Expert: Sherry Wallin - 9/11/2009

Question
QUESTION: prove that (sinA.sin2A+sin3A.sinA6A)/(sinA.cos2A+sin3A.cos6A)=5tanA

(sinA.sin2A+sin3A.sinA6A)/(sinA.cos2A+sin3A.cos6A)
=(2sinA.sin2A+2sin3A.sinA6A)/(sinA.cos2A+2sin3A.cos6A)
=(cos(2A-A)-cos(2A+A)+cos(6A+3A)-cos(6A+3A))/(sin(2A+A)-sin(2A-A)+sin(6A+3A)-sin(6A-3A))
=(cosA-cos3A+cos3A-cos9A)/(sin3A-sinA+sin9A-sin3A)
=(cosA-cos9A)/(sin9A-sinA)
=(2sin5A.sin4A)/(2cos5A.sin4A)=tan5A

please explain how this statement (sinA.sin2A+sin3A.sinA6A)/(sinA.cos2A+sin3A.cos6A) became (2sinA.sin2A+2sin3A.sinA6A)/(sinA.cos2A+2sin3A.cos6A).please reply

ANSWER: Hi Pratap you have a typo: sinA6A, is this suppose to be sin6A? And what are you using the periods for? Is it multiplication? If so you can write sinAsin2A 0r sinA*sin2A and that is more universal.  Math Prof

---------- FOLLOW-UP ----------

QUESTION: (sinA*sin2A+sin3A*sinA6A)/(sinA*cos2A+sin3A*cos6A)
=(2sinA*sin2A+2sin3A*sin6A)/(sinA*cos2A+2sin3A*cos6A)
=(cos(2A-A)-cos(2A+A)+cos(6A+3A)-cos(6A+3A))/(sin(2A+A)-sin(2A-A)+sin(6A+3A)-sin(6A-3A))
=(cosA-cos3A+cos3A-cos9A)/(sin3A-sinA+sin9A-sin3A)
=(cosA-cos9A)/(sin9A-sinA)
=(2sin5A sin4A)/(2cos5A sin4A)=tan5A

its the multiplication operator.please explain how this statement (sinA.sin2A+sin3A.sin6A)/(sinA.cos2A+sin3A.cos6A) became (2sinA.sin2A+2sin3A.sin6A)/(sinA.cos2A+2sin3A.cos6A).please reply  

Answer
Pratap~
    You can't multiply the top of a fraction by 2 and not multiply the bottom by 2 also. The 2 expressions are not equivalent. Someone has made an error. If you like I can provide a correct proof of your problem:

(sinA*sin2A+sin3A*sin6A)/(sinA*cos2A+sin3A*cos6A)
= (1/2)[cos(A-2A)-cos(A+2A)+cos(3A-6A)-cos(3A+6A)]/(1/2)[sin(A+2A)+sin(A-2A)+sin(3A+6A)+sin(3A-6A)]
= [cos(A-2A)-cos(A+2A)+cos(3A-6A)-cos(3A+6A)]/[sin(A+2A)+sin(A-2A)+sin(3A+6A)+sin(3A-6A)] (1/2) cancel
= [cos(-A)-cos(3A)+cos(-3A)-cos(9A)]/[sin(3A)+sin(-A)+sin(9A)+sin(-3A)]
= [cos(A)-cos(3A)+cos(3A)-cos(9A)]/[sin(3A)-sin(A)+sin(9A)-sin(3A)]

Note the sign changes. The cos(-x) = cos(x) because the cosine function is an even function and sin(-x) = -sin(x) because the sine function is an odd function.

= [cos(A)-cos(9A)]/sin(9A)-sin(A)]  just simplification
= -2sin((A+9A)/2)sin((A-9A)/2)/2sin((9A-A)/2)cos((9A+A)2/)]
= -sin(5A)sin(-4A)/sin(4A)cos(5A)  just simplification
= --sin(5A)sin(4A)/sin(4A)cos(5A) brought the minus sign out of sin(-4A) as -sin(4A)
= sin(5A)sin(4A)/sin(4A)cos(5A)  two minus signs next to each other become a positive
= sin(5A)/cos(5A)  sin(4A) cancel
= tan(5A) because the tangent of angle equal the sine of the angle over the cosine of the angle

Note I used Product-to-sum formulas in the first part and Sum-to-product formulas in the 2nd half

Math Prof