# A + b + c = 270 potom cos2a + cos2b + cos2c

Dec 20, 2016 · For any given B (or B/2) there will be only one correct sign, which you already know from the diagram that we explored back in Functions of Any Angle. Example: If B = 280°, then B/2 = 140°, and you know that sin 140° is positive because the angle is in Quadrant II (above the axis).

sinB. sinC 50. Buktikanlah bahwa cos 2 A + cos 2 B + cos 2 C = 2 – 2.sinA. sinB. sinC dimana Feb 28, 2012 · I am sure there needs to be a condition on a, b, and c.

= 2.cos(180°-C).cos(A-B)-2cos^2 C. +1. = - 2.cosC.cos(A-B)-2.cos^2 C 11/30/2020 We have,2sin2B+4cosA+B sinA sinB+cos2A+B=1-cos2B+cos2A+B+4cosA+B sinA sinB=1+cos2A+B-cos2B+4cosA+B sinA sinB=1-2sinAsinA+2B+4cosA+B sinA sinB ∵ cosC-cosD=-2sinC+D2sinC-D2=1-2sinAsinA+2B-2sinBcosA+B=1-2sinAsinA+2B-sinB+A+B+sinB-A+B ∵ 2sinCcosD=sinC+D+sinC-D=1-2sinAsinA+2B-sinA+2B+sin-A=1-2sinAsinA=1-2sin2A=cos2A. Q23. Answer : (c) cosec θ ( I —cos2A) b 2 c 2 sin2A— b2c2 16 sin A —abc( — b c sin A abc abc (s— sin — (s— sin ( ) ákJÎE " ( moduli space) o ( Alexandria ) ( Eratosthenes , 284— 192 B.C. ) 7, 270 , 6,378 15%! ( syene ) ( Aswan Dam ) ákJFfiŒYU 10 , ( 23 2 cosa cosb cos c + cos2b cos2 c) } — (cos2 a + cos2 b + cos2c ) + 2 cosa cos b cos c 3/4/2009 The same thing may be proved by forming the square of the same determinant according to the ordinary rule; when if we write cos a a"cos" cos "y" + cos COs = cos a, &c., we get 1, cosc, cosb cos c, 1, cos a cos b, cos a, 1, which expanded is 1 + 2 cos a cos b cos c - cos2a - cos2b - cos2c, which is known to have the value in question.

## 10. If A + B + C = 270°, then cos2A + cos2B + cos2C + 4sinAsinBsinC = …. [EAMCET 2003] 1) 0 2) 1 3) 2 4)3 Ans: 2 Sol. cos2A cos2B cos2C++=−1 4sinAsinBsinC (or) Put A = B = C= 90° 11. cos 76 cos 16 cos76 cos1622°+ °− ° °= [EAMCET 2002] 1) 1 2 2) 0 3) 1 4 − 4) 3 4 Ans: …

In triangle A B C, a = 9, b = 8, and c = 4 then prove that cos B − 2 cos C = − 3 4 View solution If a,b,c are the lengths of the opposite sides respectively to the angles Considering the angles A,B,C as the interior angles of a triangle, hence, the sum A+B+C is of 180^o . A+B+C = pi => A+B = pi - C => (A+B)/2 = pi/2 - C /2 Aug 03, 2012 · xét dạng ΔABC thoả mãn điều kiện: Cos2A + Cos2B + Cos2C + 1 = 0 Một cách hỏi khác: Chứng minh rằng tam giác ABC vuông nếu thoả mãn điều kiện. Dec 20, 2016 · For any given B (or B/2) there will be only one correct sign, which you already know from the diagram that we explored back in Functions of Any Angle.

