Inter 1st Year Maths Trigonometric Functions Solutions Exercise 3c

Practicing the AP Board Solutions Class 11 Maths and Chapter 3 Inter 1st Year Maths Trigonometric Functions Solutions Exercise 3c Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Trigonometric Functions Solutions Exercise 3c

Trigonometric Functions Exercise 3c Solutions

Trigonometric Functions Class 11 Exercise 3c Solutions – Trigonometric Functions 3c Exercise Solutions

I.

Question 1.
Find the value of
(i) sin 75°
(ii) tan 15°
(iii) cot 15°
(iv) cos 75°
(v) sin 105°
(vi) tan 75°
(vii) cot 75°
(viii) cos 105°
(ix) tan 105°
(x) cot 105°
(xi) cos 15°
(xii) sin 15°
Solution:
(i) It can be written as sin(45° + 30°)
Using the formula sin(x + y) = sin x cos y + cos x sin y
sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°
= \(\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\)
= \(\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}\)
= \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

(ii) It can be written as (tan 45° – 30°)
Using formula tan(x – y) = \(\frac{\tan x-\tan y}{1+\tan x \tan y}\)

(iv) cos 75° = cos (45° + 30°)
= cos 45° cos 30° – sin 45° sin 30° [∵ cos (A + B) = cos A cos B – sin A sin B]
= \(\frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\)
= \(\frac{\sqrt{6}-\sqrt{2}}{4}\)
= 0.2585

(v) sin 105° = sin (60° + 45°)
= sin 60° cos 45° + cos 60° sin 45° [∵ sin (A + B) = sin A cos B + cos A sin B]
= \(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}+\frac{1}{2} \cdot \frac{\sqrt{2}}{2}\)
= \(\frac{\sqrt{6} \cdot \sqrt{2}}{4}\)
= 0.9659

(vi) tan 75° = tan (45° + 30°)
Tan (A + B) = \(\frac{{Tan} A+{Tan} B}{1-{Tan} A {Tan} B}\)

(xi) cos 15° = cos (45° – 30°)
= cos 45° cos 30° + sin 45° sin 30° [∵ cos (A – B) = cos A cos B + sin A sin B]
= \(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\)
= \(\frac{\sqrt{6}+\sqrt{2}}{4}\)
= 0.9659

(xii) sin 15° = sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30° [∵ sin (A – B) = sin A cos B – cos A sin B]
= \(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\)
= \(\frac{\sqrt{6}-\sqrt{2}}{4}\)
= 0.2588

Question 2.
Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x.
Solution:
L.H.S = sin (n + 1) x sin (n + 2)x + cos (n + 1)x cos (n + 2)x
By multiplying and dividing by 2
= \(\frac {1}{2}\) [2 sin (n + 1)x sin (n + 2)x + 2 cos (n + 1)x cos (n + 2)x]
Using the formula
-2 sin A sin B = cos (A + B) – cos (A – B)
2 cos A cos B = cos (A + B) + cos (A – B)
= \(\frac {1}{2}\) [cos {(n + 1)x – (n + 2)x} – cos {(n + 1)x + (n + 2)x} + cos {(n + 1)x + (n + 2)x} + cos {(n + 1)x – (n + 2)x}]
= \(\frac {1}{2}\) × 2 cos {(n + 1)x – (n + 2)x}
= cos (-x)
= cos x
= R.H.S

II. Prove that:

Question 1.
\(\sin ^2 \frac{\pi}{6}+\cos ^2 \frac{\pi}{3}-\tan ^2 \frac{\pi}{4}=-\frac{1}{2}\)
Solution:

Question 2.
\(2 \sin ^2 \frac{\pi}{6}+{cosec}^2 \frac{7 \pi}{6} \cos ^2 \frac{\pi}{3}=\frac{3}{2}\)
Solution:

Question 3.
\(\cot ^2 \frac{\pi}{6}+{cosec} \frac{5 \pi}{6}+3 \tan ^2 \frac{\pi}{6}=6\)
Solution:

Question 4.
\(2 \sin ^2 \frac{3 \pi}{4}+2 \cos ^2 \frac{\pi}{4}+2 \sec ^2 \frac{\pi}{3}=10\)
Solution:

Question 5.
Prove the following:
\(\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)\) – \(\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)\) = sin(x + y)
Solution:

Question 6.
\(\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}=\left(\frac{1+\tan x}{1-\tan x}\right)^2\)
Solution:

Question 7.
\(\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^2 x\)
Solution:

Question 8.
\(\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x\)
Solution:

Question 9.
sin² 6x – sin² 4x = sin 2x sin 10x
Solution:
Consider
L.H.S = sin² 6x – sin² 4x
Using the formula
sin A + sin B = \(2 \sin \left(\frac{\mathrm{~A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\)
sin A – sin B = \(2 \cos \left(\frac{\mathrm{~A}+\mathrm{B}}{2}\right) \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\)
So, we get (sin 6x + sin 4x) (sin 6x – sin 4x)
By further calculation
\(\left[2 \sin \left(\frac{6 x+4 x}{2}\right) \cos \left(\frac{6 x-4 x}{2}\right)\right]\) \(\left[2 \cos \left(\frac{6 x+4 x}{2}\right) \sin \left(\frac{6 x-4 x}{2}\right)\right]\)
We get (2 sin 5x cos x) (2 cos 5x sin x)
It can be written as (2 sin 5x cos 5x) (2 sin x cos x)
= sin 10x sin 2x
= R.H.S

