Trace-the-changes-in-the-sign-and-magnitude-of-sin-3-cos-2-as-the-angle-increases-from-0-to-pi-2-also-find-its-minimum-and-maximum-values-

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

\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 \dots

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