TS Inter 1st Year Maths 1A Trigonometric Ratios upto Transformations Solutions Exercise 6(b)

I.
Question 1.
Find the periods for the given 1 – 5 functions.
(i) cos (3x + 5) + 7

Answer:
Let f(x) = cos (3x + 5) + 7
We have period of cos x is 2π ∀ x ∈ R
∴ f (x) is periodic and period of f is = $\frac{2 \pi}{|3|}=\frac{2 \pi}{3}$
(or) f (x + p) = f(x)
⇒ cos (3x + 3p + 5) + 7 = cos (2π+ 3x + 5) + 7
3x + 3p + 5 = 2π + 3x + 5
⇒ 3x = 2π
⇒ x = $\frac{2 \pi}{3}$

Question 2.
tan x

Answer:
The function tan x is periodic with period π
∴ f(x) = tan 5x is periodic and its period is
$\frac{\pi}{|5|}=\frac{\pi}{5}$

Question 3.
cos$\left(\frac{4 x+9}{5}\right)$ (Mar. ’14)

Answer:
The function f(x) = cos x ∀ x ∈ R has the period 2π
∴ f(x) = cos$\left(\frac{4 x+9}{5}\right)$ is periodic and period of f is $\frac{2 \pi}{\frac{4}{5}}=\frac{5 \pi}{2}$

Question 4.
|sin x|

Answer:
The function sin x has period 2π ∀ x ∈ R
But f(x) = |sin x| is periodic and its period is π
[∵ f(x + π) = |sin(x + π)| = |-sin x| = sinx]

Question 5.
tan (x + 4x + 9x + …. + n²x) (n any positive integer) (March 2015-A.P&T.S)

Answer:
tan [1 + 2² + 3² + ……… + n²) x
= tan$\left[\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\right]$x
Period = $\frac{6 \pi}{n(n+1)(2 n+1)}$

Question 6.
Find a sine function whose period is $\frac{2}{3}$.

Answer:
$\frac{2 \pi}{\mathrm{k}}=\frac{2}{3}$ ⇒ 3π = |k| ∴ sin kx = sin (3n x)

Question 7.
Find a cosine function whose period is 7. (March 2013)

Answer:
f(x) = cos[$\frac{2 \pi}{7}$ .x] ($\frac{2 \pi}{k}$ = 7 ⇒ $\frac{2 \pi}{7}$ = k)

II. Sketch the graph of the following functions
Question 1.
tan x between 0 and $\frac{\pi}{4}$.

Answer:

Question 2.
cos 2x in the interval [0, π]

Answer:

Question 3.
sin 2x in the interval (0, π).

Answer:

Question 4.
sin x in the interval [-π, +π]. (May 2014)

Answer:

Question 5.
cos²x in [0, π].

Answer:

Question 6.
Sketch the region enclosed by y = sin x, y = cos x and X – axis in the interval [0, π].

Answer:
y = sin x