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

Guide :

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

whithout assuming a+b+c=180

Research, Knowledge and Information :

Is [math]\sin(a) \sin(b) \sin(c) + \cos(a) \cos(b) \cos(c ...

[math]\sin a \sin b \sin c + \cos a \cos b \cos c[/math] [math]\qquad\leq \sin a \sin b + \cos a \cos b[/math] ... [math]\sin(a)\sin(b)\sin(c)+\cos(a)\cos(b)\cos(c ...
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The Law of Sines - Math is Fun - Maths Resources

The Law of Sines. The Law of Sines (or Sine Rule) is very useful for solving triangles: ... Law of Sines: a/sin A = b/sin B = c/sin C : Put in the values we know:
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sin A - sin B + sin C = 4 sin (A/2) cos (B/2) sin (C/2)

I'll do a similar but different problem. If a+b+c=π = 180°, prove that the identity is true. sin a + sin b + sin c=4cos(a/2)cos(b/2)cos(c/2) LHS = sina+sinb +sinc
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S = 2R²sin(A)·sin(B)·sin(C) - Cut-the-Knot

Relations between various elements of a ... (b+c) Applying the sine area formula to triangles ABL a and ACL a and then to the ... sin(B/2)sin(C/2) + sin(A/2)cos ...
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How to expand Sin (A-B+C) and tan (A+B-C) - Quora

Before anything, consider [math]\sin(A+B+C) = \sin(A + (B+C)) [/math] [math]= \sin A \cos(B+C) + \cos A\sin(B+C) [/math] [math]= \sin A(\cos B \cos C - \sin B \sin C ...
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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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trigonometry - Prove $\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A ...

If $A+B+C=180$ degrees, then prove $$ \sin^2(A)+\sin^2(B)-\sin ... sin^2A + \sin^2B - \sin^2(A + B) = -2\sin A\sin B \cos(A+B)$$ Inserting the sine and cosine ...
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Suggested Questions And Answer :

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?

sin x + cos y =1, so cos y = 1-sin x, so sin y = sqrt(1-(1-sin x)^2)=sqrt(2sin x - sin^2x) If we differentiate with respect to x we get cos x - sin y.dy/dx=0. So we can write dy/dx=(cos x)/(sin y). Differentiate again and we get -sin x - cosy(dy/dx)(dy/dx) - sin y.d2y/dx2=0 [d2y/dx2 represe
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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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if sinx+sin^2x=1 then what is the value of cos^12x+cos^10x+cos^8x+cos^6x

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^
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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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