Question 10.
cos² 2x – cos² 6x = sin 4x sin 8x
Solution:
Consider
L.H.S = cos² 2x – cos² 6x
Using the formula
cos A + cos B = \(2 \cos \left(\frac{\mathrm{~A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\)
cos A – cos B = \(-2 \sin \left(\frac{\mathrm{~A}+\mathrm{B}}{2}\right) \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\)
So, we get (cos 2x + cos 6x) (cos 2x – cos 6x)
By further calculation
\(\left[2 \cos \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right]\) \(\left[-2 \sin \left(\frac{2 x+6 x}{2}\right) \sin \left(\frac{2 x-6 x}{2}\right)\right]\)
We get [2 cos 4x cos (-2x)] [-2 sin 4x sin (-2x)]
It can be written as [2 cos 4x cos 2x] [-2 sin 4x (-sin -2x)]
So, we get [2 sin 4x cos 4x] [2 sin 2x cos 2x]
= sin 4x sin 8x
= R.H.S

Question 11.
\(\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}\)
Solution:

Question 12.
\(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}\) = tan 4x
Solution:

so, we get
= \(\frac{\sin 4 x}{\cos 4 x}\)
= tan 4x
= R.H.S

Question 13.
\(\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}\)
Solution:

Question 14.
\(\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}\) = tan 2x
Solution:

Question 15.
\(\frac{\sin x-\sin 3 x}{\sin ^2 x-\cos ^2 x}\) = 2 sin x
Solution:

Question 16.
tan 4x = \(\frac{4 \tan x\left(1-\tan ^2 x\right)}{1-6 \tan ^2 x+\tan ^4 x}\)
Solution:
Consider
L.H.S. = tan 4x = tan 2(2x)
By using the formula

Question 17.
cos 4x = 1 – 8 sin²x cos²x
Solution:
Consider
L.H.S = cos 4x
We can write it as cos 2(2x)
Using the formula
cos 2A = 1 – 2 sin²A = 1 – 2 sin² 2x
Again, by using the formula
sin 2A = 2 sin A cos A = 1 – 2(2 sin x cos x)²
So, we get 1 – 8 sin²x cos²x = R.H.S.

III. Prove the following.

Question 1.
\(\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\) \(\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]\) = 1
Solution:
Consider
L.H.S = \(\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\) \(\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]\)
It can be written as sin x cos x (tan x + cot x)
So, we get
sin x cos x \(\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)\) (sin x cos x) \(\left[\frac{\sin ^2 x+\cos ^2 x}{\sin x \cos x}\right]\)
= 1
= R.H.S

Question 2.
sin 2x + 2 sin 4x + sin 6x = 4 cos²x sin 4x
Solution:
Consider
L.H.S = sin 2x + 2 sin 4x + sin 6x
= [sin 2x + sin 6x] + 2 sin 4x
Using the formula
sin A + sin B = \(2 \sin \left(\frac{\mathrm{~A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\)
\(\left[2 \sin \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right]\) + 2 sin 4x
By further simplification
2 sin 4x cos (-2x) + 2 sin 4x
It can be written as
2 sin 4x cos 2x + 2 sin 4x
Taking common terms
2 sin 4x (cos 2x + 1)
Using the formula
= 2 sin 4x (2 cos²x – 1 + 1)
= 2 sin 4x (2 cos²x)
= 4 cos²x sin 4x
= R.H.S

Question 3.
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Solution:
Consider
L.H.S = cot 4x (sin 5x + sin 3x)
It can be written as
\(\frac{\cos 4 x}{\sin 4 x}\left[2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\right]\)
Using formula
sin A + sin B = \(2 \sin \left(\frac{\mathrm{~A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\)
\(\left(\frac{\cos 4 x}{\sin 4 x}\right)\) [2 sin 4x cos x]
So, we get 2 cos 4x cos x
Similarly, R.H.S = cot x (sin 5x – sin 3x)
It can be written as
\(\frac{\cos x}{\sin x}\left[2 \cos \left(\frac{5 x+3 x}{2}\right) \sin \left(\frac{5 x-3 x}{2}\right)\right]\)
Using the formula
sin A – sin B = \(2 \cos \left(\frac{\mathrm{~A}+\mathrm{B}}{2}\right) \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\)
\(\frac{\cos x}{\sin x}\) [2 cos 4x sin x]
So, we get 2 cos 4x cos x
Hence, L.H.S = R.H.S

Question 4.
\(\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}\) = cot 3x
Solution:

Question 5.
cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
Solution:
Consider
L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x
It can ce written as
cot x cot 2x – cot 3x (cot 2x + cot x) = cot x cot 2x – cot(2x + x) (cot 2x + cot x)
Using the formula
cot(A + B) = \(\frac{\cot A \cot B-1}{\cot A+\cot B}\)
cot x cot 2x – \(\left[\frac{\cot 2 x \cot x-1}{\cot x+\cot 2 x}\right]\) (cot 2x + cot x)
so, we get cot x cot 2x – (cot 2x cot x – 1)
= 1
= R.H.S

Question 6.
cos 6x = 32 cos⁶x – 48 cos⁴x + 18 cos²x – 1
Solution:
Consider
LH.S = cos 6x
It can be written as cos 3(2x)
Using the formula
cos 3A = 4 cos³A – 3 cos A
cos 3(2x) = 4 cos³ 2x – 3 cos 2x
Again, by using the formula
cos 2x = 2 cos²x – 1
= 4[(2 cos²x – 1)3 – 3(2 cos²x – 1)]
By further simplification
4[(2 cos²x)³ – (1)³ – 3(2 cos²x)² + 3(2 cos²x)] – 6 cos²x + 3
we get 4[8 cos⁶x – 1 – 12 cos⁴x + 6 cos²x] – 6 cos²x + 3
By multiplication
32 cos⁶x – 4 – 48 cos⁴x + 24 cos²x – 6 cos²x + 3
On further calculation
32 cos⁶x – 48 cos⁴x + 18 cos²x – 1 = R.H.S