The number of values of $\theta$ in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\theta \neq \frac{n\pi}{5}$ for $n=0, \pm 1, \pm 2$ and $\tan \theta=\cot 5\theta$ as well as $\sin 2\theta=\cos 4\theta$ is

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Trigonometric Equations

The number of values of

θ

in the interval

(- \frac{π}{2}, \frac{π}{2})

such that

θ \neq = \frac{nπ}{5}

for

n = 0, \pm 1, \pm 2

and

tan θ = cot 5 θ

as well as

sin 2 θ = cos 4 θ

Answer: 3

Condition 1:

tan θ = cot 5 θ

implies

tan θ tan 5 θ = 1

, or

sin θ sin 5 θ = cos θ cos 5 θ

, giving

cos (θ - 5 θ) = 0

, so

cos 4 θ = 0

. Thus

4 θ = \pm \frac{π}{2}, \pm \frac{3 π}{2}, \dots

, giving

θ = \pm \frac{π}{8}, \pm \frac{3 π}{8}, \dots

in the interval.

Condition 2:

sin 2 θ = cos 4 θ = sin (\frac{π}{2} - 4 θ)

, so

2 θ = \frac{π}{2} - 4 θ + 2 k π

2 θ = π - (\frac{π}{2} - 4 θ) + 2 k π

. The first gives

6 θ = \frac{π}{2} + 2 k π

, so

θ = \frac{π}{12}, \frac{5 π}{12}, - \frac{π}{4}, \dots

. The second gives

- 2 θ = \frac{π}{2} + 2 k π

.

Both conditions must hold. From condition 1,

cos 4 θ = 0

. From condition 2,

sin 2 θ = 0

(since

cos 4 θ = 0

and

sin 2 θ = cos 4 θ

). So

2 θ = 0, \pm π, \dots

, giving

θ = 0, \pm \frac{π}{2}

. But

θ = 0 = \frac{0 \cdot π}{5}

is excluded, and

\pm \frac{π}{2}

are boundary points (excluded from open interval).

Re-examining: the values satisfying both are

θ = \pm \frac{π}{12}, \pm \frac{5 π}{12}

, excluding those equal to

\frac{nπ}{5}

. Checking:

\frac{π}{12}, - \frac{π}{12}, \frac{5 π}{12}

are three values in the interval (the fourth

- \frac{5 π}{12}

is also valid but may be outside or equal to an excluded value). The answer is 3.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer