Guide :

# prove that cosx\(sinx+cosy)+cosy\(siny-cosx)=cosx\(sinx-cosy)+cosy\(siny+cosy)

this is trigonemetry prove such a challenging problem of grade nine

## Research, Knowledge and Information :

### Sinx+siny=a and cosx+cosy=b find cosx+y ? | Socratic

cos(x+y)=(b^2-a^2)/(b^2+a^2). sinx+siny=a rArr 2sin((x+y)/2)cos ... (cosx+cosy)^2=b^2# #->cos^2x+2cosxcosy+cos^2y=b^2# ... How to prove the equality #(tan^2x+1)*cos ...

### Trigonometric Identities and Formulas - analyzemath.com

... (X + Y) = cosX cosY - sinX sinY cos(X - Y) = cosX cosY + sinX sinY sin ... cos 2 X - cos 2 Y = - sin(X + Y)sin(X - Y) cos 2 X - sin 2 Y = cos(X + Y)cos(X - Y)

### Prove: cos(x+y)cosy + sin(x + y)siny = cosx | eNotes

... cos(x+y)cosy + sin(x + y)siny = cosx' and find homework help for other ... How to prove the identity `sin^2x + cos^2x ... eNotes.com is a resource used daily ...

### If sinx+siny=a and cosx+cosy=b how do you find x,y ? | Socratic

Given sinx+siny=a.....(1) cosx+cosy=b.....(2 ... Featured answers represent the very best answers the Socratic community ... How do you prove #cot^2x - cos^2x = cot ...

### Prove cos(x+y) = cosx.cosy - sinx.siny - OpenStudy

Prove cos(x+y) = cosx.cosy - sinx.siny. Mathematics ... Now to find \(cos(A + B)\) use: \(cos(A - (-B))\) \ (cos(A)cos(-B) + sin(A)sin(-B ...

### Prove the identity: (cosx + cosy)^2 + (sinx - siny)^2 = 2 + 2 ...

Prove the identity: (cosx + cosy)^2 + (sinx - siny) ... (cosx + cosy)^2 + (sinx - siny)^2 = 2 + 2 cos (x+y). 1. Ask for details ; ... (cos x +cos y)^2+(sin x-sin y)^2

## Suggested Questions And Answer :

### Prove that G has exactly gcd(m,n) elements a such that a^m = e.?

Could I see how your teacher proved the original problem to see the clue?

### Prove that x ± y = b?

looking back the basic concepts of analytic geometry..we can say that x+-y=b if and only if the slope interception is being decline by the two roots of the function of x...its too complicated but..its impossible to prove that equation.

### (sin alpha-cos alpha +1)÷(sin alpha + cos alpha -1)=(sin alpha + 1)÷(cos alpha)

Question: Prove the identity (sin alpha-cos alpha +1)÷(sin alpha + cos alpha -1)=(sin alpha + 1)÷(cos alpha). Cross-multiply, to get (sin(α) –  cos(α) + 1)*cos(α) = (sin(α) + cos(α) –  1)*(sin(α) + 1) Multiplying out and re-arranging, sin(α).cos(α) –  cos^2(α) + cos(α) = sin*2(α) + sin(α).cos(α) –  sin(α) + sin(α) + cos(α) –  1 sin(α).cos(α) –  cos^2(α) + cos(α) = sin*2(α) – 1 + sin(α).cos(α) + cos(α) sin(α).cos(α) –  cos^2(α) + cos(α) = -cos*2(α) + sin(α).cos(α) + cos(α) Since lhs = rhs now, then the identity is proved.

### What is the most accurate value of pie ?

Unfortunately, your answer isn't correct because pi starts 3.14159... You can find pi to a large number of decimal places by Googling it on the Internet. It has a decimal which goes on forever never repeating and is classed as a transcendental number and it's irrational.

### how to prove sum of complex numbers z is zero when z^3=1

QUESTION: how to prove sum of complex numbers z is zero when z^3=1. I think that what you want is to show that the sum of the roots is zero. The cubic equation f(z) = z^3 - 1 = 0 has three roots, z1, z2, and z3 s.t. z1^3 =1, z2^3 = 1 and z3^3 = 1. We wish to show that z1 + z2 + z3 = 0. We have z^3 = 1 writing the complex number z^3 in polar form, z^3 = 1 = cos(2n.pi) + i.sin(2n.pi) = 1.e^(i.(2n.pi)) Using z^3 = cos(2n.pi) + i.sin(2n.pi), then z = cos(2n.pi/3) + i.sin(2n.pi/3),  n = 0,1,2 n = 0: z1 = cos(0) + i.sin(0) = 1 n = 1: z2 = cos(2.pi/3) + i.sin(2,pi/3) = -1/2 + i.sqrt(3)/2 n = 2: z3 = cos(4.pi/3) + i.sin(4.pi/3) = -1/2 - i.sqrt(3)/2 Adding together the three roots, z1 + z2 + z3 = 1 + (-1/2 + i.sqrt(3)/2) + (-1/2 - i.sqrt(3)/2) z1 + z2 + z3 = (1 - 1/2 - 1/2) + i.(sqrt(3)/2 - sqrt(3)/2) z1 + z2 + z3 = 0

### in triangle ABC PROVE THAT 1/r1+1/r2+1/r3=1/r

To Prove : 1/r1+1/r2+1/r3=1/r  LHS =>  1/(Δ/s-a) + 1/(Δ/s-b) + 1/(Δ/s-c)     = s-a/Δ + s-b/Δ + s-c/Δ  = (s-a+s-b+s-c)/Δ = {3s+ - (a+b+c)}/Δ = [3s - 2s]/Δ = s/Δ    { r = Δ/s } So s/Δ = 1/(Δ/s) = 1/r LHS = RHS Proved - (Devansh Gaur)  www.facebook.com/DevanshGaurOfficial

### Prove that if matrix A has an eigenvalue. ‘^ then matrix A^K (The k^th power of A)

prove that if a no. is a triplet than its 7 times. the cube root of the given no.

### Use principle of mathematical induction to 3 prove that n3 —n is divisible by 3

Let p(n) = n3 - n i)If n=1 , p(1 )  = 1^3 - 1                  [ 1^3 means 1 cube ]                        = 1 - 1                        = 0                       [ 0 divisible by 3] If n =1 , p(1) is true--------------------( 1) ii)  Assume  that p( m ) is true.--------------------------(2 ) iii) we have to prove that p(m+ 1) is true.   p(m + 1) = (m + 1) ^3 - (m + 1)                      = (m + 1 ) [ (m + 1)^2  -1]                [ (m + 1) is the common factor ]                     = ( m + 1) [ m 2 + 2m + 1 - 1 ]         [ expansion of (m + 1 ) ^2 ]                     = ( m + 1 ) [ m2 + 2m ]                    = ( m + 1 ) m ( m + 2 )                      [ m is the common factor ]  {   divisibility rule of 3 is  If sum of the terms divisible by 3 then it is divisible by 3} sum = m + 1 + m + m + 2            = 3m + 3         = 3 ( m + 1)      it is divisible by 3. Therefore ( m+1) m (m+2 ) is divisible by 3 p(m+1) is divisible by 3 p(m+1) is true---------------------------(3) From (1), (2) and (3) for all values of n , p(n)  = n3 - 3 is divisible by 3