prove that sin6A+sin2A+ sin8A+sin4A /cos6A+cos2A+ cos8A+ cos4A=tan5A
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Answer:
sin5a/cos5a=tan5a
Step-by-step explanation:
2sin(2A+4A/2)cos(2A-4A/2) + 2sin(6A+8A/ 2)cos(6A-8A/2)/2cos(2A+4A/2)cos(2A-4A/ 2)+2cos(6A+8A/2)cos(6A-8A/2)
------using identity sinA + sinB=2sin(A+B/ 2)cos(A-B/2) cosA+cosB= 2cosA+B/2 cos A-B/2
2 sin3Acos(-A)+2sin7Acos(-A)/2cis3Acis(-A)
+2cos7Acos(-A)
2cos(-A) (sin3A+sin7A)/2cos(-A) (cos3A+cos7A)
sin3a+sin7a/cos3a+cos7a
2sin3a+7a/2 cos3a-7a/2/2cos3a+7a/
2cos3a-7a/2
sin5a cos -4a/cos5a cos-4a
sin5a/cos5a=tan5a
#BrainLock .
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