If sina+sinb=x and cosa+cosb=y, find the values of tan(a+b)/2 and tan(a-b)/2
Answers
Therefore the value of tan(a + b)/2 = x/y and tan(a - b)/2 = √(4 - x² - y²)/√(x² + y²)
Given : two trigonometric equations are : sin a + sin b = x and cos a + cos b = y
To find : the value of tan(a + b)/2 and tan(a - b)/2
solution : sin a + sin b = x
using formula, sinC + sinD = 2sin(C + D)/2 cos(C - D)/2
⇒2sin(a + b)/2 cos(a - b)/2 = x ........(1)
again, cos a + cos b = y
using formula, cosC + cosD = 2cos(C + D)/2.cos(C - D)/2
⇒2cos(a + b)/2 cos(a - b)/2 = y ........(2)
from equations (1) and (2) we get,
[2sin(a + b)/2.cos(a - b)/2 ]/[2cos(a + b)/2.cos(a - b)/2] = x/y
⇒tan(a + b)/2 = x/y
so sin(a + b)/2 = x/√(x² + y²)
now from equation (1) we get,
2 × x/√(x² + y²) × cos(a - b)/2 = x
⇒cos(a - b)/2 = √(x² + y²)/2
so tan(a - b)/2 = √{2² - x² - y²}/√(x² + y²)
⇒tan(a - b)/2 = √(4 - x² - y²)/√(x² + y²)
Step-by-step explanation:
prove that
Sin(A+B)= 2ab/ a^2+b^2
Sin(A+B) = sin A cos B +CosA SinB
Cos(A+B) = cos A cosB + SinA SinB
Finding a^2 +b^2 (1)
a^2 +b^2= sin^2A +sin^2B +2sinA SinB + Cos^2A +Cos^2B + 2Cos A Cos B
= 1+1 +2 ( CosA CosB + SinA SinB )
=2+2( CosA cos B +sinA SinB)
a^2+b^2 --2 =2 Cos(A-B) )
Cos (A-B) = (a^2+b^2 )/2 --1 (1)
Finding values of ab (2)
2ab =2 (sinA +SinB ) (CosA +CosB)
= 2 [ sinA CosA + SinA cos B + SinB Cos A + SinB Cos B]
2ab= 2 (1+ Cos A CosB + SinA SinB )
ab= [1+ cos(A --B)]
ab=1 +(a^2+b^2)/2 --1 from (1)
ab=( a^2+b^2)/2
Hence sin(A+B) = 2ab /a^2+b^2
LH S =R H S