Trignometry Question: @ =angle 2 = square prove sin2@/1-cos@ - cos2@/1-sin@ =cos@-sin@?
L.H.S. can be written like this sin2@ * (1 + cos@)/[(1+cos@)(1-cos@)] - cos2@ * (1 + sin@)/[(1 + sin@)(1-sin@) [Multiplying and dividing both terms by 1+cos@ and 1+ sin@ respective] = sin2@(1 + cos@)/(1-cos2@) - cos2@(1 + sin@) / (1 - sin2@) => now using the identity cos2@ + sin2@ = 1 => cos2@ = 1- sin2@ and sin2@ = 1- cos2@ => sin2@(1+ cos@)/sin2@ - cos2@(1+sin@)/cos2@ => 1 + cos@ - (1 + sin@) => 1 + cos@ - 1 - sin@ => cos@ - sin@ = R.H.S.
How can I solve the integration of, e^x (1+ sin x / 1+ cos x) dx?
[math]\large\displaystyle\star[/math] A2AEvaluate:[math]\large\displaystyle I = \large\displaystyle \int e^x \left(\frac{1 + \sin x}{1 + \cos x}\right) \, dx[/math][math]\implies\large\displaystyle I = \large\displaystyle\int e^x \left(\frac{1 + 2 \sin \left(\frac{x}{2} \right) \cos \left(\frac{x}{2}\right)}{2 \cos^2 \left(\frac{x}{2} \right)}\right) \, dx[/math][math]\implies\large\displaystyle I = \large\displaystyle\int e^x \left(\frac{1}{2} \sec^2 \left(\frac{x}{2} \right) + \tan \left(\frac{x}{2}\right)\right) \, dx[/math][math]\large\displaystyle\star[/math] We know that:[math]\large\displaystyle\int e^x [f(x) + f'(x)] \, dx = \large\displaystyle e^x f(x) + C[/math][math]\implies\large\displaystyle I = \large\displaystyle \int e^x \left(\large\displaystyle\underbrace{\tan \left(\frac{x}{2}\right)}_{\large\displaystyle f(x)} + \large\displaystyle\underbrace{\frac{1}{2} \sec^2 \left(\frac{x}{2} \right)}_{\large\displaystyle f'(x)} \right) \, dx[/math][math]\implies\boxed{\boxed{\large\displaystyle I = \large\displaystyle e^x \tan \left(\frac{x}{2}\right) + C}}[/math]Thanks![math]\large\displaystyle\boxed{\huge{\huge{\displaystyle\ddot\smile}}}[/math]
How do you simplify the expression 1/1-sin x multiplied by 1/1+sin x?
LHS is by diff. of squares, RHS is using the pythagorean theorem (sin^2(x)+cos^2(x)=1,1-sin^2(x)=cos^2(x))
How do I solve the equation (1 - cos x) / (1 + cos x) = 2/5?
How do I solve the equation (1 - cos x) / (1 + cos x) = 2/5?First, multiplying both sides of the equation by 1 + cos(x), we have:1 - cos(x) = 0.4[1 + cos(x)] = 0.4 + 0.4cos(x)Adding cos(x)-0.4 to both sides of the equation, we have:0.6 = 1.4cos(x)Dividing both sides by 1.4, we have: cos(x) = 3/7This gives us the approximate solutions:66.623° + 360k°293.377° + 360k°where k is an integer
Prove (1-sinx)(secx+tanx)=cosx?
(1-sinx)(secx+tanx) =sec(x) + tan(x) - sin(x)sec(x) - sin(x)tan(x) .......1.......... sin(x)................. 1.................. sin(x) =------------ + ---------- - sin(x)(----------) - sin(x)(------------) ... cos(x)........ cos(x)........ cos(x)............... cos(x) = sec(x) - sin^2(x)sec(x) =sec(x)[1 - sin^2(x)] =sec(x)cos^2(x) =cos^2(x) x 1/cos(x) = cos(x) prove//
Sin x + cos x = a, then how can you show that |sinx - cosx| = [math]\sqrt{2-a^{2}}[/math]?
Thanks for the A2A..here given data is: sinx + cosx = a,we want to find modulus of sinx-cosx right?let us deal with it,first of all let us take the data and square it on both the sides,ie after evaluating,According to trigonometry (I presume you must be having the knowledge about this formula), (sinx + cosx)^2 = sin^2x + 2sinxcosx + cos^2x..From the data we can write,a^2 = 1 + 2sinxcosx.. (because sin^2x + cos^2x = 1)thus, sinxcosx = (a^2 - 1)/2 ———-> Equation-1now let us evaluate the square of (sinx - cosx),(sinx - cosx)^2 = sin^2x - 2sinxcosx + cos^2x(sinx - cosx)^2 = 1 - 2sinxcosx(sinx - cosx)^2 = 1 - (2 * (a^2 - 1)/2) (from equation-1)(sinx - cosx)^2 = 2 - a^2While we remove squares, we have to apply square root on the other side,and we know that the value on the right should always positive,hence the result |sinx - cosx| = Square root of (2 - a^2) is obtained..Hope it helps :)
Please help me verify any of these trig identities!?
1. cscx / (cotx + tanx) = cosx always switch to sines and cosines replace cscx with 1/sinx, cotx with cosx/sinx and tanx with sinx/cosx so you get (1/sinx)/(cosx/sinx +sinx/cosx) = cosx this is a complex fraction with LCD of sinxcosx multiply each piece by sinxcosx cosx/(cos^2(x) + sin^2(x)) = cosx but cos^2(x) + sin^2(x) = 1 so you get cosx = cosx 2. cos/(1-sinx) - cosx/(1+sinx) = 2tanx LCD of left side is (1-sinx)(1 + sinx) which is 1-sin^2x making equivalent fractions gives you cosx(1+sinx)/1-sin^2x - cosx(1 - sinx)/1-sin^2x = 2sinx/cosx multiply out and add to get 2cosxsinx/1-sin^2x = 2sinx/cosx but 1-sin^2x = cos^2x so you have 2cosxsinx/cos^2x = 2sinx/cosx cancel out a cosx 2sinx/cosx = 2sinx/cosx
What is the maximum value of sin x + cos x?
The ans is √22 methods I m aware of..Differentiation of y=sin x +cos xdy/dx=Cos x -sin xSince maxima is needed,dy/dx=0Sin x = cos xX= 45 degrees. (2nπ+45)Put this value of x in y and u get ur answerMethod 2Multiply and divide both terms by (√1+1) ie√2(√2)(1/√2. Sin x. +1/√2. Cos x)The second bracket forms an identity of sin (45+x) whose max value can be 1So the ans is 1(√2)√2If a doubt remains… feel free to ask:)Edit…Method 3Sinx+sin(pi/2-x)2sin(pi/4) cos(2x -pi/2)2^(1/2) cos(2x-pi/2)And therefore is root 2 ×12^(1÷2)
The limit of (1-cosx)/sin^2x as x approaches 0?
= 0/0 Using l hopital rule: limit of (1-cosx) / sin^2 x , x-->0 = limit of sinx / 2sinxcosx , x-->0 = limit of sinx / sin(2x) , x-->0 = 0/0 l hopital rule : = limit of cosx / 2cos(2x) , x-->0 = 1/2
How do I prove that [math]\left( \dfrac{1+\sin(x) +i\cos(x)}{1+\sin(x)-i\cos(x)} \right)^n = e^{inx}[/math]?
Input:Exact result:Alternate forms: