Trigonometric Functions

52. L.H.S. = 2 cos $\frac{\pi }{13}$ cos $\frac{9\pi }{13}$ + cos $\frac{3\pi }{13}$ + cos $\frac{5\pi }{13}$ =∂ .

= 2 cos $\frac{\pi }{13}$ cos $\frac{9\pi }{13}$ + 2 cos $\begin{array}{l}\left(\frac{3\pi }{13}+\frac{5\pi }{13}\right)\\ \overline{2}\end{array}$ cos $\frac{\left(\frac{3\pi }{13}-\frac{5\pi }{13}\right)}{2}$

$\left[?cosA+cosB=2cos\frac{A+B}{2}cos\frac{A-B}{2}\right]$

= 2 cos $\frac{\pi }{13}$ cos $\frac{9\pi }{13}$ + 2. cos $\left(\frac{8\pi /13}{2}\right)$ cos $\left(\frac{-2\pi /13}{2}\right)$

= 2 cos $\frac{\pi }{13}$ cos $\frac{9\pi }{13}$ + 2 cos $\frac{4\pi }{13}$ cos $\frac{\pi }{13}$ [ $?$ cos (x) = cosx].

= 2 cos $\frac{\pi }{13}$ $\left[cos\frac{9\pi }{13}+cos\frac{4\pi }{13}\right].$

= 2 cos $\frac{\pi }{13}\left[2\cdot cos\left(\frac{\frac{9\pi }{13}+\frac{4\pi }{13}}{2}\right)cos\left(\frac{\frac{9\pi }{13}-\frac{4\pi }{13}}{2}\right)\right]$

= 2 cos $\frac{\pi }{13}$ $\left[2·cos\left(\frac{13\pi /13}{2}\right)cos\left(\frac{5\pi /13}{2}\right)\right]$

= 2 cos $\frac{\pi }{3}$ * 2 *cos $\frac{\pi }{2}$ * cos $\frac{5\pi }{26}$ .

= 2 cos $\frac{\pi }{2}$ 2* 0*cos $\frac{5\pi }{26}$

= 0

= R.H.S.

51. We have,

sinx + sin 3x + sin 5x = 0.

(sinx + sin 5x) + sin 3x = 0.

Using sin A + sin B = 2 sin $\frac{A+B}{2}$ cos $\frac{A-B}{2}$ .

2 sin $\left(\frac{x+5x}{2}\right)$ cos $\left(\frac{x-5x}{2}\right)$ + sin 3x = 0.

2 sin $\frac{6x}{2}$ cos$\frac{-4x}{2}$  + sin 3x = 0.

2 sin 3xcos (-2x) + sin 3x = 0.

sin 3x [2 cos 2x + 1] = 0 [ $?$ cos (-x) = cosx].

sin 3x = 0 or 2 cos 2x + 1 = 0.

3x = nπ, n∈z. or cos 2x = $\frac{-1}{2}$ = -cos $\frac{\pi }{3}$ = cos  π -$\frac{\pi }{3}$= cos $\frac{2\pi }{3}$

x= $\frac{n\pi }{3}$ , n∈z or 2x = 2nπ± $\frac{2\pi }{3}$ .

x = nπ±$\frac{\pi }{3}$ , n∈z.

50. We have,

sec² 2x = 1 tan 2x

1 + tan² 2x = 1 tan 2x [ sec²x = 1 + tan²x]

tan² 2x + tan 2x = 0.

tan 2x (tan 2x + 1) = 0.

tan 2x = 0         or         tan 2x + 1 = 0.

2x = nπ, x∈z or tan 2x = -1 = -tan $\frac{\pi }{4}$ = tan π-$\frac{\pi }{4}$  = tan $\frac{3\pi }{4}.$

x= $\frac{n\pi }{2}$ , n∈z or 2x = nπ + $\frac{3\pi }{4}$ , n∈z.

x = $\frac{n\pi }{2}+\frac{3\pi }{8},$ n∈z

49. We have,

sin 2x + cosx = 0.

