Math, asked by Jevana, 1 year ago

PT Cos3a-sin3a/cosa+sina=1+2cos2a


kinkyMkye: look at my solution consider a=30 then the question is invalid hope i could help

Answers

Answered by kinkyMkye
3

cos 3A= 4 cos³ A - 3 cos A

Sin 3A = 3 Sin A - 4 sin³A

cos 3A-Sin 3A= 4 cos³ A - 3 cos A -3 Sin A + 4 sin³ A
= 4(cos³A+sin³A)-3(cos A+sinA)

consider the identity a³+b³=(a+b)³-3ab(a+b)=(a+b)(a²+b²-ab)
cos 3A-Sin 3A=4(cos³A+sin³A)-3(cos A+sinA)
= 4((cosA+sinA)(cos²A+sin²A-cosAsinA))-3(cos A+sinA)
= (cos A + sin A)(4(1-cosASinA)-3) =1-2sin2A

so Cos3a-sin3a/cosa+sina=1-2sin2A

consider A=30° and the above is true
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