UrbanPro
Login Ask & Answer Success Stories Paid Courses Free Classes Tuition & Classes Fees Write a Review All Categories Help Center
Post your Learning Need Signup as a Tutor
✕

Popular Cities

Bangalore

Hyderabad

Delhi

Mumbai

Chennai

Pune

Kolkata

Gurgaon

Ahmedabad

Jaipur

Noida

✕

Take Class 10 Tuition from the Best Tutors

  • Affordable fees
  • 1-1 or Group class
  • Flexible Timings
  • Verified Tutors

Book a Free Demo

Class 10 > Learn Class IX-X Tuition > IF A+B+C = Pie Then prove that Sin2A + Sin2B + Sin2C...
✕

Search in

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

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

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

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

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

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

How to create an interest in maths and science for students?
Teach them with enthu...
Ravi Ranjan
i need classes for IX and x. please help me out near vasai
You can contact me.
Badri
0 0
8
For x students do properly Reflection and refraction carbon and its compound these chapter are easy and scoring both
That's true reflection and refraction will cover most of your physics part and carbon and its compounds will cover chemistry. One more thing Heredity and reproduction chapters will cover biology.
Abdhesh
Why moon and stars are not seen during at night?
your question is wrong it should be day, during the day sunlight rays is brighter then the sky and we can"t see it, and also due to the the effect of our eye intesities
Sainath
0 0
5
Name the organ which shows visible response in the unit of nervous system.
The basic units of the nervous system are nerve cells.
Kumar Upendra Akshay
2 0
9

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

Ask a Question

Related Lessons

Keep writing a habbit
Whenever you start study take some notes of it , make points describing that topic and start using highlighter so next you can revise the main points on the go.
L

Leena S.

0 0
0
Some formulas of algebra
Algebra: 1. (a + b) = a + 2ab + b 2. (a – b) = a – 2ab + b 3. (a + b) (a – b) = a – b 4. (x + a)(x + b) = x + (a + b)x + ab 5. (x + a)(x – b) = x + (a – b)x –...
K

Kanchan

2 0
0
Vedic Mathematics for Multiplication of two 2-digited special numbers
In case of two 2-digited numbers whose digits in 10's place are same and sum of the digits in unit places is 10 For example:- 38 and 32 (10's place digits are equal and 8+2=10) 38×32 can be found...
N

Naimish K.

0 0
1
Vedic Mathematics for finding square of a number whose unit place carries 5
As we learned from my previous lesson i,e; (In case of two 2-digited numbers whose digits in 10's place are same and sum of the digits in unit places is 10 For example:- 38 and 32 (10's place digits...
N

Naimish K.

0 0
0
Get a huge Spanish Vocabulary in 5 minutes
Here are 4 simple rules that work 99% of the time. Remember them and think over examples yourself and you will have mastery in 5 minutes. Rule 1: Many English words that end with TY can be made into...

Maria C.

1 0
0

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
X

Looking for Class 10 Tuition Classes?

The best tutors for Class 10 Tuition Classes are on UrbanPro

  • Select the best Tutor
  • Book & Attend a Free Demo
  • Pay and start Learning

Take Class 10 Tuition with the Best Tutors

The best Tutors for Class 10 Tuition Classes are on UrbanPro

This website uses cookies

We use cookies to improve user experience. Choose what cookies you allow us to use. You can read more about our Cookie Policy in our Privacy Policy

Accept All
Decline All

UrbanPro.com is India's largest network of most trusted tutors and institutes. Over 55 lakh students rely on UrbanPro.com, to fulfill their learning requirements across 1,000+ categories. Using UrbanPro.com, parents, and students can compare multiple Tutors and Institutes and choose the one that best suits their requirements. More than 7.5 lakh verified Tutors and Institutes are helping millions of students every day and growing their tutoring business on UrbanPro.com. Whether you are looking for a tutor to learn mathematics, a German language trainer to brush up your German language skills or an institute to upgrade your IT skills, we have got the best selection of Tutors and Training Institutes for you. Read more

ThinkVidya Learning Pvt Ltd © 2010-2025All Rights Reserved