# If cosθ + cos^2 θ = 1, then sin^12 θ + 3 sin^10 θ + 3 sin^8 θ+ sin^6 θ + 2 sin^4 θ + 2 sin^2 θ – 2 =?

7,610 results-
## 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 + -
## 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: -
## 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 -
## 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 + -
## Trigonometry

Solve the equation for solutions in the interval 0<=theta<2pi Problem 1. 3cot^2-4csc=1 My attempt: 3(cos^2/sin^2)-4/sin=1 3(cos^2/sin^2) - 4sin/sin^2 = 1 3cos^2 -4sin =sin^2 3cos^2-(1-cos^2) =4sin 4cos^2 -1 =4sin Cos^2 - sin=1/4 (1-sin^2) - sin =1/4 -
## 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 -
## math

Eliminate the parameter (What does that mean?) and write a rectangular equation for (could it be [t^2 + 3][2t]?) x= t^2 + 3 y = 2t Without a calculator (how can I do that?), determine the exact value of each expression. cos(Sin^-1 1/2) Sin^-1 (sin 7pi/6) -
## 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 -
## Calculus

Consider the function f(x)=sin(1/x) Find a sequence of x-values that approach 0 such that (1) sin (1/x)=0 {Hint: Use the fact that sin(pi) = sin(2pi)=sin(3pi)=...=sin(npi)=0} (2) sin (1/x)=1 {Hint: Use the fact that sin(npi)/2)=1 if n= 1,5,9...} (3) sin -
## Calculus

Consider the function f(x)=sin(1/x) Find a sequence of x-values that approach 0 such that (1) sin (1/x)=0 {Hint: Use the fact that sin(pi) = sin(2pi)=sin(3pi)=...=sin(npi)=0} (2) sin (1/x)=1 {Hint: Use the fact that sin(npi)/2)=1 if n= 1,5,9...} (3) sin -
## calculus

Find complete length of curve r=a sin^3(theta/3). I have gone thus- (theta written as t) r^2= a^2 sin^6 t/3 and (dr/dt)^2=a^2 sin^4(t/3)cos^2(t/3) s=Int Sqrt[a^2 sin^6 t/3+a^2 sin^4(t/3)cos^2(t/3)]dt =a Int Sqrt[sin^4(t/3){(sin^2(t/3)+cos^2(t/3)}]dt=a Int -
## Calculus

Hello, Could somebody kindly check my answer for the following question? Find the derivative of the following function: h(x)=3e^(sin(x+2)) h'(x)=3'(e^(sin(x+2))+3(e^(sin(x+2))' h'(x)=0(e^(sin(x+2))+3(e^(sin(x+2))(cos(1)) h'(x)=3cos1(e^(sin(x+2)) I would -
## 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. -
## math

Prove that for all real values of a, b, t (theta): (a * cos t + b * sin t)^2 <= a^2 + b^2 I will be happy to critique your work. Start on the left, square it, (a * cos t + b * sin t)^2 = a^2 (1 - sin^2t) + 2ab sin t cost+ b^2 (1 - cos^2 t)= a^2 + b^2 - -
## Calculus

Consider the function f(x)=sin(1/x) Find a sequence of x-values that approach 0 such that sin (1/x)=0 sin (1/x)=1 sin (1/x)=-1 Is sin sin (1/x)=0 and sin (1/x)=-1 does not exist. What is sin (1/x)=1 then. -
## 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 -
## Calculus

Use a Riemann sum with n = 3 terms and the right endpoint rule to approx. ∫(1, 2) sin(1/x)dx. My teacher just needs the terms written out, no need to add or multiply. This is a problem she did up on the board, so here's her answer: sin(4/3)(1/3) + -
## 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 -
## 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 - -
## pre calc trig check my work please

sin x + cos x -------------- = ? sin x sin x cos x ----- + ----- = sin x sin x cos x/sin x = cot x this is what i got, the problem is we have a match the expression to the equation work sheet and this is not one of the answers. need to figure out what im -
## Trigonometry

