prove that:(b+c)cosA+(c+a)cosB+(a+b)cosC=a+b+c?
970971
we know \n\na = b cos C + c cos B\nb= a cos C + c cos A\nc= b cos A + a cos B\nadding the 3 and combining cos terms together we get\na+b+c = (a+b) cos C + (b+c) cos A + (c+a) cos B\n\nproved
785851
2cos^2x+cosx-1=0\n2cos^2x+2cosx-cosx-1=0\n(2cosx-1)(cosx+1)=0\n2cosx-1=0\ncosx=1/2=cospi/3\nx=2npi+/-pi/3\ncosx+1=0\ncosx=-1=cospi\nx=2npi+-pi
165476
How about (a+c)/(b+d)  ? more interesting!!\n\nbtw this only work when a/b != c/d
788468
let a be cos2x  | a | <= 1\n4a^2 - 4a + 1 = 0\n<=> (2a - 1)^2 = 0\n<=> 2a -1 = 0 \n<=> a=1/2\n<=> cos2x = cos (pi/3)\n<=> 2x = +- pi/3 + k2pi (k is a subset of Z )\n<=> x = +- pi/6 + kpi\naccording to the problem, we have these following roots :\nx = pi/6 or x = 7pi/6 or x = -pi/6  or x = 5pi/6
576001
By the transitive property, a=c
411302
cos^2 (2x) is equal to 1-sin^2(2x), remember, sin^2(x)+cos^2(x)=1
558731
distributive property;\n\na(b+c) = ab + ac
672272
I have taken the liberty of calling it the (a+b)^2=a^2+b^2+2ab ; (a+b+c)^2=a^2+b^2+c^2+2ab+2ac+... ; (a+b+c+d)^2=a^2+b^2+c^2+d^2+2a...\n+2cd etc... theorem. A bit long winded but exact.
676730
Opening brackets \nLHS=SinA+ SinA. TanA+ CosA+ CosA.CotA\n=SinA+Sin^2A/CosA+CosA+Cos^2A/SinA\nGrouping\n=(SinA+Cos^2A/SinA)+(CosA+Sin^2A?CosA)\n=(Sin^2A+Cos^2)/sinA+(Cos^2A+Sin^2A)/CosA\n=1/sinA+1/CosA\n=SecA+CosecA\n\nEasy Pesy
950350
LHS=sin^2x-cos^2x=(sinx+cosx)(sinx-cosx)\nRHS=1-2sinxcosx\n=(sin^2x+cos^2x)-2sinxcosx as sin^2x+cos^2x=1\n=(sinx-cosx)^2\nLHS and RHS can be equal only ifsinx+cosx=sinx-cosx\nso it is not an identity\nan identity must hole for all values
464966
Its actually easy:\n\ncos(a+b) = cos a cos b - sin a sin b\ncos(a-b) = cos a cos b + sin a sin b\n\nso your lhs = cos^2 a cos^2 b - sin^2 a sin^2 b\n=cos^2 a ( 1 - sin^2 b) - ( 1 - cos^2 a) sin^2 b\n=cos^2 a - cos^2 a sin^2 b - sin^2 b + cos^2 a sin^2 b\n=cos^2 a - sin^2 b\n= (cos a - sin b)(cos a + sin b)\n=rhs\n\nHope I am clear here.
880753
we have cot A/2 \n\n= (cos A/2)/(sin A/2) = 2 cos^2A/2/sin A = (cos A+1)/ Sin A\n\nwe need to prove cosA(cos A +1)/sin A + (cos C+1)(cos C)/ sin C \n= 2 cos B(1+ cos B)/sin B\n\n1st term =  cos^2 A/sin A + cos A/sin A = (1-sin ^2A)/sin A+ cos A/sin A\n= (1+ cos A)/sin A - sin A\nas sin A sin B and sin C are in AP we need to show that\n\n(1+cos A)/Sin A, (1+ cos B)/sin B, (1+ cos C)/sin C are in AP\n\n1+ cos A = 1+ (b^2+c^2-a^2)/2bc\n= (2bc+b^2+c^2-a^2)/2bc\nso (1+cos A)/sin A = (2bc +( b^2+c^2-a^2)/2abc k\n\nas sin A/ a= sin B/b = sin C/c = k\n(1+Cos C)/sin C = (2ab+(b^+a^2-c^2)/2abck\n\nBy additing you should be able to get
139883
(a+b)(a-b) \n= a^2 - ab + ba - b^2\n= a^2 - b^2 -ab +ab\n= a^2 - b^2
546258
cos2x=cos(x+x)=cos^2x-sin^2x=1-sin^2x-sin^2x=1-2sin^2x\ncos2x=1-2sin^2x  \ncos2x+2sin^2x-sinx=1-2sin^2x+2sin^2x-sinx=1-sinx=0\nsinx=1\nx=2kπ+π/2  ,   k=...,-1,0,1,...
40893
Remember this identity?\ncos A - cos B = - 2 sin ½(A + B) sin ½(A - B)\n\nIn this case, A is 5X and B is 5X\n3sin(3X)sin(2X) = -1.5(-2sin(½(5X+X))sin(½(5X-X)))\n\nUsing the identity I gave at first you get:\n-1.5(cos(5X)-cos(X)) = 1.5cos(X)-1.5cos(5X)\n\nYou could also write this using only additions as:\n1.5cos(X) + 1.5cos(5X+pi)
356153
using trig identities\nsin^2x+cos^2x=1, so 1-sin^2x=cos^2x and sin^2x=1-cos^2x\nsubstitue\ncos^2x(cosx)-2sin^2cosx\ncos^3x-2sin^2xcosx\ncosx(cos^2x-2*(1-cos^2x))\ncosx(cos^2x-2+2cos^2x)\ncosx(3cos^2x-2)
741518
1+(cotx)^2=(cscx)^2\n\nLeft hand side = 1 + cot²x\n= 1 + cos²x/sin²x\n= (sin²x + cos²x)/sin²x\n= 1/sin²x\n=csc²x\n= right hand side ......... QED