IF A+B+C = Pie Then prove that Sin2A + Sin2B + Sin2C = 4SinASinBSinC U.K.AKSHAY

Asked by Last Modified  

14 Answers

Learn Class IX-X Tuition

Follow 14
Answer

Please enter your answer

Rajesh N V

Home Tutor

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) - cos(A+B) ) = 2sinC . 2sin A sin B =4 sinA sin B sin C
Comments
Dislike Bookmark

Professional Tutor with 15 years of experience.

LHS = sin 2A + sin 2B + sin 2C = sin (pie -- 2A) + sin (pie --2B) + sin (pie --2C) = 2sin(A+B)cos(A-B) + 2sinCcosC = 2sin(pie - C)cos(A-B) + 2sinCcosC = 2sinCcos(A - B) + 2sinCcosC = 2sinC (cos (A - B)+ cos C) = 2sinC (2cos (A...
read more
LHS = sin 2A + sin 2B + sin 2C = sin (pie – 2A) + sin (pie –2B) + sin (pie –2C) = 2sin(A+B)cos(A-B) + 2sinCcosC = 2sin(pie - C)cos(A-B) + 2sinCcosC = 2sinCcos(A - B) + 2sinCcosC = 2sinC (cos (A - B)+ cos C) = 2sinC (2cos (A - B + C)/2 cos (A - B - C)/2) = 2sinC (2cos (pie - 2B)/2 cos(2A - pie)/2 = 4sinAsinBsinC = RHS Hence Proved read less
Comments
Dislike Bookmark

Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

sin2A + sin2B + sin2C = 2 sin (2A + 2B)/2 . cos (2A - 2B)/2 + sin2C = 2sin(A + B).cos(A - B) + 2 sinC.cosC = 2sin(A + B).cos(A - B) + 2 sin (Pie - (A + B)) cos (Pie - (A + B)) = 2 sin(A + B) (cos (A - B) - cos (A + B)) = 2 sin(A + B).2sinA.sinB = 4 sinA.sinB.sinC
Comments
Dislike Bookmark

Chess player

Sin 2 is a common so, take those common factor out sin2(A+B+C), we know that A+B+C=pie apply on that so, sin2pie is equal to in value 0.1096
Comments
Dislike Bookmark

Tutor

LHS 2sinAcosA+2sinBcosB+2sinCcosC....(1) A= pi-(B+C), B= pi-(A+C), C= pi-(B+A). So, cosA= cos = -cos = - ......(2) cosB= cos = -cos = - cosC= cos =.-cos = - . Putting these values in (1) and combining common terms -2 + 6sinAsinBsinC. putting value for cosBcosC from (2) in...
read more
LHS 2sinAcosA+2sinBcosB+2sinCcosC....(1) A= pi-(B+C), B= pi-(A+C), C= pi-(B+A). So, cosA= cos[pi-(B+C)] = -cos[B+C] = - [cosBcosC-sinBsinC] ......(2) cosB= cos[pi-(A+C)] = -cos[A+C] = - [cosAcosC-sinAsinC] cosC= cos[pi-(B+A)] =.-cos[B+A] = - [cosBcosA-sinBsinA]. Putting these values in (1) and combining common terms -2[sinAcosBcosC+sinBcosCcosA+sinCcosBcosA] + 6sinAsinBsinC. putting value for cosBcosC from (2) in first term and taking cosA common from second and 3rd term = -2[sinA [sinBsinc-cosA] + cosA [sin (B+C)] ] + 6sinAsinBsinC. As A= pi-(B+C) sinA=sin(B+C) we get after cancellation = -2 [sinAsinBsinC] + 6sinAsinBsinC. = 4sinAsinBsinC.= RHS Hence proved read less
1 Comments
Dislike Bookmark

