sec x -1)- (cos x/ tan^2x)=cot^2 x 4. sin^3 z cos^2 z= sin^3 z - sin^5 z -
Trigonometry
Hello all, In our math class, we are practicing the trigonometric identities (i.e., sin^2(x)+cos^2(x)=1 or cot(x)=cos(x)/sin(x). Now, we are working on proofs that two sides of an equation are equal (for example, sin(x)*csc(x)=1; -
Pre Calculus
Multiply; then use fundamental identities to simplify the expression below and determine which of the following is not equivalent. (sin x + cos x) ^2 a. 1+2sinxcosx b. sec^2x−tan^2x+2cosxsinx c.sec x + 2 sin x/sec x d. sin^2x+cos^2x e. 1+2cos (pi / 2 - -
Trig
Solve in terms of sine and cosine: sec(x) csc(x)- sec(x) sin(x) so far I have: 1/cos(x) 1/sin(x) - 1/cos(x) sin(x) I am not sure where to go to from there. The book says the answer is cot(x) or cos(x)/sin(x) Thank you in advance. -
verifying trigonometric identities
How do I do these problems? Verify the identity. a= alpha, b=beta, t= theta 1. (1 + sin a) (1 - sin a)= cos^2a 2. cos^2b - sin^2b = 2cos^2b - 1 3. sin^2a - sin^4a = cos^2a - cos^4a 4. (csc^2 t / cot t) = csc t sec t 5. (cot^2 t / csc t) = csc t = sin t -
tigonometry
expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by -b and using that cos(-b)= cos(b) sin(-b)= -sin(b) gives: sin(a-b) = sin(a)cos(b) - cos(a)sin(b) Add the two equations: -
Alg2/Trig
Find the exact value of the trigonometric function given that sin u = 5/13 and cos v = -3/5. (Both u and v are in Quadrant II.) Find csc(u-v). First of all, I drew the triangles of u and v. Also, I know the formula of sin(u-v) is sin u * cos v - cos u * -
algebra
Can someone please help me do this problem? That would be great! Simplify the expression: sin theta + cos theta * cot theta I'll use A for theta. Cot A = sin A / cos A Therefore: sin A + (cos A * sin A / cos A) = sin A + sin A = 2 sin A I hope this will -
Precalculus
Use one of the identities cos(t + 2πk) = cos t or sin(t + 2πk) = sin t to evaluate each expression. (Enter your answers in exact form.) (a) sin(19π/4) (b) sin(−19π/4) (c) cos(11π) (d) cos(53π/4) (e) tan(−3π/4) (f) cos(π/4) (g) sec(π/6+ 2π) -
Math:)
1. Simplify the expression. [csc^2(x-1)]/[1+sin x] a. csc x+1 b. csc x(csc x-1) c. sin^2 x-csc x**** d. csc^2 x-cos xtan x 2. Which of the following expressions can be used to complete the equation below? sec x/1+cot^2 x a. tan x b. tan^2 x c. tan x cos x -
trig
Reduce the following to the sine or cosine of one angle: (i) sin145*cos75 - cos145*sin75 (ii) cos35*cos15 - sin35*sin15 Use the formulae: sin(a+b)= sin(a) cos(b) + cos(a)sin(b) and cos(a+b)= cos(a)cos(b) - sin(a)sin)(b) (1)The quantity = sin(145-75) = sin -
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 w sin w + cos w tan w = -
Pre Calculus
Use one of the identities cos(t + 2ðk) = cos t or sin(t + 2ðk) = sin t to evaluate each expression. (Enter your answers in exact form.) (a) sin(17ð/4) (b) sin(−17ð/4) (c) cos(17ð) (d) cos(45ð/4) (e) tan(−3ð/4) (f) cos(7ð/4) (g) sec(ð/6+2ð) -
Pre Calculus
Use one of the identities cos(t + 2ðk) = cos t or sin(t + 2ðk) = sin t to evaluate each expression. (Enter your answers in exact form.) (a) sin(17ð/4) (b) sin(−17ð/4) (c) cos(17ð) (d) cos(45ð/4) (e) tan(−3ð/4) (f) cos(7ð/4) (g) sec(ð/6+2ð) -
Trig
The question is: Set up a 2 column proof to show that each of the equations is an identity. Transform the left side to become the right side. a. (tan + cot)^2 = sec^2 + csc^2 I'm having trouble with this. b. (cos + sin)/cos + (cos - sin)/sin = csc sec I'm -
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 - -
Math
Prove the identity of the following equation: (cos 2x)/(1/(cos x)) * (sin(pi + x))/(tan x) = (sec(x) - csc(x)) * (csc(x))/(sec^2 x) * (1 - cos^2 x) Show steps to accomplish the answers -
