Trigonometric Functions: Important Questions

Very Short Answer Questions [1 Mark Questions]

Ques. Find the value of sin 30° and cos 30°

Ans. sin 30° = \(\frac{1}{\sqrt{2}}\)

cos 30° = \(\frac{\sqrt{3}}{2}\)

Ques. Find the sign of sin 300°

Ans. For the given angle, we need to find the coterminal angle which lies between 0° and 360°.

270° < 300° < 360°

Therefore 300° angle lies in the fourth quadrant.

Thus sin 300° is negative.

Ques. What is sin 15°

Ans. sin 15° can also be written as sin (45° – 30°)

= sin 45° cos 30° – cos 45° sin 30°

= (\(\frac{1}{\sqrt{2}}\)) × (\(\frac{\sqrt{3}}{2}\)) – (\(\frac{1}{\sqrt{2}}\)) × (\(\frac{1}{2}\)) 

= \(\frac{\sqrt{3} - 1}{2\sqrt{2}}\)

Ques. Find the sign of cos 400°

Ans. For the given angle, we need to find coterminal angle which lies between 0° and 360°.

400° = 360° + 40°

400° and 40° are coterminal angles hence their trigonometric ratios are the same.

Since 40° lies in the first quadrant, 400° also lies in the first quadrant

Thus cos 400° is positive.

Ques. Solve cos 30° × cos 60° + sin 30° × sin 60°

Ans. cos 30° × cos 60° + sin 30° × sin 60°

= \(\frac{\sqrt{3}}{2}\) × \(\frac{1}{2}\) + \(\frac{1}{2}\) × \(\frac{\sqrt{3}}{2}\)

= \(\frac{\sqrt{3}}{2}\)

Ques. Find the sign of cot (- 206°)

Ans.

-206° = -360° + 154°

154° and - 206° are coterminal angles.

Since 154° lies in the second quadrant, therefore cot (- 206°) is negative.

Ques. Solve 4 cos 30° – cos² 45°

Ans. 4 cos 30° – cos² 45°

= 4 (\(\frac{\sqrt{3}}{2}\)) – (\(\frac{1}{\sqrt{2}}\))²

= \(\frac{4\sqrt{3}}{2}\) – \(\frac{1}{2}\)

= \(\frac{4\sqrt{3}-1}{2}\)

Ques. Solve sin²0 – sin²\(\frac{\pi}{6}\)

Ans. sin²0 – sin²\(\frac{\pi}{6}\)

= (0)² – (\(\frac{1}{2}\))²

= -\(\frac{1}{4}\)

Short Answer Questions [2 Marks Questions]

Ques. Find the value of sin \(\frac{41\pi}{4}\)

Ans. We know the sine function is periodic with period 2π

Therefore,

sin \(\frac{41\pi}{4}\) = sin \(\left( 10\pi + \frac{\pi}{4}\right)\)

= sin \(\frac{\pi}{4}\)

= \(\frac{1}{\sqrt{2}}\)

Ques. Find the value of cos 765°

Ans. We know that cosine function is periodic with period 2π

Therefore cos 765° = cos (720° + 45°)

= cos (2 × 360° + 45°)

= cos 45°

= \(\frac{1}{\sqrt{2}}\)

Ques. sin π + 2 cos π + 3 sin \(\frac{3\pi}{2}\) + 4 cos \(\frac{3\pi}{2}\) – 5 sec π – 6 cosec \(\frac{3\pi}{2}\)

Ans. sin π + 2 cos π + 3 sin \(\frac{3\pi}{2}\) + 4 cos \(\frac{3\pi}{2}\) – 5 sec π – 6 cosec \(\frac{3\pi}{2}\)

= 0 + 2(-1) + 3(-1) + 4(0) – 5(-1) – 6(-1)

= 0 – 2 – 3 + 0 + 5 + 6

= 6

Ques. Find all the trigonometric functions of the angle made by OP with X-axis where P is (-5, 12)

Ans. Let θ be the measure of the angle in standard position whose terminal arm passes through P (-5, 12).