### The same thing may be proved by forming the square of the same determinant according to the ordinary rule; when if we write cos a a"cos" cos "y" + cos COs = cos a, &c., we get 1, cosc, cosb cos c, 1, cos a cos b, cos a, 1, which expanded is 1 + 2 cos a cos b cos c - cos2a - cos2b - cos2c, which is known to have the value in question.

sin(- B)] = 1 – 4 sin A sin B cos Get an answer for 'If tan a = b/c prove that c*cos2a + b*sin2a = c .' and find homework help for other Math questions at eNotes cos 90 Ccos A B 1 2sin2 C Rumus Rumus Trigonometri 11 2sinCcos A B 2sin 2C 1 from MATH 46196 at SMAN 96 JAKARTA Apr 21, 2011 · làm ơn chứng minh cho mình: cos2A + cos2B + cos2C = 1 – 2cosAcosBcosC (cos2A: cos bình phương A)? Nov 08, 2017 · If #cosA+cosB+cosC=0# then prove that #cos3A+cos3B+cos3C=12cosAcosBcosC#? Mar 09, 2018 · LHS=cos^2A+sin^2A*cos2B =1/2[2cos^2A+2sin^2A*cos2B] =1/2[1+cos2A+(1-cos2A)*cos2B =1/2[1+cos2A+cos2B-cos2A*cos2B =1/2[{1+cos2B}+{cos2A(1-cos2B)}] =1/2[2cos^2B+2sin^2B Mar 08, 2020 · Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students. Sep 27, 2017 · prove that cosa cosb cosc cos a b c 4cosa b 2cosb c 2cosc a 2 8zep81ff -Mathematics - TopperLearning.com If A, B, C are the angles of a triangle and sin 3 θ = sin (A − θ) sin (B − θ) sin (C − θ), prove that cot θ = cot A + cot B + cot C and conversely.

In triangle A B C, a = 9, b = 8, and c = 4 then prove that cos B − 2 cos C = − 3 4 View solution If a,b,c are the lengths of the opposite sides respectively to the angles Considering the angles A,B,C as the interior angles of a triangle, hence, the sum A+B+C is of 180^o . A+B+C = pi => A+B = pi - C => (A+B)/2 = pi/2 - C /2 Aug 03, 2012 · xét dạng ΔABC thoả mãn điều kiện: Cos2A + Cos2B + Cos2C + 1 = 0 Một cách hỏi khác: Chứng minh rằng tam giác ABC vuông nếu thoả mãn điều kiện. Dec 20, 2016 · For any given B (or B/2) there will be only one correct sign, which you already know from the diagram that we explored back in Functions of Any Angle. Example: If B = 280°, then B/2 = 140°, and you know that sin 140° is positive because the angle is in Quadrant II (above the axis). [TEX]cos^2 A +cos^2 B + cos^2 C = 1- 2cosAcosBcosC[/TEX] Từ vế trái ta sử dụng công thức hạ bậc : [TEX]= (1 + cos2A)/2 + (1 + cos2B)/2 + (1 + cos2C)/2[/TEX] If cos2A + cos2B + cos2C = 1 then ABC is a (a) Right angle triangle (b) Equilateral triangle (c) All the angles are acute (d) None of these Asked In Maths (7 years ago) Unsolved Read Solution (2) Is this Puzzle helpful? (7) (5) Submit Your Solution Trigonometry cos 2A + cos 2B - cos 2C = 2 cos (A+B) cos (A-B) - cos 2C = 2 cos (180°-C) cos (A-B) - cos 2C = - 2 cos C cos (A-B) - (2 cos^2 C - 1) = 1 - 2 cos C {cos (A-B) + cos Answer: sin2A + sin 2B +sin2C = 2 sin(A+B)cos(A-B) + 2sinC cosC=2sinC cos(A-B)+2sinC cosC=2sinC (cos(A-b) + cos C) =2sin C(cos(A-B) menu menu best neet coaching center | best iit jee coaching institute | best neet, iit jee coaching institute search e) cos2A+cos2B+cos2c=-1-4cosAcosBcosC f) cos2A-cos2B+cos2C=1-4sinAcosBsinC g) cos 2A + cos2B –cos 2C =1-4 sinA sinB sinC h) tanA+tanB+tanC=tanAtanBtanC i) tan2A+tan2B+tan2C=tan2Atan2Btan2C j) tan tan 5 +tan5 tan @ +tan@ tan =1 k) A+ B+ @ =2+2cosAcosBcosC l) A- B - @ =-2cosAsinBsinC 16) if A+B+C =270, then prove cos2A+cos2B+cos2C=1 Jan 11, 2021 · 1 Answer to if A+B+C= pi prove that (cosA)^2+(cosB)^2+(cosC)^2 = 1 - 2cosAcosBcosC. LHS: cos²A + cos²B +cos²C = (1/2)(1+cos2A) + (1/2)(1+cos2B) +cos²(A+B) [ SInce Combine $$\cos(2a)+\cos(2b)+\cos(2c)=-4\cos(a)\cos(b)\cos(c)-1$$ and $$\cos(a)\cos(b)\cos(c) \leq \frac{1}{8}.$$ Both formulas can be derived by using elementary methods.

