Prove that:
sin 3A cos 4A - sin A cos 2A/ sin 4A sin A + cos 6A cos A = tan 2A
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To prove, cos^6A-cos^4A+8cos^2A=4
we will take given equation, and we will try to convert it in whole cos term equation
so we have,
==> sinA+sin^3A =cos^2A , squaring both side we get
==> (sinA+sin^3A)^2 =(cos^2A)^2
==> sin^2A +2sin^2sinA*sin^3A +(sin^3A)^2=cos^4A
==> sin^2A +2sin^2sinA*sin^3A +(sin^3A)^2 = cos^4A
we have ,1-sin^2A =cos^2A;
and some math formula like,
(a-b)^2=a^2-2ab+b^2
(a-b)^3=a^3 - 3a^2 *b+3a*b^2 -b^3
(sin^3A)^2 =(sin^2A)^3=(sin^6)
by using this we get,
==>sin^2A +2(sin^2A)^2 +(sin^2A)^3 =cos^4A
==> 1-cos^2A +2(1-cos^2A)^2 +(1-cos^2A)^3 =cos^4A
==> 1-cos^2A +2(1–2cos^2A +cos^4A)+(1–3cos^2A +3cos^4A -cos^6A)=cos^4A
==>1-cos^2A + 2–4cos^2A +2cos^4A +1–3cos^2A +3cos^4A -cos^6A -cos^4A=0
==>4–8cos^2A +4cos^4A -cos^6A
so we rearranging the term we get
==> –8cos^2A +4cos^4A -cos^6A=-4
==> cos^6A -4cos^4A+8cos^2A=4
hence this is a proof.
we will take given equation, and we will try to convert it in whole cos term equation
so we have,
==> sinA+sin^3A =cos^2A , squaring both side we get
==> (sinA+sin^3A)^2 =(cos^2A)^2
==> sin^2A +2sin^2sinA*sin^3A +(sin^3A)^2=cos^4A
==> sin^2A +2sin^2sinA*sin^3A +(sin^3A)^2 = cos^4A
we have ,1-sin^2A =cos^2A;
and some math formula like,
(a-b)^2=a^2-2ab+b^2
(a-b)^3=a^3 - 3a^2 *b+3a*b^2 -b^3
(sin^3A)^2 =(sin^2A)^3=(sin^6)
by using this we get,
==>sin^2A +2(sin^2A)^2 +(sin^2A)^3 =cos^4A
==> 1-cos^2A +2(1-cos^2A)^2 +(1-cos^2A)^3 =cos^4A
==> 1-cos^2A +2(1–2cos^2A +cos^4A)+(1–3cos^2A +3cos^4A -cos^6A)=cos^4A
==>1-cos^2A + 2–4cos^2A +2cos^4A +1–3cos^2A +3cos^4A -cos^6A -cos^4A=0
==>4–8cos^2A +4cos^4A -cos^6A
so we rearranging the term we get
==> –8cos^2A +4cos^4A -cos^6A=-4
==> cos^6A -4cos^4A+8cos^2A=4
hence this is a proof.
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