Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.3

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.3

Question 1.
Find the values of
(i) sin(480°)
(ii) sin(-1110°)
(iii) cos(300°)
(iv) tan(1050°)
(v) cot(660°)
(vi) tan \(\left(\frac{19 \pi}{3}\right)\)
(vii) sin \(\left(-\frac{11 \pi}{3}\right)\)
Solution:
(i) sin(480°) = sin(360° + 120°) = sin 120°
= sin(90° + 30°) = cos 30° = \(\sqrt{3}\)/2

(ii) sin(-1110°) = -sin(1110°)
= – sin (360° × 3 + 30°)
= -sin 30° = -1/2

(iii) cos(300°) = cos(270° + 30°) = sin 30° = 1/2

(iv) tan(1050°) = tan [3(360°) – 30°]

(v) cot(660°) = cot (360° × 2 – 60°)

Question 2.
\(\left(\frac{5}{7}, \frac{2 \sqrt{6}}{7}\right)\) is a point on the terminal side of an angle θ is standard position. Determine the trigonometric function values of angle θ .
Solution:

Question 3.
Find the values of other five trigonometric functions for the following:

(i) cos θ = \(-\frac{1}{2}\); θ lies in the III quadrant.
Solution:
Taking the Numerical values

(ii) cos θ = \(\frac{2}{3}\) ; θ lies in the I quadrant
Solution:

(iii) sin θ = –\(\frac{2}{3}\) ; θ lies in the IV quadrant
Solution:

(iv) tan θ = -2; θ lies in the II quadrant
Solution:

(v) sec θ = \(\frac{13}{5}\) ; θ lies in the IV quadrant
Solution:

Question 4.

Solution:
cot(180° + θ) = cot θ
sin (90° – θ) = cos θ
cos(-θ) = cos θ
sin (270 + θ) = – cos θ
tan(-θ) = -tan θ
cosec (360° + θ) = cosec θ

Question 5.
Find all the angles between 0° and 360° which satisfy the equation sin² θ = \(\frac{3}{4}\)
Solution:

Question 6.

Solution:
LHS = sin² 10° + sin² 20° + sin² 70° + sin² 80°
= sin² 10° + sin² (90° – 10°) + sin² 20° + sin²(90° – 20°)
= sin² 10° + (cos 10°)² + sin² 20° + (cos 20°)²
= (sin² 10+ cos² 10) + sin² 20° + cos² 20°
= 1 + 1 = 2 = RHS

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.3 Additional Questions Solved

Question 1.
Prove that: sin 600°. tan (-690°) + sec 840°. cot (-945°) = \(\frac{3}{2}\)
Solution:

Question 2.
Prove that sin (270° – θ) sin (90° – θ) – cos (270° – θ) cos (90° + θ) + 1 = 0
Solution:
LHS = sin (270° – θ) sin (90° – θ) – cos (270° – θ) cos (90° + θ) + 1
Now, sin (270° – θ) = sin {180°+ (90°- θ)}
= – cos (90° – θ) = – sin θ
LHS = – cos θ . cos θ – (- sin θ) (- sin θ) + 1
= – cos² θ – sin² θ + 1
= – (cos² θ + sin² θ) + 1 = -1 + 1 = 0 = RHS

Question 3.
Prove that cos 24° + cos 55° + cos 125° + cos 204° + cos 300° = \(\frac{1}{2}\)
Solution:
cos 204° = cos (180°+ 24°) = – cos 24°
cos 125° = cos (180° – 55°) = – cos 55°
LHS = cos 24° + cos 55° + (- cos 55°) + (- cos 24°) + cos 300°
= cos 24° + cos 55° – cos 55° – cos 24° + cos 300°

Question 4.

Solution:

Question 5.

Solution:

LHS = [(1 +cot α) + cosec α][(1 + cot α) – cosec α]
= (1 + cot α)² – cosec² α
= 1 + cot²α + 2 cot α – cosec² α
[∵ 1 + cot²α = cosec²α]
= cosec²α + 2 cot α – cosec²α
= 2 cot α = RHS

Question 6.

Solution:

= sec (450° – θ)
= sec[360° + (90° – θ)]
sec (90° – θ) = cosec θ

Question 7.

Solution:
cos (90° + θ) = – sin θ
sec (- θ) = sec θ
tan (180° – θ) = – tan θ
sec (360° – θ) = sec θ
sin (180° + θ) = – sin θ
cot (90° + θ) = – tan θ

Question 8.
Find x from the equation cosec (90° + A) + x cos A cot (90° + A) = sin (90° + A).
Solution: