Question

Sin105°cos15°+sin15°sin165°+tg225°;​

Answer 1

$Sin105^{o}Cos15^{o}+Sin15^{o}Sin165^{o}+tg225^{o}=Sin(90^{o}+15^{o})Cos15^{o} +Sin15^{o}Sin(180^{o} -15^{o})+tg(180^{o}+45^{o} )=Cos15^{o}Cos15^{o}+Sin15^{o}Sin15^{o}+tg45^{o}=Cos^{2}15^{o}+Sin^{2}15^{o}+tg45^{o}=1+1=2$

Answer 2

Let's first simplify each term individually:

sin105°cos15° = (sin(60° + 45°)cos15°) = (sin60°cos45° + cos60°sin45°)cos15° = ((√3/2)(√2/2) + (1/2)(√2/2))cos15° = ((√6 + √2)/4)cos15°

sin15°sin165° = sin15°sin(180° - 15°) = sin15°sin165° = (1/2)(cos150° - cos30°) = (1/2)(-√3/2 - √3/2) = -√3/4

tg225° = tan(180° + 45°) = tan45° = 1

Now we can substitute these values into the original expression and simplify:

sin105°cos15° + sin15°sin165° + tg225° = ((√6 + √2)/4)cos15° - √3/4 + 1 ≈ 0.8519

Therefore, the value of the expression is approximately 0.8519.