I need help with I just can't seem to get anywhere. this is as far as I have got: Solve for b arcsin(b)+ 2arctan(b)=pi arcsin(b)=pi-2arctan(b) b=sin(pi-2arctan(b)) Sub in Sin difference identity let 2U=(2arctan(b)) sin(a-b)=sinacosb-cosasinb -
## 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? -
## math

find all solutions in the interval [0,2 pi) sin(x+(3.14/3) + sin(x- 3.14/3) =1 sin^4 x cos^2 x Since sin (a+b) = sina cosb + cosb sina and sin (a-b) = sina cosb - cosb sina, the first problem can be written 2 sin x cos (pi/3)= sin x The solution to sin x = -
## 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 + -
## 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) -
## 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. sin2ƒÆ = (sqrt 3)/2 2. sin^2ƒÆ = cos^2ƒÆ + 1/2 3. sin 2x -
## 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 -
## calculus

Find the points on the curve y= (cos x)/(2 + sin x) at which the tangent is horizontal. I am not sure, but would I find the derivative first: y'= [(2 + sin x)(-sin x) - (cos x)(cos x)]/(2 + sin x)^2 But then I don't know what to do or if that is even -
## maths

Prove: sin^212+sin^221+sin^239+sin^248=1+sin^29+sin^218 -
## 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 -
## d/dx

d/dx( ln |sin(pi/x)| ) = ? Thanks. If those are absolute value signs, the derivative will not exist when sin (pi/x) = 0, because of the sign change that occurs there. Assume sin (pi/x) > 0 Let u(x) = pi/x and v(x) = sin x, and use the chain rule. d/dx -
## Calculus

Consider the function f(x)=sin(1/x) Find a sequence of x-values that approach 0 such that sin (1/x)=0 sin (1/x)=1 sin (1/x)=-1 Is sin sin (1/x)=0 and sin (1/x)=-1 does not exist. What is sin (1/x)=1 then. How would I show the sequence of values, any help -
## 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 -
## 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 -
## trigonometry HELP pleasE!

these must be written as a single trig expression, in the form sin ax or cos bx. a)2 sin 4x cos4x b)2 cos^2 3x-1 c)1-2 sin^2 4x I need to learn this!! if you can show me the steps and solve it so I can learn I'd be grateful!!! 1) apply the formula for sin -
## Limit Calculas

Evaluate lim->4 sin(2y)/tan(5y) Here is what I have so far. I am not sure the next steps. Can someone help me? 1. sin(2y)/(sin(5y)*cos(5y)) 2. (sin(2y)*cos(5y))/sin(5y) -
## math

Given that sin x + sin y = a and cos x + cos y =a, where a not equal to 0, express sin x + cos x in terms of a. attemp: sin x = a - sin y cos x = a - cos y sin x + cos x = 2A - (sin y + cos y) -
## Math

Show that for real x that {[cos x + 2 sin x + 1]/[cos x + sin x] } cannot have a value between 1 and 2. Let y = [(cos x+2 sin x + 1)/(cos x + sin x) ] y(cos x + sin x) = (cos x + 2 sin x + 1) sin x(y-2) + cos x(y-1)=1 , I just feel that this isn't the way -
## 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/Trig

Suppose you wish to express sin(3t) in terms of sint and cost. Apply the sum formula to sin(3t) = sin(t+2t) to obtain an expression that contains sin(2t)=sin(t+t) and cos(2t)=cos(t+t). Apply the sum formulas to those two expressions. Enter the resulting -
## Calculus

Evaluate the integral. S= integral sign I= absolute value S ((cos x)/(2 + sin x))dx Not sure if I'm doing this right: u= 2 + sin x du= 0 + cos x dx = S du/u = ln IuI + C = ln I 2 + sin x I + C = ln (2 + sin x) + C Another problem: S ((sin (ln x))/(x)) dx I -
## Pre calc

sin(θ − ϕ); tan(θ) = 5/12 θ in Quadrant III, sin(ϕ) = − sqaure root10/10 ϕ in Quadrant IV. I used the sin equation sin(a)cos(b)-sin(a)cos(b) However I am still getting the wrong answer -
## Trigonometry

Prove the identity sin(x+y+z)+sin(x+y-z)+sin(x-y+z)+ sin(x-y-z) = 4 sin(x)cos(y)cos(z) This identity is so long and after i tried to expand the left side and it just looked something crap Thanks for you help :) -
## 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 -
## Public High School Pre Calculus HH

Ok I do not know how to do this problem. I know that csc is simply sin^-1 sin (- pi/12) csc ( (37 pi)/12 ok now I know this also sin (- pi /12 ) = - sin ( pi/12) not really sure how that helps I don't know were to go from here ( - sin (pi/12) )/(sin ( (25 -
## calculus

Evaluate lim->4 sin(2y)/tan(5y) Here is what I have so far. I am not sure the next steps. Can someone help me? 1. sin(2y)/(sin(5y)*cos(5y)) 2. (sin(2y)*cos(5y))/sin(5y) -
## Trig.

tan^2BeatacscBeta-tan^2 (simplify) (sin/cos)^2Beta times 1/sin-(sin/cos)^2 (sin^2/cos^2)-(sin^2/cos^2)=-sin/cos Is this correct? -
## MathsSs triG

Consider sin(x-360)sin(90-x)tan(-x)/cos(90+x) 1.A.SIMPLIFY sin(x-360)sin(90-x)tan(-x)/cos(90+x) to a single trigonometric ratio B.hence or otherwise without using a calculator,solve for X if 0<X<360. sin(x-360)sin(90-x)tan(-x)/cos(90+x) =0,5 -
## 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

Create sketches by hand of the following sine functions, showing at least one full period and state the domain and range. Submit them to your teacher using the drop box above. It may be easiest for you to scan your hand drawn sketches and submit the scan -
## 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? -
## Trigonometry

Does anyone have a good website that shows the proofs for these equations? sin(u+v) = sin(u)cos(v) + sin(v)cos(u) cos(u+v) = cos(u)cos(v) + sin(v)sin(u) Thanks! -
## MATH

1.)Find the exact solution algebriacally, if possible: (PLEASE SHOW ALL STEPS) sin 2x - sin x = 0 2.) GIVEN: sin u = 3/5, 0 < u < ï/2 Find the exact values of: sin 2u, cos 2u and tan 2u using the double-angle formulas. 3.)Use the half-angle formulas -
## 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; -
## maths 2

The original question I had was write arcsin4 in the form a+ib. I manage and understand how to get so far BUT How do I get from cosacoshb-isinasinhb=4 to 2m(pi)+/- iarccosh4 arcsin4 = a + b i ---> 4 = sin(a + bi) sin(a + bi) = sin(a)cos(bi) + cos(a) -
## math

Determine all the possible values of x where 0 deg is more than or equal to x and x is more or equal to 360 deg such that i'm not sure the solution, please correct it, sin x sec x - 3 sin x = 0 the solution: sin x 1/cos x - 3 sin x = 0 sin x - 3 sin x = -
## 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 -
## Geometry One multiple choice question!

Write the ratios for sin A and cos A. {picture is of a right triangle, ABC. segment AC is 8, segment AB is 17, and segment CB is 15.} sin A=15/17, cos A=8/17 sin A=15/8, cos A=8/17 sin A=15/17, cos A=8/15 sin A=8/17, cos A=15/17 -
## antiderivatives

i need the antiderivative of sinxcosx dx for intergral pie/2 to 0 thanks. sin(x)cos(x)dx = sin(x)d[sin(x)] = 1/2d[sin^2(x)] -
## Math, Pre-Calc

the original problem was: Solve: sin(3x)-sin(x)=cos(2x) so far i've gooten to: sin(x)(2sin(x)cos(x)-1)=cos^2(x)-sin^2(x) Where would I go from here? -
## Integration by Parts

integral from 0 to 2pi of isin(t)e^(it)dt. I know my answer should be -pi. **I pull i out because it is a constant. My work: let u=e^(it) du=ie^(it)dt dv=sin(t) v=-cos(t) i integral sin(t)e^(it)dt= -e^(it)cos(t)+i*integral cost(t)e^(it)dt do integration by -
## Trig

If angle A is 45 degrees and angle B is 60 degrees. Find sin(A)cos(B), find cos(A)sin(B), find sin(A)sin(B), and find cos(A)cos(B) The choises for the first are: A. 1/2[sin(105)+sin(345)] B. 1/2[sin(105)-sin(345)] C. 1/2[sin(345)+cos(105)] D. -
## 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!

