Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.3 Text Book Back Questions and Answers, Notes.

## 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 \(\left(\frac{-11 \pi}{3}\right)\)

Answer:

(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 (1050°) = tan(12 × 90 – 30°)

= – tan30° = – \(\frac{1}{\sqrt{3}}\)

(v) cot 660°

cot 660° = cot (7 × 90 + 30°)

= – tan 30° = – \(\frac{1}{\sqrt{3}}\)

(vi) tan \(\left(\frac{19 \pi}{3}\right)\)

(vii) sin \(\left(\frac{-11 \pi}{3}\right)\)

Question 2.

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

Answer:

Given \(\left(\frac{5}{7}, \frac{2 \sqrt{6}}{7}\right)\) is a point on the terminal side of an angle θ in standard position.

Question 3.

Find the values of the other five trigonometric functions of the following

(i) cos θ = –\(\frac{1}{2}\), θ lies in the III quadrant

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

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

(iv) tan θ = – 2, θ lies in the II quadrant

(v) sec θ = \(\frac{13}{5}\), θ lies inthe IVquadrant

Answer:

(i) cos θ = –\(\frac{1}{2}\), θ lies in the III quadrant

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

We know that cos^{2}θ + sin^{2}θ = 1

\(\left(\frac{2}{3}\right)^{2}\) + sin^{2}θ = 1

\(\frac{4}{9}\) + sin^{2}θ = 1

Since θ lies in the I quadrant all trigonometric functions are positive.

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

We know that cos^{2}θ + sin^{2}θ = 1

cos^{2}θ + \(\left(-\frac{2}{3}\right)^{2}\) = 1

cos^{2}θ + \(\frac{4}{9}\) = 1

Since θ lies in the fourth quadrant cos θ is positive.

(iv) tan θ = – 2, θ lies in the II quadrant

We know that sec^{2}θ – tan^{2}θ = 1

sec^{2}θ – (-2)^{2} = 1

sec^{2}θ – 4 = 1

sec^{2}θ = 1 + 4 = 5

sec θ = ± √5

Since θ lies in the second quadrant sec θ is negative.

(v) sec θ = \(\frac{13}{5}\), θ lies inthe IVquadrant

We know that sec^{2}θ – tan^{2}θ = 1

\(\left(\frac{13}{5}\right)^{2}\) – tan^{2}θ = 1

\(\frac{169}{25}\) – 1 = tan^{2}θ

Question 4.

Prove that

Answer:

Question 5.

Find all the angles between 0° and 360° which satisfy the equation sin^{2}θ = \(\frac{3}{4}\)

Answer:

sin^{2}θ = \(\frac{3}{4}\) ⇒ sin θ = ± \(\frac{\sqrt{3}}{2}\)

sin 60° = \(\frac{\sqrt{3}}{2}\)

sin 120° = sin (180° – 60°)

= sin 60° = \(\frac{\sqrt{3}}{2}\)

∴ θ = 60° and 120°

Question 6.

Show that

Answer:

= sin^{2} 10° + sin^{2} 20° + [cos 20°]^{2} + [cos 10°]^{2}

= sin^{2} 10° + sin^{2} 20° + cos^{2} 20° + cos^{2} 10°

= sin^{2} 10° + cos^{2} 10° + sin^{2} 20° + cos^{2} 20°

= 1 + 1 = 2