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 centre of the circle, O. BO=AO=OC=r, radius of the circle. Angle AOC is twice ABC because the angle at the centre is twice the angle at the circumference for sector ABC. AB=BC=12m. Angle ABC is bisected by the side bisector on AC. If AC's bisector is at X on AC, then AX=XC=10m and sinABX=AX/AB=10/12=5/6. Angle AOX=twice angle ABX [or AOX=180-twice angle ABX], because angle AOC=2ABC and ABC=2ABX and AOC=2AOX. It follows that angle AOX=ABC=2sin^-1(5/6). AX/AO=10/r=sinAOX.  Trig identities: sin2y=2sinycosy and sin(y)=sin(180-y). Applying the first of these identities: sinAOX=sinABC=sin2ABX=2sinABXcosABX=2*5/6*sqrt(1-(5/6)^2). Therefore r=10/(2sinABXcosABX)10/(2*5/6*sqrt(1-(5/6)^2))=10.8544m approx. (36/sqrt(11) or 36sqrt(11)/11). (In actual fact AOC as the angle at the centre is a reflex angle and ABC is an obtuse angle. This does not alter the logic because the sine of an angle is the same as the sine of its complement. BO, the radius, is longer than BX. Angle AOX=180-2ABX, so AX/AO=sin(180-2ABX)=sin(2ABX).)
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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)=(3/2)(cos(4x)+1), the integral of which is (3/2)(sin(4x)/4+x)=(3/8)sin(4x)+3x/2.] ∫cos^3(2x)dx=∫cos^2(2x)cos(2x)dx. Integrate by parts: dv=cos(2x)dx, v=sin(2x)/2; u=cos^2(2x), du=-4cos(2x)sin(2x)dx. ∫cos^3(2x)dx=cos^2(2x)sin(2x)/2+2∫sin^2(2x)cos(2x)dx. [d(uv)/dx=vdu/dx+udv/dx; so udv/dx=d(uv)/dx-vdu/dx; therefore, ∫udv=uv-∫vdu, hence integrating by parts using u and v.] Let p=sin(2x), dp=2cos(2x)dx so 2∫sin^2(2x)cos(2x)dx=∫p^2dp=p^3/3=sin^3(2x)/3. So (1/8)∫cos^3(2x)dx=cos^2(2x)sin(2x)/16+sin^3(2x)/24. ∫sin^6(x)dx=x/8-(3/16)sin(2x)+(3/64)sin(4x)+3x/16-cos^2(2x)sin(2x)/16-sin^3(2x)/24+k= 5x/16-(3/16)sin(2x)+(3/64)sin(4x)-cos^2(2x)sin(2x)/16-sin^3(2x)/24+k, where k is constant of integration. This further reduces to: 5x/16-(3/16)sin(2x)+(3/64)sin(4x)-(1-sin^2(2x))sin(2x)/16-sin^3(2x)/24+k= 5x/16-(3/16)sin(2x)+(3/64)sin(4x)-sin(2x)/16+sin^3(2x)/16-sin^3(2x)/24+k. ∫sin^6(x)dx=5x/16-(1/4)sin(2x)+(3/64)sin(4x)+(1/48)sin^3(2x)+k, where k is constant of integration. By carrying out the conversion to single angles below, I can confirm that this solution using multiple angles is correct. The multiple angles can be further reduced to single angles, although the question hasn't so requested: sin(2x)=2sin(x)cos(x); sin(4x)=2sin(2x)cos(2x)=2(2sin(x)cos(x))(1-2sin^2(x))=4sin(x)cos(x)-8sin^3(x)cos(x); sin^3(2x)=(2sin(x)cos(x))^3=8sin^3(x)cos(x)(1-sin^2(x))=8sin^3(x)cos(x)-8sin^5(x)cos(x). Applying these: ∫sin^6(x)dx=5x/16-sin(x)cos(x)/2+(3/16)(sin(x)cos(x)-2sin^3(x)cos(x))+(1/6)(sin^3(x)cos(x)-sin^5(x)cos(x))+k; ∫sin^6(x)dx=5x/16-(5/16)sin(x)cos(x)-(5/24)sin^3(x)cos(x)-(1/6)sin^5(x)cos(x)+k. Using differentiation, I can confirm that this answer is correct.
