Question

prove the identities pls,,,,,,,

Answer 1

Answer :

{ Making the corrections, solutions are added }

1)

Now, sin(60° + A) sinA sin(60° - A)

= (sin60° cosA + cos60° sinA) (sin60° cosA - cos60° sinA) sinA

= (sin²60° cos²A - cos²60° sin²A) sinA

= ($\frac{3}{4}$ cos²A - $\frac{1}{4}$ sin²A) sinA

= $\frac{1}{4}$ (3 cos²A - sin²A) sinA

= $\frac{1}{4}$ (3 - 3 sin²A - sin²A) sinA

= $\frac{1}{4}$ (3 sinA - 4 sin³A)

= $\frac{1}{4}$ sin3A

2)

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

= ($\frac{1}{4}$ cos²A - $\frac{3}{4}$ sin²A) cosA

= $\frac{1}{4}$ (cos²A - 3 sin²A) cosA

= $\frac{1}{4}$ (cos²A - 3 + 3 cos²A) cosA

= $\frac{1}{4}$ (4 cos³A - 3 cosA)

= $\frac{1}{4}$ cos3A

Hence, proved.

3)

Now, tan(60° + A) tanA tan(60° - A)

$=( \frac{tan60 + tanA}{1 - tan60 \: tanA} )(\frac{tan60 - tanA}{1 + tan60 \: tanA})tanA \\ \\ = (\frac{ {tan}^{2} 60 - {tan}^{2} A}{1 - {tan}^{2}60 \: {tan}^{2} A } )tanA \\ \\ = ( \frac{3 - {tan}^{2} A}{1 - 3 {tan}^{2}A } ) \: tanA\\ \\ = \frac{3tanA - {tan}^{3} A}{1 - 3 {tan}^{2}A } \\ \\ = tan3A$

Hence, proved.

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Answer 2

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)