prove the identities please,,,, remaining two identities ,,,,,.
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Answer :
1)
Now, cos(60° - A) cosA cos(60° + A)
= {cos(60° - A) cos(60° + A)} cosA
= (cos60° cosA + sin60° sinA) (cos60° cosA - sin60° sinA) cosA
= (cos²60° cos²A - sin²60° sin²A) cosA
= ( cos²A - sin²A) cosA
= (cos²A - 3 sin²A) cosA
= (cos²A - 3 + 3 cos²A) cosA
= (4 cos³A - 3 cosA)
= cos3A
Hence, proved.
2)
Now, tan(60° + A) tanA tan(60° - A)
Hence, proved.
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1.
sin(a)sin(60 - a)sin(60 + a) =>
sin(a) * (sin(60)cos(a) - sin(a)cos(60)) * (sin(60)cos(a) + sin(a)cos(60)) =>
sin(a) * (sin(60)^2 * cos(a)^2 - sin(a)^2 * cos(60)^2) =>
sin(a) * ((3/4) * cos(a)^2 - (1/4) * sin(a)^2) =>
(1/4) * sin(a) * (3cos(a)^2 - sin(a)^2) =>
(1/4) * sin(a) * (3 - 3sin(a)^2 - sin(a)^2) =>
(1/4) * (3sin(a) - 4sin(a)^3) =>
(1/4) * sin(3a)
2.
formula for cox( x+y) cos(x-y) = cos ^2 (y) - sin^2( x)
here cos (60-A) cos (60+A) = cos^2(A)- sin^2 (60)
= cos ^2 (A) - 3/4
now cos A cos (60-A) cos (60+A = cos A ( cos^2 (A) - 3/4)
\ = 1/4 ( 4 cos ^3 (a) - 3 cos A)
{since 4 cos ^3 (a) - 3 cos A = cos 3A }
= 1/4 ( cos 3A)
hence proved
sin(a)sin(60 - a)sin(60 + a) =>
sin(a) * (sin(60)cos(a) - sin(a)cos(60)) * (sin(60)cos(a) + sin(a)cos(60)) =>
sin(a) * (sin(60)^2 * cos(a)^2 - sin(a)^2 * cos(60)^2) =>
sin(a) * ((3/4) * cos(a)^2 - (1/4) * sin(a)^2) =>
(1/4) * sin(a) * (3cos(a)^2 - sin(a)^2) =>
(1/4) * sin(a) * (3 - 3sin(a)^2 - sin(a)^2) =>
(1/4) * (3sin(a) - 4sin(a)^3) =>
(1/4) * sin(3a)
2.
formula for cox( x+y) cos(x-y) = cos ^2 (y) - sin^2( x)
here cos (60-A) cos (60+A) = cos^2(A)- sin^2 (60)
= cos ^2 (A) - 3/4
now cos A cos (60-A) cos (60+A = cos A ( cos^2 (A) - 3/4)
\ = 1/4 ( 4 cos ^3 (a) - 3 cos A)
{since 4 cos ^3 (a) - 3 cos A = cos 3A }
= 1/4 ( cos 3A)
hence proved
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