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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 represents the second derivative] -sin x - (1-sin x)cos^2x/sin^2y - sin y.d2y/dx2=0 when we make some substitutions. From this we get the second derivative: sin y.d2y/dx2=-(sin x + (1-sin x)cos^2x/(2sin x-sin^2x)=-(2sin^2x-sin^3x + (1-sin x)cos^2x)/(2sin x-sin^2x) The final substitution gives d2y/dx2=-(2sin^2x-sin^3x + (1-sin x)cos^2x)/(2sin x-sin^2x).sqrt(2sin x - sin^2x) =-(2sin^2x-sin^3x+(1-sin x)(1-sin^2x))/(sin x(2-sin x))^(3/2) =-(2sin^2x-sin^3x+1-sin^2x-sin x+sin^3x)/(sin x(2-sin x))^(3/2) =-(sin^2x-sin x+1)/(sin x(2-sin x))^(3/2) or (sin x-sin^2x-1)/(sin x(2-sin x))^(3/2)  
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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 in the right order to reduce the notation. So we have sin(X+Y+Z)=scc-sss+ccs-csc+ccs. (2) sin(X-(Y+Z))=cosXsin(Y+Z)-sinXcos(Y+Z)=csc+ccs-scc+sss. If we subtract (2) from (1), some terms cancel and others double up: (3) 2scc-2sss. (4) sin(X+(Y-Z))=sinXcos(Y-Z)+cosXsin(Y-Z)=scc+sss+ccs-csc. (5) sin(X-(Y-Z))=cosXsin(Y-Z)-sinXcos(Y-Z)=ccs-csc-scc-sss. Subtract (5) from (4): (6) 2scc+2sss. Add (3) and (6): (7) 4scc=4sinXcosYcosZ=4cosYcosZsinX, where we can put Y=A, Z=B, X=C. Therefore: 4cosAcosBsinC=(3)+(6)=(1)-(2)+(4)-(5)=sin(X+Y+Z)-sin(X-Y-Z)+sin(X+Y-Z)-sin(X-Y+Z)= sin(A+B+C)-sin(C-A-B)+sin(C+A-B)-sin(C-A+B). We have 3 terms on the left side of the equation so one of the above terms must be zero. Let's start with C-A-B=n(pi), where n is an integer, then C=A+B+n(pi). The equation then becomes: sin(2A+2B+n(pi))+sin(2A+n(pi))-sin(2B+n(pi)=4cosAcosBsinC. So if 2A+2B+n(pi)=2A, then B=-n(pi)/2 and C=A-n(pi)/2. sin(2A+n(pi)) must equal sin2B so 2B=2A+n(pi)=-n(pi), B=-n(pi)/2=A+n(pi)/2, so A=-n(pi) and C=A-n(pi)/2=-3n(pi)/2. Yes, this fits the original equation. Are there any more solutions? Next try A+B+C=n(pi) so C=n(pi)-(A+B) and we have: -sin(n(pi)-2A-2B)+sin(n(pi)-2B)-sin(n(pi)-2A)=sin(2A+2B-n(pi))-sin(2B-n(pi))+sin(2A-n(pi))=4cosAcosBsinC. This time 2A+2B-n(pi)=2A, so B=n(pi)/2 and C=n(pi)/2-A, 2B=n(pi)=2A-n(pi), A=3n(pi)/2 and C=n(pi)/2-A=-n(pi). Try C+A-B=n(pi), so C=B-A+n(pi). We have: sin(2B+n(pi))-sin(n(pi)-2A)-sin(2B-2A+n(pi))=sin(2B+n(pi))+sin(2A-n(pi))-sin(2B-2A+n(pi)). So 2C=2B-2A+n(pi), C=B-A+n(pi)/2; but C=B-A+n(pi) and n(pi)/2 does not equal n(pi), and this path reveals no solution. The solution of all zero angles is only one of many. If we put n=1 or -1 we get a sample of other solutions given here as ordered triplets, (A,B,C) in degrees: (180,90,270), (270,90,-180). Put n=2 or -2: (360,180,540), (540,180,-360).  
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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(x). sin^3(x)=(1-sin(x))sin(x)=sin(x)-sin^2(x)=sin(x)-1+sin(x)=2sin(x)-1. The expression cos^12(x)+cos^10(x)+cos^8(x)+cos^6(x) becomes: sin^6(x)+sin^5(x)+sin^4(x)+sin^3(x)= sin^3(x)(sin^3(x)+sin^2(x)+sin(x)+1)=(2sin(x)-1)(2sin(x)-1+1-sin(x)+sin(x)+1)= (2sin(x)-1)(2sin(x)+1)=4sin^2(x)-1=4-4sin(x)-1=3-4sin(x)=5±2√5 (see solution of quadratic earlier) =9.4721 or 0.5279 (roots of X^2-10X+5=0).