Physics teacher

LHS = sin 2A + sin 2B + sin 2C = sin (pie -- 2A) + sin (pie --2B) + sin (pie --2C) = 2sin(A+B)cos(A-B) + 2sinCcosC = 2sin(pie - C)cos(A-B) + 2sinCcosC = 2sinCcos(A - B) + 2sinCcosC = 2sinC (cos (A - B)+ cos C) = 2sinC (2cos (A - B + C)/2 cos (A - B - C)/2) = 2sinC (2cos (pie - 2B)/2 cos(2A...
read more
LHS = sin 2A + sin 2B + sin 2C = sin (pie – 2A) + sin (pie –2B) + sin (pie –2C) = 2sin(A+B)cos(A-B) + 2sinCcosC = 2sin(pie - C)cos(A-B) + 2sinCcosC = 2sinCcos(A - B) + 2sinCcosC = 2sinC (cos (A - B)+ cos C) = 2sinC (2cos (A - B + C)/2 cos (A - B - C)/2) = 2sinC (2cos (pie - 2B)/2 cos(2A - pie)/2 = 4sinAsinBsinC read less
Comments
Dislike Bookmark

Teaching is my passion!!!!

sin2A + sin2B + sin2C = 2 sin (2A + 2B)/2 . cos (2A - 2B)/2 + sin2C = 2sin(A + B).cos(A - B) + 2 sinC.cosC = 2sin(A + B).cos(A - B) + 2 sin (Pie - (A + B)) cos (Pie - (A + B)) = 2 sin(A + B) (cos (A - B) - cos (A + B)) = 2 sin(A + B).2sinA.sinB = 4 sinA.sinB.sinC
Comments
Dislike Bookmark

Experienced mathematics faculty

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) - cos(A+B) ) = 2sinC . 2sin A sin B =4 sinA sin B sin C
Comments
Dislike Bookmark

Excellence Award Winner ( 7 times )

2sinAcosA+2sinBcosB+2sinCcosC....(1) A= pi-(B+C), B= pi-(A+C), C= pi-(B+A). So, cosA= cos = -cos = - ......(2) cosB= cos = -cos = - cosC= cos =.-cos = - . Putting these values in (1) and combining common terms -2 + 6sinAsinBsinC. putting value for cosBcosC from (2) in first...
read more
2sinAcosA+2sinBcosB+2sinCcosC....(1) A= pi-(B+C), B= pi-(A+C), C= pi-(B+A). So, cosA= cos[pi-(B+C)] = -cos[B+C] = - [cosBcosC-sinBsinC] ......(2) cosB= cos[pi-(A+C)] = -cos[A+C] = - [cosAcosC-sinAsinC] cosC= cos[pi-(B+A)] =.-cos[B+A] = - [cosBcosA-sinBsinA]. Putting these values in (1) and combining common terms -2[sinAcosBcosC+sinBcosCcosA+sinCcosBcosA] + 6sinAsinBsinC. putting value for cosBcosC from (2) in first term and taking cosA common from second and 3rd term = -2[sinA [sinBsinc-cosA] + cosA [sin (B+C)] ] + 6sinAsinBsinC. As A= pi-(B+C) sinA=sin(B+C) we get after cancellation = -2 [sinAsinBsinC] + 6sinAsinBsinC. = 4sinAsinBsinC.= RHS - read less
Comments
Dislike Bookmark

Mathmatics Tutor

Left Hand Side 2sinAcosA+2sinBcosB+2sinCcosC....(1) A= pi-(B+C), B= pi-(A+C), C= pi-(B+A). So, cosA= cos = -cos = - ......(2) cosB= cos = -cos = - cosC= cos =.-cos = - . Putting these values in (1) and combining common terms -2 + 6sinAsinBsinC. putting value for cosBcosC...
read more
Left Hand Side 2sinAcosA+2sinBcosB+2sinCcosC....(1) A= pi-(B+C), B= pi-(A+C), C= pi-(B+A). So, cosA= cos[pi-(B+C)] = -cos[B+C] = - [cosBcosC-sinBsinC] ......(2) cosB= cos[pi-(A+C)] = -cos[A+C] = - [cosAcosC-sinAsinC] cosC= cos[pi-(B+A)] =.-cos[B+A] = - [cosBcosA-sinBsinA]. Putting these values in (1) and combining common terms -2[sinAcosBcosC+sinBcosCcosA+sinCcosBcosA] + 6sinAsinBsinC. putting value for cosBcosC from (2) in first term and taking cosA common from second and 3rd term = -2[sinA [sinBsinc-cosA] + cosA [sin (B+C)] ] + 6sinAsinBsinC. As A= pi-(B+C) sinA=sin(B+C) we get after cancellation = -2 [sinAsinBsinC] + 6sinAsinBsinC. = 4sinAsinBsinC. = Right Hand Side read less
Comments
Dislike Bookmark

