Evaluate cos^2(pi/12).
a)1/4
b)(2+sqrt3)/4**
c)3/2
d)(2-sqrt3)/4
Please help :(
recall cos(2A) = 2cos^2 A - 1
cos^2 A = (cos(2A) + 1)/2
now π/12 = 15°
so we can consider 2A to be 30° and A = 15°
cos^2 (π/12) = (cos π/6 + 1)/2 = (cos 30° + 1)/2
= ...
yup, you are correct
π rad = 180°
π / 12 = 180° / 12 = 15°
15° = 45° - 30°
cos ( A - B ) = sin A ∙ sin B + cos A ∙ cos B
sin 45° = cos 45° = √2 / 2 = 1 / √2
cos ( 45° - 30° ) = cos 15° = sin 45° ∙ sin 30° + cos 45° ∙ cos 30° =
( 1 / √2 ) ∙ 1 / 2 + ( 1 / √2 ) ∙ √3 / 2 =
( 1 / √2 ) ∙ ( 1 / 2 + √3 / 2 ) = ( 1 / √2 ) ∙ ( 1 / 2 ) ∙ ( 1 + √3 ) =
1 / ( √2 ∙ 2 ) ( √3 + 1 ) = ( √3 + 1 ) / ( 2 ∙ √2 ) = ( √3 + 1 ) / 2 √2
cos² 15° = [ ( √3 + 1 ) / 2 √2 ]² = ( √3 + 1 )² / ( 2 √2 )² = [ ( √3 )² + 2 ∙ √3 ∙1 + 1² ] / [ 2² ∙ ( √2 )² ] =
( 3 + 2 √3 + 1 ) / ( 4 ∙ 2 ) = ( 4 + 2 √3 ) / ( 4∙ 2 ) = 2 ∙ ( 2 + √3 ) / ( 4 ∙ 2 ) = ( 2 + √3 ) / 4
cos² ( π / 12 ) = cos² 15° = ( 2 + √3 ) / 4
To evaluate cos^2(pi/12), we need to use the double-angle identity for the cosine function, which states that cos(2theta) = 2cos^2(theta) - 1.
In this case, we have theta = pi/12. So, let's find cos(2(pi/12)).
Using the double-angle formula, we have:
cos(2(pi/12)) = 2cos^2(pi/12) - 1
But we know that cos(pi/6) = sqrt(3)/2, and we can rewrite pi/12 as (pi/6) / 2.
Therefore, cos(2(pi/12)) can be rewritten as cos(pi/6).
cos(pi/6) = sqrt(3)/2
Now, let's substitute this value back into the double-angle identity equation:
sqrt(3)/2 = 2cos^2(pi/12) - 1
Rearranging the equation:
2cos^2(pi/12) = sqrt(3)/2 + 1
Adding 1 to both sides:
2cos^2(pi/12) = (sqrt(3) + 2)/2
Finally, dividing both sides by 2:
cos^2(pi/12) = (sqrt(3) + 2) / 4.
Therefore, the correct option is b) (2+sqrt3)/4.