Trace the changes in the sign and magnitude of ((sin 3θ)/(cos 2θ)) as the angle increases from 0 to (π/2). also find its minimum and maximum values.


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Question Number 58025 by Kunal12588 last updated on 16/Apr/19

$T r a c e t h e c h a n g e s i n t h e s i g n a n d m a g n i t u d e$ $o f \frac{\sin 3 θ}{\cos 2 θ} a s t h e a n g l e i n c r e a s e s f r o m 0 t o \frac{π}{2} .$ $a l s o f i n d i t s m i n i m u m a n d m a x i m u m v a l u e s .$
Commented by Kunal12588 last updated on 16/Apr/19

	Answered by tanmay last updated on 16/Apr/19
	$(π/2)≥θ≥0 π≥2θ≥0 ((3π)/2)≥3θ≥0 A) sin3θ=+ve (π/6)≥θ≥0 cos2θ=+ve so ((sin3θ)/(cos2θ))=+ve when (π/6)≥θ≥0 B) sin3θ=+ve [when (π/4)> θ≥(π/6) cos2θ=+ve ((sin3θ)/(cos2θ))=++ve c)sin3θ= +ve when (π/3)≥θ>(π/4) cos2θ=−ve ((sin3θ)/(cos2θ))=−ve d)sin3θ= −ve (π/2)≥θ≥(π/3) cos2θ=−ve so ((sin3θ)/(cos2θ))=+ve pls check... f(θ)=((sin3θ)/(cos2θ)) (df/dθ)=((cos2θ×3cos3θ+2sin3θ×sin2θ)/(cos^2 2θ)) (df/dθ)=((cos2θ×cos3θ+2cos(3θ−2θ))/(cos^2 2θ)) (df/dθ)=((cos3θ×cos2θ+2cosθ)/(cos^2 2θ)) for max/min (df/dθ)=0 cos3θ×cos2θ+2cosθ (4x^3 −3x)(2x^2 −1)+2x 8x^5 −4x^3 −6x^3 +3x+2x 8x^5 −10x^3 +5x x(8x^4 −10x^2 +5) x{2(4x^4 −5x^2 )+5} x[2{(2x^2 )^2 −2×2x^2 ×(5/4)+((25)/(16))−((25)/(16))}+5] x[2(2x^2 −(5/4))^2 −((25)/8)+5] x[2(2x^2 −(5/4))^2 +((15)/8)] so [2(2x^2 −(5/4))^2 +((15)/8)]≠0 hence 8x^5 −10x^3 +5x=0 when x=0 cosθ=0=cos(π/2) [θ=(π/2)] f((π/2))=((sin(((3π)/2)))/(cos(((2π)/2))))=((−1)/(−1))=1(maximum value) f(0)=((sin(3×0))/(cos(2×0)))=0$
	$\frac{π}{2} ⩾ θ ⩾ 0$ $π ⩾ 2 θ ⩾ 0$ $\frac{3 π}{2} ⩾ 3 θ ⩾ 0$ $A) s i n 3 θ = + v e \frac{π}{6} ⩾ θ ⩾ 0$ $c o s 2 θ = + v e$ $s o \frac{s i n 3 θ}{c o s 2 θ} = + v e w h e n \frac{π}{6} ⩾ θ ⩾ 0$ $B) s i n 3 θ = + v e [w h e n \frac{π}{4} > θ ⩾ \frac{π}{6}$ $c o s 2 θ = + v e$ $\frac{s i n 3 θ}{c o s 2 θ} = + + v e$ $c) s i n 3 θ = + v e w h e n \frac{π}{3} ⩾ θ > \frac{π}{4}$ $c o s 2 θ = - v e$ $\frac{s i n 3 θ}{c o s 2 θ} = - v e$ $d) s i n 3 θ = - v e \frac{π}{2} ⩾ θ ⩾ \frac{π}{3}$ $c o s 2 θ = - v e$ $s o \frac{s i n 3 θ}{c o s 2 θ} = + v e$ $p l s c h e c k . . .$ $f (θ) = \frac{s i n 3 θ}{c o s 2 θ}$ $\frac{d f}{d θ} = \frac{c o s 2 θ \times 3 c o s 3 θ + 2 s i n 3 θ \times s i n 2 θ}{{c o s}^{2} 2 θ}$ $\frac{d f}{d θ} = \frac{c o s 2 θ \times c o s 3 θ + 2 c o s (3 θ - 2 θ)}{{c o s}^{2} 2 θ}$ $\frac{d f}{d θ} = \frac{c o s 3 θ \times c o s 2 θ + 2 c o s θ}{{c o s}^{2} 2 θ}$ $f o r m a x / m i n \frac{d f}{d θ} = 0$ $c o s 3 θ \times c o s 2 θ + 2 c o s θ$ $(4 x^{3} - 3 x) (2 x^{2} - 1) + 2 x$ $8 x^{5} - 4 x^{3} - 6 x^{3} + 3 x + 2 x$ $8 x^{5} - 10 x^{3} + 5 x$ $x (8 x^{4} - 10 x^{2} + 5)$ $x {2 (4 x^{4} - 5 x^{2}) + 5}$ $x [2 {{(2 x^{2})}^{2} - 2 \times 2 x^{2} \times \frac{5}{4} + \frac{25}{16} - \frac{25}{16}} + 5]$ $x [2 {(2 x^{2} - \frac{5}{4})}^{2} - \frac{25}{8} + 5]$ $x [2 {(2 x^{2} - \frac{5}{4})}^{2} + \frac{15}{8}]$ $s o [2 {(2 x^{2} - \frac{5}{4})}^{2} + \frac{15}{8}] \neq 0$ $h e n c e 8 x^{5} - 10 x^{3} + 5 x = 0 w h e n x = 0$ $c o s θ = 0 = c o s \frac{π}{2} [θ = \frac{π}{2}]$ $f (\frac{π}{2}) = \frac{s i n (\frac{3 π}{2})}{c o s (\frac{2 π}{2})} = \frac{- 1}{- 1} = 1 (m a x i m u m v a l u e)$ $f (0) = \frac{s i n (3 \times 0)}{c o s (2 \times 0)} = 0$
	Commented by Kunal12588 last updated on 17/Apr/19

	$t h a n k y o u s i r$
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