Question

prove that cos(x-π/3)cos(x+π/3)=cos3x/4cosx

Answer 1

★ Question:

prove that cos(x-π/3)cos(x+π/3)=cos3x/4cosx

★ Answer :

we know that ;

cos ( A + B) = CosA cos B - sin A sin B

and

cos ( A- B ) = CosA cos B + sin A sin B

then ,

L HS = cos(x-π/3)cos(x+π/3)

= [ cos x cos π/3 + sin X sin π / 3 ] ×

[ cos x cos π/3 - sin X sin π / 3 ]

$= (\cos(x) \times \frac{1}{2} + \sin(x) \times \frac{ \sqrt{3} }{2} ) \times ( \cos(x) \times \times \frac{1}{2} - \sin(x) \times \frac{ \sqrt{3} }{2} )$

use (A+ B ) ( A-B ) = A ² - B ²

then ,

= [cos ² x/4 - 3sin² x / 4 ]

= ( cos ² x - 3sin² x )/4

= (4cos ² x-3 )/4

now multiply and divide by cos x

$= \frac{ \cos(x) }{ \cos(x) } \frac{(4 \cos {}^{2} (x) - 3)}{4}$

$= \frac{(4 \cos {}^{3} (x) - 3cosx) }{csx}$

= cos3x/4 cosx

= RHS

hence proved !

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