622) sin 7pi/6 = - sin pi/6
sin 5pi/6 = sin pi/6
so the expression = 3 sin4(pi/6) + sin4(pi/2)
= 3/16 (have u given the question correctly?)
1708\hspace{-16}(7)::$Let $y=10\cos^2x-6\sin x.\cos x+2\sin^2 x$\\\\ $=2+8\cos^2 x-6\sin x.\cos x$\\\\ $=2+4.\left(2\cos^2 x\right)-6.\sin x.\cos x$\\\\ $=2+4.\left(1+\cos 2x\right)-3.\sin 2x$\\\\ $=2+4+4\cos 2x-3\sin 2x$\\\\ $=6+\left(4\cos 2x-3\sin 2x\right)$\\\\ Using The fact $-\sqrt{a^2+b^2}\leq \left(a.\sin x+b.\cos x\right)\leq \sqrt{a^2+b^2}$\\\\ So $6-5\leq 6+\left(4\cos 2x-3\sin 2x\right)\leq 6+5$\\\\ So $1\leq 6+\left(4\cos 2x-3\sin 2x\right)\leq 11$\\\\ $1\leq y\leq 11\Leftrightarrow y\in\mathbf{\left[1,11\right]}$
3Q1)
tan9 + tan81 - tan63 - tan27
tan9 + tan81 = 1/(sin9*cos9)
tan63 + tan27 = 1/(sin27*cos27)
so we have to solve -
1/(sin9*cos9) - 1/(sin27*cos27)
or, 2/sin18 - 2/cos36
now put the values of sin18 and cos36. you will get 4.
3
h4hemang
·2011-10-18 00:49:07
Q3)
4cos20 - √3 cot20
or, 4cos20 - (√3cos20)/sin20
or, (4sin20*cos20 - √3cos20)/sin20
or, cos20(4sin20-√3)/sin20
tan40 = tan(60-20)/(1+tan60*tan20)
so, tan60 - tan20 = 4sin20
so, tan60 = 4sin20 + tan20
using this we get,
cos20(4sin20 - 4sin20 - tan20)/(sin20) since √3 = tan60
or, -1. answer.
3
h4hemang
·2011-10-18 00:58:15
Q5)
sin212 + sin221 + sin239 + sin248
or, 2-cos9(cos33 + cos87)
or, 2-cos9*cos27
now RHS =
1 + sin29 + 1 - cos218 = 2 -(cos218 - sin29)
= 2 - cos(27)cos(9).
hence proved....
