Question

prove that:
cos^2π/4-sin^2π/12=√3/4
inappropriate solutions will be reported.​

Answer 1

Answer:

√3 / 4.

Step-by-step explanation:

We are given with :

   cos^2 ( π / 4 ) - sin^2 ( π / 12 )

From the properties of trigonometric ratios :

     · cos( π / 4 ) = ( 1 / √2 )

     · sin( π / 6 ) = 1 / 2

     · sin2A = 2sinAcosA

 Using sin( π / 6 ) = 1 / 2:

⇒ sin( π / 6 ) = 1 / 2

⇒ sin( π / 12 + π / 12 ) = 1 / 2

⇒ 2sin( π / 12 ).cos( π / 12 ) = 1 / 2

    Let π / 12 = A

⇒ 2sinA.cosA = 1 / 2

      Square on both sides :

⇒ 4sin²A.cos²A = 1 / 4

⇒ 16sin²A.cos²A = 1

⇒ 16sin²A( 1 - sin²A ) = 1

    Let sin²A = a

⇒ 16a( 1 - a ) = 1

⇒ 16a - 16a² = 1

⇒ 16a² - 16a + 1 = 0

⇒ a = ( 2 ± √3 ) / 4    { using qua. eq. }

⇒ sin²A = ( 2 - √3 ) / 4           { leaving + }

⇒ sin²( π / 12 ) = ( 2 - √3 ) / 4

  Therefore, now

⇒ cos^2 ( π / 4 ) - sin^2 ( π / 12 )

 substituting values from above:

⇒ ( 1 / √2 )^2 - ( 2 - √3 ) / 4

⇒ 1 / 2 - ( 2 - √3 ) / 4

⇒ ( 2 - 2 + √3  ) / 4

⇒ √3 / 4

       Hence proved.