Question 14 of 18beginner💡 UnderstandShort Answer3 marks

State whether the following are true or false. Justify your answer. (i) sin⁡(A+B)=sin⁡A+sin⁡B\sin(A + B) = \sin A + \sin B. (ii) The value of sin⁡θ\sin \theta increases as θ\theta increases. (iii) The value of cos⁡θ\cos \theta increases as θ\theta increases. (iv) sin⁡θ=cos⁡θ\sin \theta = \cos \theta for all values of θ\theta. (v) cot⁡A\cot A is not defined for A=0∘A = 0^\circ.

(i)
Answer

False. Taking A = 30° and B = 60°: sin(30° + 60°) = sin 90° = 1, but sin 30° + sin 60° = 1/2 + 3\sqrt{3}/2 ≠\neq 1. Hence, sin(A + B) ≠\neq sin A + sin B.

(ii)
Answer

True. From the values: sin 0° = 0, sin 30° = 1/2, sin 45° = 1/2\sqrt{2}, sin 60° = 3\sqrt{3}/2, sin 90° = 1. The value of sin θ increases from 0 to 1 as θ increases from 0° to 90°.

(iii)
Answer

False. From the values: cos 0° = 1, cos 30° = 3\sqrt{3}/2, cos 45° = 1/2\sqrt{2}, cos 60° = 1/2, cos 90° = 0. The value of cos θ decreases from 1 to 0 as θ increases from 0° to 90°.

(iv)
Answer

False. sin θ = cos θ only when θ = 45° (since sin 45° = cos 45° = 1/2\sqrt{2}). For other values like θ = 30°: sin 30° = 1/2 ≠\neq cos 30° = 3\sqrt{3}/2. So sin θ ≠\neq cos θ for all values of θ.

(v)
Answer

True. cot A = cos A / sin A. For A = 0°: cos 0° = 1 and sin 0° = 0. Therefore, cot 0° = 1/0, which is not defined (division by zero).

Explanation

This question tests understanding of trigonometric identities and behavior of trigonometric functions. Students should verify statements using known values of trigonometric ratios for specific angles (0°, 30°, 45°, 60°, 90°) and fundamental definitions. For statement (i), a counterexample disproves the identity. Statements (ii) and (iii) require understanding the variation of sin θ and cos θ with θ. Statement (iv) is false as equality holds only at θ = 45°. Statement (v) uses the definition cot A = cos A/sin A and the fact that sin 0° = 0.

PreviousNext