Math(Please check)
Use the fundamental identities to simplify the expression. tan^2 Q / sec^2 Q sin^2/cos^2 / 1/cos^2 = sin^2 / cos^2 times cos^2 / 1 = The cos^2 cancels out so sin^2 is left. Is this correct? -
Mathematics - Trigonometric Identities
Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + -
Studying for math test
Multiply; then use fundamental identities to simplify the expression below and determine which of the following is not equivalent. (sin x + cos x) ^2 a. 1+2sinxcosx b. sec^2x−tan^2x+2cosxsinx c. sec x + 2 sin x/sec x d. sin^2x+cos^2x e. 1+2cos (pi/2 -x) -
calculus II
∫ tan^2 x sec^3 x dx If the power of the secant n is odd, and the power of the tangent m is even, then the tangent is expressed as the secant using the identity 1 + tan^2 x = sec^2 x I thought that since tan is even and sec is odd, we have to put this in -
trigonometry help me
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 w sin w + cos w tan w = cos w-cot w cos w 23.Find a counterexample to shows that the equation sec a� – cos -
trignonmetry
6. Prove that tan λ cos^2 λ + sin^2λ/sin λ = cos λ� + sin �λ 10. Prove that 1+tanθ/1-tanθ = sec^2θ+2tanθ/1-tan^2θ 17.Prove that sin^2w-cos^2w/tan w sin w + cos w tan w = cos w-cot w cos w 23. Find a counterexample to shows that the equation sec -
Trigonometry desperate help, clueless girl here
2. solve cos 2x-3sin x cos 2x=0 for the principal values to two decimal places. 3. solve tan^2 + tan x-1= 0 for the principal values to two decimal places. 4. Prove that tan^2(x) -1 + cos^2(x) = tan^2(x) sin^2 (x). 5.Prove that tan(x) sin(x) + cos(x)= -
Trigonometry
I need to prove that the following is true. Thanks. csc^2(A/2)=2secA/secA-1 Right Side=(2/cosA)/(1/cosA - 1) = (2/cosA)/[(1-cosA)/cosA] =2/cosA x (cosA)/(1-cosA) =2/(1-cosA) now recall cos 2X = cos^2 X - sin^2 X and we could say cos A = cos^2 A/2 - sin^2 -
trig help much appreciated! :))
1. Find the complete exact solution of sin x = . 2. Solve cos 2x – 3sin x cos 2x = 0 for the principal value(s) to two decimal places. 3. Solve tan2 x + tan x – 1 = 0 for the principal value(s) to two decimal places. 4. Prove that tan2 � – 1 + cos2 � -
Trigonometry
Verify the identities. 1.) SIN[(π/2)-X]/COS[(π/2)-X]=COT X 2.) SEC(-X)/CSC(-X)= -TAN X 3.) (1 + SIN Y)[1 + SIN(-Y)]= COS²Y 4.) 1 + CSC(-θ)/COS(-θ) + COT(-θ)= SEC θ (Note: Just relax through verifying/solving these nice fun looking math problems! -
Precal
I do not understand how to do this problem ((sin^3 A + cos^3 A)/(sin A + cos A) ) = 1 - sin A cos A note that all the trig terms are closed right after there A's example sin A cos A = sin (A) cos (A) I wrote it out like this 0 = - sin^6 A - cos^6 A + -
trig
For each expression in column I, choose the expression from column II to complete an identity: Column I Column II 1. -tanxcosx A. sin^2x/cos^2x 2. sec^2x-1 B. 1/sec^2x 3. sec x/cscx C. sin(-x) 4. 1+sin^2x D.csc^2x-cot^2x+sin^2x 5. cos^2 x E. tanx I figured -
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 26
simplify to a constant or trig func. 1. sec ²u-tan ²u/cos ²v+sin ²v change expression to only sines and cosines. then to a basic trig function. 2. sin(theta) - tan(theta)*cos(theta)+ cos(pi/2 - theta) 3. (sec y - tan y)(sec y + tan y)/ sec y combine -
precalc
prove each identity 1) (sec - cos) / tan = sin 2) (csc^2 - 1) / csc^2 = cos^2 3) (sin/csc) - 1 = -cos/sec -
Math
Eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve. x = sec Q y = cos Q x^2 + y^2 = 1/cos^2 + sin^2/cos^2 = x^2(1 +sin^2) = x^2(2-cos^2) x^2(2-1/x^2) = 2x^2 - 1 x^2 - y^2 = 1 My teacher said to use -
Precalculus check answers help!
