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

Answer is infinity....or we can add using AM upto n values and then n tends to infinity....gives value to infinity....
go through below for more infinity series details, then you will understand the concept Infinite series An infinite series is an expression like this: S = 1 + 1/2 + 1/4 + 1/8 + ... The dots mean...
Rishi
What are the major components of population growth.
The major components of population growth are Birth Rate, Death Rate, and Migration. Birth rate is the number of live births per thousand persons in a year. It is a major component of growth. Death...
Khushbu
0 0
View Comments9
what is Biomagnification?
Biomagnification, also known as bioamplification or biological magnification, is the increasing concentration of a substance, such as a toxic chemical, in the tissues of organisms at successively higher levels in a food chain.
Dr. Anshu
0 0
View Comments8
What is Oxidising agent?
A substance that tends to bring about oxidation by being reduced and gaining electrons.
Mainak
0 0
View Comments7
What is the ongoing rate for maths and physics private tuitions in Bangalore on an hourly basis?
It's in between INR 600 to 1000.
Pramod R

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

Ask a Question

Related Lessons

* Solving Science numericals or math question
While solving Science numericals or math question always follow below steps. * Write all the parameteres given in question with unit * Think what to be find or calculated * Think how to proceed * Write...

Arvind Srivastava

0 0
View Comments0
Fastest way How to check if a number is a Perfect Square or not
Squares of all integers are known as perfect squares. In this lesson, we will discuss a very interesting Mathematical shortcut: How to check whether a number is a perfect square or not. There are some...

Debraj Paul

How to calculate 1203579 * 9999999 in 2 Sec?
203579 × 9999999 = 1203579-1 | 10000000- 1203579 =120357887964 21 Number in red is 1 less than 1203579. Number in blue is (10000000-1203579). Hence the answer. This method has to be altered...

Debraj Paul

0 0
View Comments0
Common Factors
Factors that two numbers have in common are called the common factors of those numbers. Example: What are the common factors of 20 and 25? The factors of 20 are 1, 2, 4, 5, and 20. The factors...
A

Amrtha S.

0 0
View Comments0
Nothing is impossible
If u think any subject is tough to u it only u think actually it's not tough u become it tough if any other haven't learn that subject before and u tell him it is tough than it would be tough for him also...

Gagan Shakya

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 Noida
Class 10 Tuition in Lucknow
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