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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), yet to be determined: vu'+uvcot(x)=-vsin(x)cos(x). Take the left-hand side and apply the product rule for differentiation: d(uv)/dx=udv/dx+vdu/dx=uv'+vu', implying that uv'=uvcot(x), i.e., v'=vcot(x). We can separate the variables: v'/v=cot(x)=cos(x)/sin(x). Let q=sin(x) then q'=dq/dx=cos(x), dq=cos(x)dx, which gives us on integration ln(v)=∫cos(x)dx/sin(x)=∫dq/q=ln(q)=ln(sin(x)), so v=sin(x). [Note: dv/dx=v'=cos(x); v'/sin(x)=cos(x)/sin(x)=cot(x); v'/v=cot(x).] The left-hand side is therefore d(usin(x))/dx. The right-hand side becomes -sin^2(x)cos(x). So usin(x)=∫-sin^2(x)cos(x)dx. Let p=sin(x), then dp=cos(x)dx and the integral becomes -∫p^2dp=-p^3/3+K, where K is the constant of integration. The result is usin(x)=-p^3/3+K. Now we can substitute backwards: cos^2(y)sin(x)=-sin^3(x)/3+K or 3cos^2(y)sin(x)=k-sin^3(x) where k=3K, a constant.  CHECK Differentiate: -6cos(y)sin(y)sin(x)y'+3cos^2(y)cos(x)=-3sin^2(x)cos(x); 2cos(y)sin(y)sin(x)y'-cos^2(y)cos(x)=sin^2(x)cos(x) (dividing through by -3); 2cos(y)sin(y)sin(x)y'=sin^2(x)cos(x)+cos^2(y)cos(x)=cos(x)(sin^2(x)+cos^2(y)); Dividing through by cos(x): y'tan(x)sin(2y)=sin^2(x)+cos^2(y). OK!  
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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 - sin(x))^2 + sin(x)(1 - sin(x)) + 1 1 - 3sin(x) + 3sin^2(x) - sin^3(x) + 3sin(x) - 6sin^2(x) + 3sin^3(x) + 3 - 6sin(x) + 3sin^2(x) + sin(x) - sin^2(x) + 1  2sin^3(x) - sin^2(x)  - 5sin(x)  + 5 again substitute for sin^2(x) = 1 - sin(x), 2(1-sin(x))*sin(x) - (1 - sin(x)) - 5sin(x) + 5 2sin(x) - 2sin^2(x) - 1 + sin(x) - 5sin(x) + 5 -2sin(x) - 2sin^2(x) + 4 2(1 - sin(x)) + 2 - 2sin^2(x) 2sin^2(x) + 2 - 2sin^2(x) 2 Answer: The expression evaluates to: 2
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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, the relationship is generally false and the question has been misstated. All we can do is to find out what the identity should have been using known trig identities. I therefore offer the following as a restatement of the original problem. I suspect the left-hand side is wrong because of lack of symmetry, while the right-hand side is symmetrical. 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 in the right order to reduce the notation. So we have sin(X+Y+Z)=scc-sss+ccs-csc+ccs. (2) sin(X-(Y+Z))=cosXsin(Y+Z)-sinXcos(Y+Z)=csc+ccs-scc+sss. If we subtract (2) from (1), some terms cancel and others double up: (3) 2scc-2sss. (4) sin(X+(Y-Z))=sinXcos(Y-Z)+cosXsin(Y-Z)=scc+sss+ccs-csc. (5) sin(X-(Y-Z))=cosXsin(Y-Z)-sinXcos(Y-Z)=ccs-csc-scc-sss. Subtract (5) from (4): (6) 2scc+2sss. Add (3) and (6): (7) 4scc=4sinXcosYcosZ. Put X=(A-B)/2, Y=(B-C)/2, Z=(C-A)/2 to match the right-hand side of the equation. From these we can work out combinations of X, Y and Z. X+Y+Z=0 so sin(X+Y+Z)=0; X-(Y+Z)=(A-B-B+C-C+A)/2=A-B; X+Y-Z=(A-B+B-C-C-A)/2=A-C; X-(Y-Z)=(A-B-B+C+C-A)/2=C-B. Therefore: 4sin((A-B)/2).cos((B-C)/2).cos((C-A)/2)=(3)+(6)=(1)-(2)+(4)-(5)=0-sin(A-B)+sin(A-C)-sin(C-B)= sin(B-A)+sin(A-C)+sin(B-C) or sin(B-A)+sin(B-C)-sin(C-A).  
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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.sin^5(θ) Comparing the imaginary parts of (rhs) both expressions, i.sin(5θ) = i(5.cos^4(θ).sin(θ) - 10.cos^2(θ).sin^3(θ) + sin^5(θ)) sin(5θ) = 5.(1 - sin^2(θ))^2.sin(θ) - 10.(1 - sin^2(θ)).sin^3(θ) + sin^5(θ) sin(5θ) = 5.(1 - 2.sin^2(θ) + sin^4(θ)).sin(θ) - 10.(sin^3(θ) - sin^5(θ)) + sin^5(θ) sin(5θ) = 5.(sin(θ) - 2.sin^3(θ) + sin^5(θ)) - 10.(sin^3(θ) - sin^5(θ)) + sin^5(θ) sin(5θ) = 5.sin(θ) - 10.sin^3(θ) + 5.sin^5(θ) - 10.sin^3(θ) + 10.sin^5(θ) + sin^5(θ) sin(5θ) = 5.sin(θ) - 20.sin^3(θ) + 16.sin^5(θ)
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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. Let p=sin(2x), dp=2cos(2x)dx. ∫Xdx=∫p^3(3-4p^2)dp=∫(3p^3-4p^5)dp=(3/4)p^4-(2/3)p^6. That is: (3/4)sin^4(2x)-(2/3)sin^6(2x)+C where C=constant of integration.
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