View 12 more Answers

Related Questions

Wat is efficiency of carnot engine?
When the second law of thermodynamics states that not all the supplied heat in a heat engine can be used to do work, the Carnot efficiency sets the limiting value on the fraction of the heat which can be so used.
Ravi Ranjan
0 0
View Comments7
how change profile from learner to tutor?
You will have to make a profile selecting yourself as a tutor and you should be ready to go.
Allauddin
What is the difference between lava and magma?
The magma is the substance deep inside the earth's crust, which is in the molten state. The same magma when it comes to the surface of the earth, it is called lava.
Yogi
1 0
View Comments8
Who is the finance minister of our country?
Arun Jaitley is the current finance minister of India
Soumya
The mean of 10, 15, x, 5, 15 is 15. What is the value of x?
(10+15+x+5+15)/5=15 So 45 + x = 5*15 X=75-45=30
Rayyan
0 0
View Comments9

Now ask question in any of the 1000+ Categories, and get Answers from Tutors and Trainers on UrbanPro.com

Ask a Question

Related Lessons

1 sec tricks : 95*97 and 107*104
Steps: 1) Find the difference of each number from 100: 95 x 975 32) Cross Subtract the difference (95-3)=(97-5)=92 => This is the first part of the answer 3) Multiply the differences5 x 3 = 15 =>...

Debraj Paul

0 0
View Comments0
A Interesting Magic in Maths
It's a magic trick. So, here we go: Think any three digit number and repeat itself twice like this: 123123. Now, divide this number by 13, whatever the quotient has come divide it by 11. Again divide...

Vikash

0 0
View Comments0
How To Overcome Exam Pressure?
Exam pressure is something that everyone faces atleast once in their lifetime. It is a psychological fact that generates a fear in us at the time of any exam or any other opportunity. Most of us finds...
A

Ankit Khurana

0 0
View Comments0
Magnesium ribbon
K

Kouser B.

0 0
View Comments0
Reciprocal Fractions
To find the reciprocal of a fraction, flip it over, so that the numerator becomes the denominator and the denominator becomes the numerator. That is: The reciprocal of 4/5 is 5/4 Note that...
A

Amrtha S.

0 0
View Comments0

Recommended Articles

Meet Swati, a Hindi Tutor from Bangalore

Swati is a renowned Hindi tutor with 7 years of experience in teaching. She conducts classes for various students ranging from class 6- class 12 and also BA students. Having pursued her education at Madras University where she did her Masters in Hindi, Swati knows her way around students. She believes that each student...

Read full article >

Meet Mohammad Wazid, a skilled trainer for...

Mohammad Wazid is a certified professional tutor for class 11 students. He has 6 years of teaching experience which he couples with an energetic attitude and a vision of making any subject easy for the students. Over the years he has developed skills with a capability of understanding the requirements of the students. This...

Read full article >

Quest Academy - Institute of the month

Quest Academy is a professional Bangalore based NEET and JEE (Main + Advanced) training institute. The academy was incorporated in 2015 to cater to the needs of students, who aim to crack competitive exams by connecting with the best brains around. The institute helps students enhance their skills and capabilities through...

Read full article >

Meet Raghunandan.G.H, a B. Tech Tutor from...

Raghunandan is a passionate teacher with a decade of teaching experience. Being a skilled trainer with extensive knowledge, he provides high-quality BTech, Class 10 and Class 12 tuition classes. His methods of teaching with real-time examples makes difficult topics simple to understand. He explains every concept in-detail...

Read full article >

Find Class 10 Tuition near you
Class 10 Tuition in Delhi
Class 10 Tuition in Bangalore
Class 10 Tuition in Hyderabad
Class 10 Tuition in Kolkata
Class 10 Tuition in Mumbai
Class 10 Tuition in Chennai
Class 10 Tuition in Pune
Class 10 Tuition in Lucknow
Class 10 Tuition in Noida
Class 10 Tuition in Jaipur
Online Tuition Class 10

Looking for Class 10 Tuition ?

Learn from the Best Tutors on UrbanPro

Are you a Tutor or Training Institute?

Join UrbanPro Today to find students near you