Question

If a = cos 3 & b = sin 8 then
A) $a > 0,b > 0$
B) $ab < 0$
C) $a > b$
D) $ab > 0$

Answer 1

Hint: 3 and 8 given in cosine and sine respectively are clearly in radian not in degrees and also note that 1 radian = ${57^ \circ }$ use this to change the radians in degrees and then convert them all to there principal values so that finally we can judge our answer.
Complete Step by Step Solution:
Given is $a = \cos 3\& b = \sin 8$
We know that ${1^c} = {57^ \circ }$
Therefore, ${3^c} = 171\& {8^c} = 456$
Now let us substitute the values of radians, we will get it as

a = \cos 3 = \cos {171^ \circ }\\\ b = \sin 8 = \sin {456^ \circ } \end{array}$$ Now we can write $$\begin{array}{l} a = \cos {(90 + 81)^ \circ } = - \sin {81^ \circ }\\\ b = \sin {(360 + 96)^ \circ } = \sin {96^ \circ } \end{array}$$ Now $$\sin {96^ \circ }$$ is in second quadrant and we know that sine is positive in second quadrant therefore here $$a < 0\& b > 0$$ $$\therefore ab < 0$$ Which means option B is correct. Note: Any number to the power c basically denotes radian like in this case it was $${3^c},{8^c},{1^c}$$ which means 3 radian, 8 radian and 1 radian respectively. Also option C is incorrect because as $$a < 0\& b > 0$$ means $$b > a$$