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