1.) Find an expression equivalent to sec theta sin theta cot theta csc theta. tan theta csc theta sec theta ~ sin theta 2.) Find an expression equivalent to cos theta/sin theta . tan theta cot theta ~ sec theta csc theta 3.) Simplify (tan^2 theta + -
Precalculus check answers help!
1.) Find an expression equivalent to sec theta sin theta cot theta csc theta. tan theta csc theta sec theta ~ sin theta 2.) Find an expression equivalent to cos theta/sin theta . tan theta cot theta ~ sec theta csc theta 3.) Simplify (tan^2 theta + -
Trigonometry
Equation 1: (tan^2x)/(cos^2x)+sec^2x+csc^2x=sec^4x+csc^2x*cos^2x+1 Equation 2: (csc^2x*(sin^4x+cos^2x))/(cos^2x)-cos^2x = tan^2x*csc^2x+cot^2x+sin^2x-1 Equation 3: tan^2x*sin^2x - (cot^2x)/(csc^2x) = -(sin^2x*cot^2x)/(csc^2x)- cos^4x Do the following for -
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 dx - ¡ì sec^3(x) dx = sec x tan x -
trig
how do you start this equation i've been tryng it for 20min. sec^6x(secxtanx)-sec^4x(secxtanx)=sec^5xtan^3x ec^6x(secxtanx)-sec^4x(secxtanx)=sec^5xtan^3x Factor out a sec^5 tan and divide thru. Left is sec^2 x = Tan^2 x Then this should reduce to sin^2 x = -
Mathematics
(cos 2x)/(1/(cos x)) * (sin(pi + x))/(tan x) = (sec(x) - csc(x)) * (csc(x))/(sec^2 x) * (1 - cos^2 x) My friend and I are having a debate on the true identity of this equation, however my friend learned this subject and I did not. Could someone maybe shed -
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 -
Pre-Calculus
1.) Which of the following polar equations is equivalent to the parametric equations below? x=t^2 y=2t A.) r=4cot(theta)csc(theta) B.) r=4tan(theta)sec(theta) C.) r=tan(theta)sec(theta)/4 D.) r=16cot(theta)csc(theta) 2.) Which polar equation is equivalent -
Maths
Question : Integrate [x/(1+(sin a*sin x))] from 0 to pi My first thought was to apply integrate f(x) dx= f(a-x) dx method Which simplified the integral into; 2I = integrate [pi/(1+(sin a*sin x))] dx , cancelling out x Then I made the integral into the form -
AP Calculus AB
2. For an object whose velocity in ft/sec is given by v(t) = -t^2 + 6, what is its displacement, in feet, on the interval t = 0 to t = 3 secs? 3. Find the velocity, v(t), for an object moving along the x-axis if the acceleration, a(t), is a(t) = cos(t) - -
calculus trigonometric substitution
∫ dx/ (x^2+9)^2 dx set x = 3tan u dx = 3 sec^2 u du I = 3 sec^2 u du / ( 9 tan^2 u + 9)^2 = 3 sec^2 u du / ( 81 ( tan^2 u + 1)^2 = sec^2 u du / ( 27 ( sec^2 u )^2 = du / ( 27 sec^2 u = 2 cos^2 u du / 54 = ( 1 + cos 2u) du / 54 = ( u + sin 2u / 2) / 54 = -
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 ∫ -
trig
it says to verify the following identity, working only on one side: cotx+tanx=cscx*secx Work the left side. cot x + tan x = cos x/sin x + sin x/cos x = (cos^2 x +sin^2x)/(sin x cos x) = 1/(sin x cos x) = 1/sin x * 1/cos x You're almost there. thanks so -
Calc.
Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x) (dy/dx)-(dx/du)= -
calculus
Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x) (dy/dx)-(dx/du)= -
Math
cos(tan + cot) = csc only simplify one side to equal csc so far I got this far: [((cos)(sin))/(cos)] + [((cos)(cos))/(sin)] = csc I don't know what to do next -
math
Prove that for all real values of a, b, t (theta): (a * cos t + b * sin t)^2 -
Trig
Given: cos u = 3/5; 0 < u < pi/2 cos v = 5/13; 3pi/2 < v < 2pi Find: sin (v + u) cos (v - u) tan (v + u) First compute or list the cosine and sine of both u and v. Then use the combination rules sin (v + u) = sin u cos v + cos v sin u. cos (v - u) = cos u -
precalculus
I don't understand this problem: (Tanө + cos ө)/ (sec ө + cot ө) so I start off like this: ={(sinө / cos ө)+cosө}{cos ө + (sinө/cosө)} =[(sin ө +cos^2ө) (cos^2ө +sin ө)]/ cos ө but what comes next? -
trig
Hi there! I NEED SERIOUS HELP, PLEASE!!! i have such a hard time with verifying identities! The question is: [(sin(theta/2)) / csc(theta/2)] + [(cos (theta/2) / sec(theta/2)] = 1 I have a few ideas on how to solve this, but am mainly not sure how to get -
Math
State the restrictions on the variables for these trigonometric identities. a)(1 + 2 sin x cos x)/ (sin x + cos x) = sin x + cos x b) sin x /(1+ cos x) = csc x - cot x -
Trig
sin^4t-cos^4t/sin^2t cos^2t= sec^2t-csc^2t i have =(sin^2t+cos^2t)(sin^2t+cos^2t)/sin^2tcos^2t then do i go =(sin^2t+cos^2t)/sin^2tcos^2t stumped -
Math
Solve this equation algebraically: (1-sin x)/cos x = cos x/(1+sin x) --- I know the answer is an identity, and when graphed, it looks like cot x. I just don't know how to get there. I tried multiplying each side by its conjugate, but I still feel stuck. -
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. -
Simplify
Simplify expression tan(-x)csc(x) ??? Cos x -cos x Sin x Sec x -sec x -
Math
Which of these are NOT a trigonometric identity? A.) 1-cos^2x=sin^2x B.) sin^2x+1=cos^2x C.) 1+cot^2x=csc^2x D.) sec^2x-1=tan^2x -
precal
Which of the following expressions have a value of –√3? sin(11π/6) cos(5π/3) cot(7π/6) tan(11π/6) cot(2π/3) sin(5π/3) csc(5π/3) cos(7π/6) cos(π/6) csc(2π/3) it's in Quadrant 4. Trig table and unit circle give --> 2π/3+π=5π/3 Is it sin -
trig
The expression 4 sin x cos x is equivalent to which of the following? (Note: sin (x+y) = sin x cos y + cos x sin y) F. 2 sin 2x G. 2 cos 2x H. 2 sin 4x J. 8 sin 2x K. 8 cos 2x Can someone please explain how to do this problem to me? -
Trigonometry
Please review and tell me if i did something wrong. Find the following functions correct to five decimal places: a. sin 22degrees 43' b. cos 44degrees 56' c. sin 49degrees 17' d. tan 11degrees 37' e. sin 79degrees 23'30' f. cot 19degrees 0' 25'' g. tan -
math
Can you please check my work. A particle is moving with the given data. Find the position of the particle. a(t) = cos(t) + sin(t) s(0) = 2 v(0) = 6 a(t) = cos(t) + sin(t) v(t) = sin(t) - cos(t) + C s(t) = -cos(t) - sin(t) + Cx + D 6 = v(0) = sin(0) -cos(0) -
Calculus problem
Evaulate: integral 3x (sinx/cos^4x) dx I think it's sec3 x , but that from using a piece of software, so you'll have to verify that. Using uppercase 's' for the integral sign we have S 3sin(x)/cos4dx or S cos-4(x)*3sin(x)dx If you let u = cos(x) then du = -
trig
Third time is the charm? I'll try again. Could someone show me how, (- sin (x/2) /( 2 sin (x/2) + cos (x/2)) is an alternate representation for, 1 / ( 4 tan (x/2) + 2 ) TIA Carol This doesn't require the solving of any equations. For example, ( same -