The identities cos(a-b)=cos(a)cos(b)sin(a)sin(b) and sin(a-b)=sin(a)cos(b)-cos(a)sin(b) are occasionally useful. Justify them. One method is to use rotation matricies. Another method is to use the established identities for cos(a+b) and sin (a+b). -
## Math Help

Hi! Can someone help check this for me and see if I'm doing it right? Thanks!! :) Directions: Use the Half-Angle formulas to determine the exact value of sin(pi/12). Here's what I have: π/12 = ( 180° ) / 12 = 15°. = sin ( π/12 ) = sin 15° = -
## precalc

prove the identity: cos^4 - sin^4 = cos^2 - sin^2 (cos^2 + sin^2)(cos^2 - sin^2) cos^2 + sin^2 = 1 cos^2 - sin^2 = cos^2 - sin^2 is this correct? -
## 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 -
## 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

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, calculus

Interference Two identical tuning forks are struck, one a fraction of a second after the other. The sounds produced are modeled by f1(t) = C sin ωt and f2(t) = C sin(ωt + α). The two sound waves interfere to produce a single sound modeled by the sum of -
## Geometry Please Help With One Multiple Choice

Write the ratios for sin A and cos A. {picture is of a right triangle, ABC. segment AC is 8, segment AB is 17, and segment CB is 15.} sin A=15/17, cos A=8/17 sin A=15/8, cos A=8/17 sin A=15/17, cos A=8/15 sin A=8/17, cos A=15/17 -
## 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. -
## Calculus

I need to find the exact solutions on the interval [0,2pi) for: 2sin^2(x/2) - 3sin(x/2) + 1 = 0 I would start: (2sin(x/2)-1)(sin(x/2)-1) = 0 sin(x/2)=1/2 and sin(x/2)=1 what's next? Ok, what angle has a sin equal to say 1/2 sin (x/2)=1 arc sin (1) = x/2 -
## Adv function

sin(7x)= sin(x)[cos^2(3x)-sin^2(3x)]+2cos(x)cos(3x)sin(3x) I tried for an hour but still don't know how. Plzz help -
## Math

Helle, i need help for simplify the expression: [( cos x) ( sin x) - ( sin x) ( - sin x)] / ( sin x) ² -
## Trigonometry

I have some trigonometric equations to do, but I'm pretty lost, and I have to get them done in a timely fashion, so any help would be much appreciated. "Solve the following trig equations. Give all the positive values of the angle between 0 degrees and 360 -
## Integral

That's the same as the integral of sin^2 x dx. Use integration by parts. Let sin x = u and sin x dx = dv v = -cos x du = cos x dx The integral is u v - integral of v du = -sinx cosx + integral of cos^2 dx which can be rewritten integral of sin^2 x = -sinx -
## Math

Evaluate the integral of (e^2x)*sin^3 x dx I let u = e^2x, du = (1/2)e^2x dx v= (-1/3)cos^3 x , dv =sin^3 x dx When I used integration by parts and solved it all out I got: (37/36)intgral of (e^2x)*sin^3 x dx = (-1/3)(e^2x)*cos^3 x + (1/18)(e^2x)*sin^3 x -
## 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

I'm trying to find the convolution f*g where f(t)=g(t)=sin(t). I set up the integral and proceed to do integration by parts twice, but it keeps working out to 0=0 or sin(t)=sin(t). How am I supposed to approach it? integral (sin(u)sin(t-u)) du from 0 to t. -
## 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. -
## trig integration

i'm having trouble evaluating the integral at pi/2 and 0. i know: s (at pi/2 and 0) sin^2 (2x)dx= s 1/2[1-cos(2x)]dx= s 1/2(x-sin(4x))dx= (x/2)- 1/8[sin (4x)] i don't understand how you get pi/4 You made a few mistakes, check again. But you don't need to -
## Math

Evaluate *Note - We have to find the exact value of these. That I know to do. For example sin5π/12 will be broken into sin (π/6) + (π/4) So... sin 5π/12 sin (π/6) + (π/4) sin π/6 cos π/4 + cos π/6 sin π/4 I -
## Precalc