r = OP = \(\sqrt{(-5)^2 + 12^2}\) = 13

P (x, y) = (-5, 12)

Thus x = -5 and y = 12

Sin θ = \(\frac{y}{r}\) = \(\frac{12}{13}\)

Cosec θ = \(\frac{r}{y}\) = \(\frac{13}{12}\)

Cos θ = \(\frac{x}{r}\) = \(\frac{-5}{13}\)

Sec θ = \(\frac{r}{x}\) = \(\frac{-13}{5}\)

Tan θ = \(\frac{y}{x}\) = \(\frac{-12}{5}\)

Cot θ = \(\frac{x}{y}\) = \(\frac{-5}{12}\)

Ques. For θ = 30°, Verify that sin 2θ = 2 sin θ cos θ

Ans.

Given θ = 30°, thus 2θ = 60°

Sin θ = sin 30° = \(\frac{1}{2}\)

Cos θ = cos 30° = \(\frac{\sqrt{3}}{2}\)

Sin 2θ = sin 60° = \(\frac{\sqrt{3}}{2}\)

LHS

= 2 × \(\frac{1}{2}\) × \(\frac{\sqrt{3}}{2}\)

= \(\frac{\sqrt{3}}{2}\)

= RHS

Ques. Solve sin²0 + sin²\(\frac{\pi}{6}\) + sin²\(\frac{\pi}{3}\) + sin²\(\frac{\pi}{2}\)

Ans.

= sin²0 + sin²\(\frac{\pi}{6}\) + sin²\(\frac{\pi}{3}\) + sin²\(\frac{\pi}{2}\)

= (0)² + (\(\frac{1}{2}\))² + (\(\frac{\sqrt{3}}{2}\))² + (1)²

= 0 + \(\frac{1}{4}\) + \(\frac{3}{4}\) + 1

= 2

Ques. If Sin θ = –\(\frac{3}{5}\) and 180° < θ < 270° then find all trigonometric functions.

Ans. Since and 180° < θ < 270°. θ lies in the third quadrant.

Since sin θ = -\(\frac{3}{5}\) thus cosec θ = - \(\frac{5}{3}\)

Now cos²θ = 1 – Sin²θ

Thus, cos²θ = 1 – \(\frac{9}{25}\)

= \(\frac{16}{25}\)

Then cos θ = -\(\frac{4}{5}\)

Thus, sec θ = -\(\frac{5}{4}\)

Now tan θ = \(\frac{\sin \theta}{\cos \theta}\)

Thus,

tan θ = \(\frac{3}{4}\)

And cot θ = \(\frac{4}{3}\)

Long Answer Questions [3 Marks Questions]

Ques. If sin A = \(\frac{-5}{13}\), π < A < \(\frac{3\pi}{2}\) and cos B = \(\frac{3}{5}\), \(\frac{3\pi}{2}\) < B < 2π, find cos (A – B)

Ans.

Given, sin A = \(\frac{-5}{13}\)

We know that,

Cos²A = 1 – sin²A 

= 1 – (\(\frac{-5}{13}\))²

= 1 – \(\frac{25}{169}\)

= \(\frac{144}{169}\)

Thus, Cos A = \(\pm \frac{12}{13}\)

Since π < A < \(\frac{3\pi}{2}\)

A lies in the third quadrant

Cos A < 0

Cos A = -\(\frac{12}{13}\)

Also, cos B = \(\frac{3}{5}\)

Sin²B = 1 – cos²B

= 1 – (\(\frac{3}{5}\))²

= 1 – \(\frac{9}{25}\)

= \(\frac{16}{25}\)

Thus Sin B = \(\pm \frac{4}{5}\)

Since \(\frac{3\pi}{2}\) < B < 2π

Thus B lies in the fourth quadrant.