Calculus 12th grade (double check my work please)
1.)Find dy/dx when y= Ln (sinh 2x) my answer >> 2coth 2x. 2.)Find dy/dx when sinh 3y=cos 2x A.-2 sin 2x B.-2 sin 2x / sinh 3y C.-2/3tan (2x/3y) D.-2sin2x / 3 cosh 3yz...>> my answer. 2).Find the derivative of y=cos(x^2) with respect to x. A.-sin (2x) B.-2x -
trigonometry
How do you work these out? sec u- 1 / 1-cos u = sec u sec x-cos x= sin x tan x 1/sin x - 1/csc x= csc x - sin x -
solving trig. equations
tan(3x) + 1 = sec(3x) Thanks, pretend 3x equals x so tanx + 1 = secx we know the law that 1 + tanx = secx so tanx + 1 becomes secx and... secx = secx sec(3x) = sec(3x) [just put 3x back in for x- you don't really have to change 3x to x but it kinda makes -
Trig
Find sin(s+t) and (s-t) if cos(s)= 1/5 and sin(t) = 3/5 and s and t are in quadrant 1. =Sin(s)cos(t) + Cos(s)Sin(t) =Sin(1/5)Cos(3/5) + Cos(-1/5)Sin(3/5) = 0.389418 Sin(s-t) =sin(s)cos(t) - cos(s)sin(t) =sin(-3/5)cos(1/5) - cos(1/5)sin(3/5) =Sin-3/5 -
Trig Help!
Question: Trying to find cos π/12, if cos π/6 = square root 3 over 2, how to find cos π/12 using DOUBLE angle formula? This is what I got so far.. cos 2(π/6) = cos (π/6 + π/6) = (cos π/6)(cos π/6) - (sin π/6)(sin π/6) = cos^2 π/6 - sin^2 π/6 Is -
math
Simplify the trigonometric function sin^4x-cos^4x cos^2â-sin^2â=1+2cosâ (1+cot^2x )(cos^2x )=cot^2x cot^2t/csct =(1-sin^2t)/sint (Work on both sides!) sinècscè- sin^2è=cos^2è -
trigo math
7. Prove that tan B� sin B� + cos �B = sec B�. 11. Prove that tanλ cos^2λ +sin^2λ/sinλ = cos λ� + sin λ�. 12. Prove that 1+tanθ/1+tanθ = sec^2θ+2tanθ/ 1-tan^2θ. 21. Prove that sin^2w-cos^2w/ tan w sin w + cos w tan w = cos w� – cot w� cos w�. -
Math - Solving Trig Equations
What am I doing wrong? Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) = cos(2x) sin(x) cos(x) = 2cos^2(x) - 1 cos(x) (+/-)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1 cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) -
product of function
Given: f(x)=2 cos x and g(x)=sin x. Which of these expressions is equivalent to (fxg)(π/16)? a) cos π/8 b) sin π/8 c) cos π/4 d) sin π/4 Please explain the answer, thank you. -
Trigonometry
For this question, they want me to use fundamental trig identities to simplify the expression. The problem is as follows; (tanx/csc^2x + tanx/sec^2x)(1+tanx/1+cotx) - 1/cos^2x I got as far as this; tanx(1/csc^2x + 1/sec^2x)(1+tanx/1+cotx) - sec^2x. I -
Math Help Please
What are the ratios for sin A and cos A? The diagram is not drawn to scale. Triangle Description- AB = 29 AC = 20 BC - 21 A. sin A = 20/29, cos A = 21/29 B. sin A = 21/29, cos A = 20/21 C. sin A = 21/29, cos A = 20/29****? D. sin A = 21/20, cos A = 20/21 -
Calculus re-post
Does anybody know how to solve this question? a) Find the arc length function for the curve measured from the point P in the direction of increasing t from P and then reparametrize the curve with respect to arc length starting from P. b) Find the point 4 -
Calculus - MathMate Please help
ok, i tried to do what you told me but i cant solve it for c because they cancel each others out! the integral for the first one i got is [sin(c)cos(x)-cos(c)sin(x)+sin(x)+c] and the integral for the 2nd one i got is [-sin(c)cos(x)+cos(c)sin(x)-sin(x)+c] I -
Trig
