If $s i n^{2} A = x$ then sin 3A. sin A is polynomial in x, whose degree is equal to

5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2$
If $s i n^{2} A = x$ then........................
$∵ c o s α + c o s β = b$
$2 c o s (\frac{α + β}{2}) c o s (\frac{α - β}{2}) = b . . . . (i)$
Now $s i n α + s i n β = a$
$2 s i n (\frac{α + β}{2}) c o s (\frac{α - β}{2}) = a . . . . (i i)$
Divide (ii) by (i)
$t a n (\frac{α + β}{2}) = a / b$
$\Rightarrow t a n θ = a / b$
Now $s i n 2 θ + c o s 2 θ = \frac{2 \frac{\frac{a}{b}}{a^{2}}}{b^{2}} + \frac{\frac{1 - \frac{a^{2}}{b^{2}}}{a^{2}}}{b^{2}}$
$= \frac{2 a b + b^{2} - a^{2}}{a^{2} + b^{2}}$
$\Rightarrow s i n 2 θ + c o s 2 θ = 1 - \frac{2 a (a - b)}{a^{2} + b^{2}} \Rightarrow n = - 2$
Now; $c o s e c^{n} A = x$
$\Rightarrow$ $s i n^{2} A = x$
$S i n 3 A s i n A = 3 s i n^{2} A - 4 s i n^{4} A$
$= 3 x - 4 x^{2}$
So degree $= 2$

Suggest Corrections

Similar questions

{sin}^{2} A = x

sin A sin 2 A sin 3 A sin 4 A

x

s i n^{2} A = x

sin A sin 2 A sin 3 A sin 4 A = a x^{2} + b x^{3} + c x^{4} + d x^{5}

12 a - 10 b + 8 c - 11 d

f (x)

4

x = 1

x = 2

lim x \to 0 (\frac{f (x)}{x^{2}} + 1) = 3

f (- 1)

If $s i n^{- 1} \frac{1}{3} + s i n^{- 1} \frac{2}{3} = s i n^{- 1} x,$ then x is equal to
[Roorkee 1995]

\int \frac{c o s^{7} x + c o s^{5} x}{s i n^{2} x + s i n^{4} x} d x = - \frac{A}{744} s i n^{3} x + 5 s i n x - \frac{2}{s i n x} - 12 t a n^{- 1} (s i n x) + C

Join BYJU'S Learning Program