Sin (π / 3 + α) = 1 / 3, then cos (π / 6 - α)=
According to the induction formula: sin (π / 2 - α) = cos α
The results are as follows
cos(π/6-α)
=sin[π/2-(π/6-α)]
=sin(π/3+α)
=1/3
If x belongs to [0, π / 2], we prove that 1 ≤ √ SiN x + √ cos x ≤ 2 ^ 3 / 4
Substitution of variables
Let t = root (SiNx) (0
How to prove (2-sin (x) ^ 2) - (2-cos (x) ^ 2) = cos2x?
(2-sin(x)^2)-(2-cos(x)^2)
=2-sin(x)^2-2+cos(x)^2
=cos(x)^2-sin(x)^2
=cos2x