Sin B = -\(\frac{4}{5}\)

Cos (A – B) = cosA cosB + sinA sinB

= (-\(\frac{12}{13}\))(\(\frac{3}{5}\)) + (-\(\frac{5}{13}\))(-\(\frac{4}{5}\))

= -\(\frac{36}{65}\) + \(\frac{20}{65}\)

= -\(\frac{16}{65}\)

Ques. Prove that (1 + tan²A) + (1 + \(\frac{1}{\tan^2A}\) ) = \(\frac{1}{\sin^2A- \sin^4A}\)

Ans.

LHS

= (sec²A) + (\(\frac{\sec^2 A}{\tan^2 A}\))

= \(\frac{1}{\cos^2 A}\) + \(\frac{\frac{1}{\cos^2A}}{\frac{\sin^2 A}{\cos^2 A}}\)

= \(\frac{1}{\sin^2A \cos^2A}\)

= \(\frac{1}{\sin^2A(1-\sin^2A)}\)

= \(\frac{1}{\sin^2A - \sin^4A}\)

LHS = RHS

Ques. Prove that (sin θ + sec θ)² + (cos θ  + cosec θ)² = (1 + cosec θ sec θ)²

Ans.

LHS

= (sin θ + sec θ)² + (cos θ + cosec θ)²

= (sin θ + \(\frac{1}{\cos \theta}\))² + (cos θ + \(\frac{1}{\sin \theta}\))²

= \({\left( \frac{\cos \theta \sin \theta + 1}{\cos \theta} \right)}^2\) + \({\left( \frac{\sin \theta \cos \theta + 1}{\sin \theta} \right)}^2\)

= \(\frac{(1 + \sin \theta \cos \theta)^2}{\cos^2 \theta}\) + \(\frac{(1 + \sin \theta \cos \theta)^2}{\sin^2 \theta}\)

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

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

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

= \(\left(1+\frac{1}{ \operatorname{cosec} \theta} \times \frac{1}{\sec \theta}\right)^{2} \times \sec^{2} \theta \times \operatorname{cosec}^{2} \theta\)

= \(\left(\frac{\operatorname{cosec} \theta \sec \theta+1}{\operatorname{cosec} \theta \sec \theta}\right)^{2} \times \sec^{2} \theta \times \operatorname{cosec}^{2} \theta\)

= (1 + cosec θ sec θ)²

LHS = RHS

Ques. Solve cos2A + cos2B + cos2C = -1 – 4cosAcosBcosC

Ans.

LHS

= cos2A + cos2B + cos2C

= (cos2A + cos2B) + cos2C

= 2 cos \(\frac{2A + 2B}{2}\) cos \(\frac{2A - 2B}{2}\) + cos2C

= 2 cos (A + B) cos (A – B) + cos2C

[we know that A + B + C = π

Thus A + B = π – C

cos (A + B) = cos (π - C)

cos (A + B) = - cos C]

= -2cosC × cos (A – B) + 2cos²C – 1

= -2cosC (cos (A – B) – cosC) – 1

= -2cosC (cos (A – B) – cos (A + B)) – 1

= -2cosC × 2cosA × cosB – 1

= -1 – 4cosAcosBcosC

LHS = RHS

Very Long Answer Questions [5 Marks Questions]

Ques. Prove that cosA + cosB - cosC = 4 cos\(\frac{A}{2}\) cos\(\frac{B}{2}\) sin \(\frac{C}{2}\) – 1

Ans.

LHS

= cosA + cosB – cosC

= (cosA + cosB) – cosC

= 2 cos \(\frac{A + B}{2}\) cos \(\frac{A - B}{2}\) – cosC

= 2 cos (\(\frac{\pi}{2} - \frac{C}{2}\)) cos\(\frac{A - B}{2}\) + cosC

= 2 sin\(\frac{C}{2}\) cos \(\frac{A - B}{2}\) – (1 – 2sin²\(\frac{C}{2}\))

[A + B + C = π

\(\frac{A+B+C}{2}\) = \(\frac{\pi}{2}\)

\(\frac{A+B}{2}\) = \(\frac{\pi}{2} - \frac{C}{2}\)]

= 2 sin\(\frac{C}{2}\) cos\(\frac{A - B}{2}\) + (2sin²\(\frac{C}{2}\) – 1)