Let x, y, and z be real numbers such that cos(x) + cos(y) + cos(z) = sin(x) + sin(y) + sin(z) = 0. Prove that cos(2x) + cos(2y) + cos(2z) = sin(2x) + sin(2y) + sin(2z) = 0. -
## Precalc

1. Prove sin - cos +1/ sin + cos - 1 = sin +1/cos 2. 1+ sin/ 1- sin = (sec + tan)^2 Thanks!! -
## Math

I need help solving for all solutions for this problem: cos 2x+ sin x= 0 I substituted cos 2x for cos^2x-sin^2x So it became cos^2(x)-sin^2(x) +sinx=0 Then i did 1-sin^2(x)-sin^2(x)+sinx=0 = 1-2sin^2(x)+sinx=0 = sinx(-2sinx+1)=-1 What did i do wrong?? the -
## Calculus

∫((cos^3(x)/(1-sin^(2)) What is the derivative of that integral? I have been trying to use trig identities but can't find one to simplify this equation. I can't find one for (cos^3(x) or (1-sin^(2)) My options -sin(x) + C sin(x) + C (1/4)cos^(4)(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 -
## Math (Linear Systems)

28 N + T2 sin 12 = T1 sin 42 T2 cos 12 = T1 cos 42 T2 sin 12 + T3 sin 54 = W2 T2 cos 12 = T3 cos 54 Im solving for T1,2,3 and W2 I just cant seem to get the system to work -
## Maths

Cos^4(θ)/cos^2(α) + sin^4(θ)/ sin^2(α)=1 Prove that cos^4 alpha/cos^ thetha + sin^4alpha/ sin^2thetha= 1 -
## math

how would you prove that sin^2(a)-cos^2(b)= sin^2(b)-cos^2(a). i'm not completely sure that this is right but i used the difference of two squares on it to get (sin(a)+cos(b))(sin(a)-cos(b)) then after that i am stuck. please help -
## Math - Calculus

The identity below is significant because it relates 3 different kinds of products: a cross product and a dot product of 2 vectors on the left side, and the product of 2 real numbers on the right side. Prove the identity below. | a × b |² + (a • b)² = -
## Trig

4. Asked to simplify the expression sin(180−è), Rory volunteered the following solution: sin(180−è) = sin 180−sin è, and, because sin 180 is zero, it follows that sin(180−è) is the same as −sin è. Is this answer correct? -
## 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 -
## Math

Asked to simplify the expression sin(180Ã¢Ë†â€™ÃƒÂ¨), Rory volunteered the following solution: sin(180Ã¢Ë†â€™ÃƒÂ¨) = sin 180Ã¢Ë†â€™sin ÃƒÂ¨, and, because sin 180 is zero, it follows that -
## 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 -
## trig integration

s- integral endpoints are 0 and pi/2 i need to find the integral of sin^2 (2x) dx. i know that the answer is pi/4, but im not sure how to get to it. i know: s sin^2(2x)dx= 1/2 [1-cos (4x)] dx, but then i'm confused. The indefinite integral of (1/2) [1-cos -
## calculus

Evaluate the integral of (x)cos(3x)dx A. (1/6)(x^2)(sin)(3x)+C B. (1/3)(x)(sin)(3x)-(1/3)(sin)(3x)+C C. (1/3)(x)(sin)(3x)+(1/9)(cos)(3x)+C D. (1/2)(x^2)+(1/18)(sin)^2(3x)+C -
## Math

the original problem was: Solve: sin(3x)-sin(x)=cos(2x) so far i've gotten to: sin(x)(2sin(x)cos(x)-1)=cos^2(x)-sin^2(x) Where would I go from here? -
## calculus

using the squeeze theorem, find the limit as x->0 of x*e^[8sin(1/x)] what i did was: -1<=sin(1/x)<=1 -8<=8*sin(1/x)<=8 e^(-8)<=e^[8*sin(1/x)]<=e^(8) x*e^(-8)<=x*e^[8*sin(1/x)]<=x*e^(8) lim x->0 [x*e^(-8)] = 0 lim x->0