Prove: (sin3a/sina)²-(cos3a/cosa)²=8cos2a
wherever it is defined.
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Answered by
0
Heya dear
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LHS
sin3A / sinA - cos3A / cosA
= ( sin3A ) ( cosA ) - ( cos3A ) ( sinA ) / sinAcosA
= ( 3sinA - 4sin³A ) cosA - ( 4cos³A - 3cosA ) ( sinA ) / sinAcosA
= 3sinAcosA - 4sin³cosA - 4cos³AsinA + 3sinAcosA / sinAcosA
= 6sinAcosA - 4sin³cosA - 4cos³sinA / sinAcosA
= sinAcosA ( 6 - 4sin²A - 4cos²A ) / sinAcosA
= 6 - 4sin²A - 4cos²A
= 6 - 4 ( sin²A + cos²A ) [ °.° sin²a + cos²b = 1
= 6 - 4 × 1
= 6 - 4
= 2
RHS
thanks
===================================
LHS
sin3A / sinA - cos3A / cosA
= ( sin3A ) ( cosA ) - ( cos3A ) ( sinA ) / sinAcosA
= ( 3sinA - 4sin³A ) cosA - ( 4cos³A - 3cosA ) ( sinA ) / sinAcosA
= 3sinAcosA - 4sin³cosA - 4cos³AsinA + 3sinAcosA / sinAcosA
= 6sinAcosA - 4sin³cosA - 4cos³sinA / sinAcosA
= sinAcosA ( 6 - 4sin²A - 4cos²A ) / sinAcosA
= 6 - 4sin²A - 4cos²A
= 6 - 4 ( sin²A + cos²A ) [ °.° sin²a + cos²b = 1
= 6 - 4 × 1
= 6 - 4
= 2
RHS
thanks
guptaushma2014:
The question is totally different
ohj
*Ok
Answered by
0
(cos2a+2cos^2a)^2-(co2a-sin^2a)^2=rhs
4cos^2a-4sin^2a+4cos2a(sin^a+cos^a)= rhs
4(cos2a)+4cos2a=rhs
8cos2a= rhs
lhs = rhs hence prove
4cos^2a-4sin^2a+4cos2a(sin^a+cos^a)= rhs
4(cos2a)+4cos2a=rhs
8cos2a= rhs
lhs = rhs hence prove
What formula you have used?
co^x-sin^x=cos2x
But isn't it cos3x in the question
we cab break it as cos (2x+x)
ok
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