2 sin cosx + cosx = 0 ( $?$ sin 2x = 2 sin xcosx)

cosx (2 sin x + 1) = 0.

cosx = 0 or 2 sinx + 1 = 0.

x = (2n + 1) $\frac{\pi }{2}$, n∈z or sin x = $\frac{1}{2}$ = -sin $\frac{\pi }{6}$ = sin  π + $\frac{\pi }{6}$= sin $\frac{7\pi }{6}.$

x = (2n + 1) $\frac{\pi }{2}$, x∈z or x= nπ + (-1)ⁿ $\frac{7\pi }{6},$ n∈z.

48. We have,

cos 3x + cosx-cos 2x = 0.

(cos 3x + cosx) cos 2x = 0

Using cos A + cos B = 2 cos $\frac{A+B}{2}$ cos $\frac{A-B}{2}$

2 cos $\left(\frac{3x+x}{2}\right)$ cos $\left(\frac{3x-x}{2}\right)$ - cos 2x = 0.

2 cos $\frac{4x}{2}$ cos $\frac{2x}{2}$ - cos 2x = 0

2 cos 2xcosx - cos 2x = 0

cos 2x. (2 cosx - 1) = 0.

cos 2x = 0 or 2 cosx -1 = 0.

2x = (2n + 1) $\frac{\pi }{2}$, n∈z or cosx = $\frac{1}{2}$ = cos $\frac{\pi }{3}$

x = (2n + 1) $\frac{\pi }{4}$, n∈z or x = 2nx± $\frac{\pi }{3}$, n∈z.

47. We have,

cos 4x = cos 2x.

⇒ cos 4x-cos 2x = 0.

⇒ -2 sin $\left(\frac{4x+2x}{2}\right)$ sin $\left(\frac{4x-2x}{2}\right)$ = 0.

⇒sin $\frac{6x}{2}$ sin $\frac{2x}{2}$ = 0

⇒sin 3x sin x = 0.

∴sin 3x = 0 or sin x = 0

3x = nπ or x = nπ, n∈z,

⇒x = $\frac{n\pi }{3}$ or x = nπ, n∈z

46. We have cosec x = 2 .i e, cosec x = (-)ve,

So, the principal solution lies in IIInd and IVth quadrent.

Now, cosec x = -2 = -cosec $\frac{\pi }{6}$ = cosec $(\pi +\frac{\pi }{6}$= cosec $\left(2\pi -\frac{\pi }{6}\right)$

So the principal solution are x = $\left(\pi +\frac{\pi }{6}\right)$ and $\left(2\pi -\frac{\pi }{6}\right)$

= $\left(\frac{6\pi +\pi }{6}\right)$ and $\left(\frac{12\pi -\pi }{6}\right)$

= $\frac{7\pi }{6}$ and $\frac{11\pi }{6}.$

As cosec x = cosec $\frac{7\pi }{6}$ ⇒sin x = sin $\frac{7\pi }{c}$ $\left[?cosecx=\frac{1}{sinx}\right]$

The general solution has the form,

x = nπ + (-1)n $\frac{7\pi }{6}$ , n∈z.

45. We have, cot x = -√3  i.e., cot x is negative

So, the principal solution lies in II^nd and IV^th quadrant.

So the principal solution are a = $\left(\pi -\frac{\pi }{6}\right)$ and $\left(2\pi -\frac{\pi }{6}\right)$

= $\left(\frac{6\pi -\pi }{6}\right)$ and $\left(\frac{12\pi -\pi }{6}\right)$

= $\frac{5\pi }{6}$ and $\frac{11\pi }{6}.$

As cot x = cot $\frac{5\pi }{6}$ tan x = tan $\frac{5\pi }{6}$ $\left[?cotx=\frac{1}{tanx}\right]$

The general solution has the form.

x = n+ $\frac{5\pi }{6}$ , nz

43. tanx = √3

We have, tan x = √3

Since tan x is (+) ve the principal solution lies in I^st and III^nd quadrant

Now, tan x = √3 = tan $\frac{\pi }{3}$ . = tan $\left(\pi +\frac{\pi }{3}\right)$

Principal solution are x = $\frac{\pi }{3}$  and $\left(\pi +\frac{\pi }{3}\right)$

        $\frac{\pi }{3}$ =  and $\left(\frac{3\pi +\pi }{3}\right)$

        $\frac{\pi }{3}$ = and $\frac{4\pi }{3}$ .

As tan x = tan $\frac{\pi }{3}$

The general solution is.

x = np + $\frac{\pi }{3}$ , n∈z