30. 3 ∴cos2A+cos2B+cos2C=cos2 3 +cos2 3 +cos2 3 = 4 . 4．已知一扇形的周长为 c(c＞0)，当扇形的弧长为何值时，它有最大面积？ 并求出面积的最 大值． cos 90 Ccos A B 1 2sin2 C Rumus Rumus Trigonometri 11 2sinCcos A B 2sin 2C 1 from MATH 46196 at SMAN 96 JAKARTA 3/10/2018 3/8/2020 If A + B + C = π. Prove that: cos 2A + cos 2B – cos2C = 1 – 4 sin A . sin B · cos C Answer: L.H.S. = cos2A + cos2B – cos2C = 2cos(A + B) cos(A – B) – (2 cos 2 C – 1) = -2cosC Cos(A – B) – 2 cos 2 C + 1 [∵ cosC=-cos(A+B)] = 1 – 2cos C[cos(A – B) – cos (A + B)] = 1 – 2cos C[-2 sinA .

-1-2cosC (cos (A-B) -cosC) A + B + C = 180. cos2A + cos2B + cos2C = 2cos(A + B)cos(A -- B) + 2cos^2C -- 1 = --1 + 2cos^2C + 2cos(180 -- C)cos(A -- B) = --1 + 2cos^2C -- 2cosCcos(A -- B) Hey !!! a + b + c = 3π/2 = 270° a + b = 270° - c _____1) From given : => cos2a + cos2b+ cos2c => we know a formula cosc + cosd = 2cos(c + d)/2 × cos ( c -d)/2 If A+B+C = 270 (or 3pi/2) then find cos2A + cos2B + cos2C + 4sinAsinBsinC. 10. If A + B + C = 270°, then cos2A + cos2B + cos2C + 4sinAsinBsinC = …. [EAMCET 2003] 1) 0 2) 1 3) 2 4)3 Ans: 2 Sol. cos2A cos2B cos2C++=−1 4sinAsinBsinC (or) Put A = B = C= 90° 11. cos 76 cos 16 cos76 cos1622°+ °− ° °= [EAMCET 2002] 1) 1 2 2) 0 3) 1 4 − 4) 3 4 Ans: 4 Sol. 22() 1 cos 76 1 sin 16 2cos76 cos16 2 °+ − °− ° ° ()( )() 1 In ΔABC, A + B + C = π show that cos2A +cos2B – cos2C = 1 – 2sinA sinB cosC A + B + C = 180.

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Justify your answers. (a) sin(180 + è) (b) cos(180 + è) (c) tan(180 + è) B. Show that there are at least two ways to calculate the angle formed by the vectors [cos 19, Maths. 32sin4acos2a=cos6a-2cos4a-cos2a+2prove that sin A, cos A: cos A = 1 / sec A & cos A = sin A / tan A: cos 2 A = (1 + cos 2A)/2: cos (A+B) = cos A cos B - sin A sin B cos(A-B) = cos A cos B + sin A sin B: cos 2A cos 2A+cos2B- cos 2C= 1–4.sinA.sinB.cosC.

## If A + B + C = π/2, prove that cos2A + cos2B + cos2C = 1 + 4sinAsinB cosC. Solution : cos2A + cos2B + cos2C : Let us use the formula of (cosC + cosD) for cos2A + cos2B = 2cos(A + B)cos(A - B) + cos2C = 2cos(90 - C)cos(A - B) + 1 - 2sin 2 C Trigonometric ratios of 270 degree plus theta.

If A + B + C = 270°, then cos2A + cos2B + cos2C + 4sinAsinBsinC = ….

30. 3 ∴cos2A+cos2B+cos2C=cos2 3 +cos2 3 +cos2 3 = 4 .