A. Find simpler, equivalent expressions for the following. Justify your answers. (a) sin(180 + è) (b) cos(180 + è) (c) tan(180 + è) B. Show that there are at least two ways to calculate the angle formed by the vectors [cos 19, sin 19] and [cos 54, sin -
mathematics
Simplify the expressions 1.) Cos^2x/sin^2x + csc x sin x 2.) (Sec x +1)(sec x - 1) -
precalc
Given that sin 64° = 9/10 and cos 64° = (√19)/10 find the following: 1. cot 64° = 2. cos -64° = 3. csc 64° = 4. csc -26° = 5. sec 244° = -
math;)
Show that sin(x+pi)=-sinx. So far, I used the sum formula for sin which is sin(a+b)=sin a cos b+cos a sin b. sin(x+pi)=sin x cos pi+cos x sin pi I think I am supposed to do this next, but I am not sure. sin(x+pi)=sin x cos x+sin pi cos pi If that is right -
Precalculus
Circle O below has radius 1. Eight segment lengths are labeled with lowercase letters. Six of these equal a trigonometric function of theta. Your answer to this problem should be a six letter sequence whose letters represent the segment lengths that equal -
maths
Choose the option that gives an expression for the indefinite integral ʃ (cos(4x) + 2x^2)(sin(4x) − x) dx. In each option, c is an arbitrary constant. Options A cos(4x) + 2x^2 +c B -1/8cos(4x) + 2x^2)^2 +c C 1/4 (sin(4x) − x)^2 + c D (1/(2 (sin(4x) -
Calculus
I wanted to confirm that I solved these problems correctly (we had to convert the polar curves to Cartesian equations). 1.rcos(theta)=1 x=1 2.r=2*sin(theta)+2*cos(theta) r^2=2rsin(theta)+2rcos(theta) x^2+y^2=2y+2x (a little unsure what do next if this is -
Algebra 2 math
In triangle GHI, angle H is a right angle, GH = 40, and cos G= 40/41. Draw a diagram and find each value in fraction and in decimal form. a) Sin G b. Sin I c. Cot G d. csc G e. cos I f. sec H - I have no idea how to do this -
Algebra
In triangle GHI, angle H is a right angle, GH=40, and cos G=40/41. Find each value in fraction and in decimal form. a. sin G b. sin I c. cot G d. csc G e. cos I f. sec H -
Pre Calculus.
Can someone check my answers please!!! Simplify (tan ^2 theta csc^2 theta-1)/(tan^2 theta). My answer: 1 Simplify ((cos x)/(sec x-1)) + ((cos x) /(sec x +1) My answer: 2cot^2 x Find a numerical value of one trigonometric function of x if (tan x/cot x )– -
maths
Choose the two options which are true for all values of x 1) cos (x) = cos ( x – pie/2) 2) sin (x + pie/2) = cos (x – pie/2) 3) cos (x) = sin (x – pie/2) 4) sin (x) = sin (x + 4pie) 5) sin (x) = cos (x – pie/2) 6) sin^2 (x) + cos^2 (x) = pie would -
math
Prove the trigonometric identity. tan x+cot x/csc x cos x=sec^2 x __= sec^2x __= sec^2x __ = sec^2x __= sec^2x __ = sec^2x 1/cos^2x=sec^2x sec^2x=sec^2x -
Pre-Cal (Trig) Help?
The following relationship is known to be true for two angles A and B: cos(A)cos(B)-sin(A)sin(B)=0.957269 Express A in terms of the angle B. Work in degrees and report numeric values accurate to 2 decimal places. So I'm pretty lost on how to even begin -
Math Help
1) 1+cos(3t)/ sin(3t) + sin(3t)/( 1+ cos(3t))= 2csc(3t) 2) sec^2 2u-1/ sec^2 2u= sin^2 2u 3) cosB/1- sinB= secB+ tanB -
AP Calculus
Find the velocity, v(t), for an object moving along the x-axis if the acceleration, a(t), is a(t) = cos(t) - sin(t) and v(0) = 3 v(t) = sin(t) + cos(t) + 3 v(t) = sin(t) + cos(t) + 2 v(t) = sin(t) - cos(t) + 3 v(t) = sin(t) - cos(t) + 4 -
Calculus
Find the velocity, v(t), for an object moving along the x-axis in the acceleration, a(t), is a(t)=cos(t)-sin(t) and v(0)=3 a) v(t)=sin(t) + cos(t) +3 b) v(t)=sin(t) + cos(t) +2 c) v(t)= sin(t) - cos(t) +3 d) v(t)= sin(t) - cos(t) +4