Question

cos^3a-sin^3a/sina-cos2a=2+sin2a/2
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Answer 1

Step-by-step explanation:

1. 2(4cos^3 pi/9 - 3 cos pi/9)

=2(cos(3*pi/9))

=2*cos(pi/3)

=2*1/2

=1

2.

To prove it we need a formula for

sin(3x) = sin2x cos x - sin x cos 2x

= 2 sin x cos^2 x - sin x ( 1 - 2 sin ^2 x)

2 sin x (1-sin^2 x) - sin x ( 1 - 2 sin ^2 x)

Sin3x= 3sin x - 4 sin ^3 x

and

cos 3x = cos 2x cos x - sin 2x sin x =

(2 cos^2 x - 1) cos x - 2 sin ^2 x cos x

(2 cos^2 x - 1) cos x - 2 (1 - cos^2 x) cos x=

Cos3x = 4 cos ^3 x - 3 cos x

(3sin x - 4 sin ^3 x)/sin x - (4 cos ^3 x - 3 cos x)/cos x

= 3 - 4 sin ^2 x + 4 cos ^2 x +3

= 6 -4(sin^2 x + cos^2 x)

= 6 - 4*1

= 2

3. 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)

5.

It is very simple ..

Just expand the function ..

Tan A + (tan 60 + tan A )/(1-tan60.tanA) + (tan (120) + tan A )/(1-tan120.tanA)

tan A + [root(3) + tanA]/[1-root(3).tanA] + [ tan A - root(3) ] /[1 + root(3).tanA]

tanA + [ root(3) + 3.tanA + tanA + root(3).tan^2A + tanA - root(3) -root(3).tan^2A + 3.tanA ] / [ 1 - 3tan^2A ]

tanA + [ 8tanA] / [ 1-3tan^2A ]

[9tanA - 3.tan^3A ]/[1-3tan^2A]

= 3.tan3A ( eqaul to RHS )

Hence proved