sin a + sin b + sin c - sin(a+b+c)

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Guide :

# sin a + sin b + sin c - sin(a+b+c)

whithout assuming a+b+c=180

## Research, Knowledge and Information :

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### Trigonometric and Geometric Conversions, Sin(A + B), Sin(A ...

Trigonometric and Geometric Conversions ... Sin(A + B) is not equal to sin ... The sine of angle A is 0.8 and the sine of angle B is 0.6.

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### Geometry problem - explain all steps taken to solve

The base AC, which is a chord of the circumscribed circle, can be bisected and the perpendicular bisector will meet the apex B, because triangle ABC is isosceles and the bisector splits the triangle into two congruent right-angled triangles. The bisector of chord AC also passes through the cent

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### Integrate : sin^6x dx?

cos(2x)=1-2sin^2(x), sin^2(x)=(1-cos(2x))/2; cos(4x)=2cos^2(2x)-1, cos^2(2x)=(cos(4x)+1)/2. sin^6(x)=(sin^2(x))^3=(1-cos(2x))^3/8=(1-3cos(2x)+3cos^2(2x)-cos^3(2x))/8. ∫sin^6(x)dx=∫(1-3cos(2x)+3cos^2(2x)-cos^3(2x))dx/8=x/8-(3/16)sin(2x)+(3/64)sin(4x)+3x/16-∫cos^3(2x)dx/8. [3cos^2(2x)=(

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### if sinx+cosy=1 then how will i solve its second order derivative?

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### Sin 2A + sin 2B- sin 2C = 4 cos A cos B sin C

Let's look at the sine of the sum of 3 angles and combinations of sum and differences: (1) sin(X+Y+Z) = sin(X+(Y+Z)) = sinXcos(Y+Z) + cosXsin(Y+Z) = sinXcosYcosZ - sinXsinYsinZ + cosXsinYcosZ + cosXcosYsinX. If we keep X, Y and Z in order we can use c and s

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sin^2(x)=1-sin(x). Quadratic in sin(x): sin^2(x)+sin(x)-1=0: sin(x)=(-1±√5)/2. This solution is substituted later into the solution of the main problem (see end). cos^2(x)=1-sin^2(x)=1-(1-sin(x))=sin(x). cos^12(x)=(cos^2(x))^6=sin^6(x); cos^10(x)=sin^5(x); cos^8(x)=sin^4(x); cos^6(x)=sin^3(

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### y'tanxsin2y=sin^2x + cos^2y solve using substitution

I read this as y'tan(x)sin(2y)=sin^2(x)+cos^2(y) where y'=dy/dx. 2y'tan(x)sin(y)cos(y)=sin^2(x)+cos^2(y) Let u=cos^2(y), then u'=-2sin(y)cos(y)y' and -u'tan(x)=sin^2(x)+u; u'tan(x)+u=-sin^2(x) Divide through by tan(x): u'+ucot(x)=-sin(x)cos(x). Multiply through by unknown function v(x),

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### sinx+sin^2x=1 find cos^12x+3cos^10x+3cos^8x+cos^6x+1

sinx+sin^2(x)=1. find cos^12(x)+3cos^10(x)+3cos^8(x)+cos^6(x)+1 sin(x) = 1 - sin^2(x) = cos^2(x) cos^2(x) = sin(x) The expression now reduces to, sin^6(x)+3sin^5(x)+3sin^4(x)+sin^3(x)+1 substituting for sin^2(x) = 1 - sin(x), (1 - sin(x))^3 + 3sin(x)(1 - sin(x))^2 +3(1 - si

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### prove that sin(A+2B)+sin(B+2C)-sin(C+2A)=4sin(A-B)/2.cos(B-C)/2.cos(C-A)/2

First, since the equation is an identity, it has to be true for all values of A, B and C, so let A=B=C, then we have: sin(3A)+sin(3A)-sin(3A)=4sin(0)cos(0)cos(0), which reduces to sin(3A)=0 which is not an identity but has a particular solution of A=n(pi)/3, where n is an integer. Therefore, t

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### show that ... question?

Question: show that Sin(5θ) = 5sin(θ) -20sin^3(θ) + 16sin^5(θ). By De Moivre's (cosθ + i.sinθ)^5 = cos(5θ) + i.sin(5θ) By expansion (cosθ + i.sinθ)^5 = cos^5(θ) + 5.cos^4(θ).i.sin(θ) + 10.cos^3(θ).i^2.sin^2(θ) + 10.cos^2(θ).i^3.sin^3(θ) + 5.cos(θ).i^4.sin^4(θ) + i^5.si

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### sin(2x)sin(4x)sin(6x)

Let X=sin(2x)sin(4x)sin(6x)=sin(2x)sin(4x)(sin(4x)cos(2x)+cos(4x)sin(2x))= sin(2x).2sin(2x)cos(2x)(2sin(2x)cos^2(2x)+(1-2sin^2(2x))sin(2x))= 2sin^2(2x)cos(2x)(2sin(2x)(1-sin^2(2x))+sin(2x)-2sin^3(2x))= 2sin^2(2x)(3sin(2x)-4sin^3(2x))cos(2x)=sin^3(2x)(3-4sin^2(2x)).2cos(2x)dx. We want ∫Xdx. Le

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