2\sqrt{2}sin10[\frac{sec5}{2}+\frac{cos40}{sin5}-2sin35]
calculate without using trigo tables
212\sqrt{2}\left[\frac{2sin5cos5 }{2cos5}+\frac{2sin5cos5cos40}{sin5}-2sin35sin10 \right]
=2\sqrt{2}\left[sin5+2cos40cos5-2sin35sin10 \right]
=2\sqrt{2}\left[sin5+(cos45+cos35)-(cos25-cos45) \right]
=2\sqrt{2}\left[\frac{2}{\sqrt{2} }+sin5+cos35-cos25 ]
=2\sqrt{2}\left[\frac{2}{\sqrt{2} }+sin5+(cos35-cos25) ]
=2\sqrt{2}\left[\frac{2}{\sqrt{2} }+sin5+ 2sin(\frac{35+25}{2})sin(\frac{25-35}{2}) ]
=2\sqrt{2}\left[\frac{2}{\sqrt{2} }+sin5+ 2sin(30)sin(-5) ]
=2\sqrt{2}\left[\frac{2}{\sqrt{2} }+sin5-sin5 ]
=4
11/2cos5 + cos40/sin5 - 2sin35
= 1/2cos5 + cos40 - (cos30-cos40)sin5
= 1/2cos5 + 2cos40-cos30sin5
= sin5 + 4cos40cos5 - 2cos5cos30sin10
so the total expression cools down to ("cool" too hot word right ??)
2√2 [sin5 + 2(cos45 + cos35) - (cos35 + cos25)]
=2√2[sin5 + cos35 + 2cos45 - cos25]
=2√2[sin5 - 2sin30sin5 + 2cos45]
=2√2[sin5 - sin5 + √2]
=2√2 x √2
=4
6moy we were nt fools....we were idiots nt to do it....same onus..:D
1ooooooooooh eragon....why did ya post !!!!!!!!! u downgraded my prestige
!!!!!!!!!!!!