= 2 sin\(\frac{C}{2}\) (cos\(\frac{A - B}{2}\) + 2sin\(\frac{C}{2}\)) – 1

= 2 sin\(\frac{C}{2}\) (cos\(\frac{A - B}{2}\) + sin \(\frac{\pi}{2} - \frac{(A+B)}{2}\)) – 1

= 2 sin\(\frac{C}{2}\) (cos\(\frac{A - B}{2}\) + cos\(\frac{A + B}{2}\)) – 1 … [sin (\(\frac{\pi}{2}\) – θ) = cosθ]

= 2 sin\(\frac{C}{2}\) × 2 cos \(\frac{A}{2}\) × cos \(\frac{B}{2}\) – 1

= 4 cos\(\frac{A}{2}\) cos\(\frac{B}{2}\) sin\(\frac{C}{2}\) – 1

LHS = RHS

Ques. Prove that SinA + SinB + SinC = 4 cos\(\frac{A}{2}\) cos\(\frac{B}{2}\) cos\(\frac{C}{2}\)

Ans.

LHS

= SinA + SinB + SinC

= (SinA + SinB) + SinC

= 2 sin \(\frac{A + B}{2}\) × cos \(\frac{A - B}{2}\) + SinC

[A + B + C = π

\(\frac{A+B+C}{2}\) = \(\frac{\pi}{2}\)

\(\frac{A+B}{2}\) = \(\frac{\pi}{2} - \frac{C}{2}\)

\(\frac{C}{2}\) = \(\frac{\pi}{2}\) – \(\frac{A+B}{2}\)]

= 2 sin(\(\frac{\pi}{2} - \frac{C}{2}\)) cos\(\frac{A - B}{2}\) + sinC

= 2 cos\(\frac{C}{2}\) cos\(\frac{A - B}{2}\) + 2 sin\(\frac{C}{2}\)cos \(\frac{C}{2}\)

= 2 cos\(\frac{C}{2}\) (cos\(\frac{A - B}{2}\) + sin\(\frac{C}{2}\))

= 2 cos\(\frac{C}{2}\) (cos\(\frac{A - B}{2}\) + sin \(\frac{\pi}{2} - \frac{(A+B)}{2}\))

= 2 cos\(\frac{C}{2}\) (cos\(\frac{A - B}{2}\) + cos\(\frac{A + B}{2}\))

= 2 cos\(\frac{C}{2}\) × 2 cos \(\frac{A}{2}\) × cos \(\frac{B}{2}\)

= 4 cos\(\frac{A}{2}\) cos\(\frac{B}{2}\) cos\(\frac{C}{2}\)

LHS = RHS

Ques. Sin²A + Sin²B + Sin²C = 2 + 2cosA cosB cosC.

Ans.

LHS

= Sin²A + Sin²B + Sin²C

= \(\frac{1- \cos 2A}{2}\) + \(\frac{1- \cos 2B}{2}\) + Sin²C

= \(\frac{1}{2}\) – \(\frac{\cos 2A}{2}\) + \(\frac{1}{2}\) – \(\frac{\cos 2B}{2}\) + Sin²C

= 1 – \(\frac{1}{2}\) [cos 2A +cos 2B] + Sin²C

= 1 – \(\frac{1}{2}\) 2cos \(\frac{2A + 2B}{2}\) cos \(\frac{2A - 2B}{2}\) + Sin²C

= 1 – cos (A + B) cos (A – B) + Sin²C

[A + B + C = π

A + B = π – C]

= 1 – cos (π + C) cos (A – B) + Sin²C

= 1 – cosC × cos (A – B) + 1 – Cos²C

= 2 + cosC × cos (A – B) – Cos²C

= 2 + cosC (cos (A – B) – CosC)

= 2 + cosC (cos (A – B) – Cos π – (A + B))

= 2 + cosC (cos (A + B) + cos (A + B))

= 2 + cosC × 2cosA × cosB

= 2 + 2cosA cosB cosC

LHS = RHS

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