Question

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

Answer 1

(i) sin(A + B) = sin A + sin B 
Let A = 30° and B = 60° 
sin (A + B) = sin (30° + 60°) 
= sin 90° 
= 1

sin A + sin B = sin 30° + sin 60°
\(=\frac{ 1}{2} + \frac{\sqrt3}{2}\)

\(=\frac{ (1 + \sqrt3)}{2}\)
Clearly, sin (A + B) \(≠\) sin A + sin B 
Hence, the given statement is false.

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(ii) The value of sin \(\text{θ}\) increases as \(\text{θ}\) increases in the interval of 0° \(< \text{θ} <\) 90° as 
sin 0° = 0
sin 30° = \(\frac{1}{2}\) = 0.5
sin 45° = \(\frac{1}{\sqrt2}\) = 0.707

sin 60° =\(\frac{ \sqrt3}{2}\) = 0.866
sin 90° = 1
Hence, the given statement is true.

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(iii) cos 0° = 1
cos 30° = \(\frac{\sqrt3}{2}\) = 0.866

cos 45° = \(\frac{1}{\sqrt2}\) = 0.707
cos 60° =\(\frac{ 1}{2}\)= 0.5
cos 90° = 0
It can be observed that the value of cos \(\text{θ}\) does not increase in the interval of 0°\(<   \text{θ}  <\) 90°.
Hence, the given statement is false.

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(iv) sin \(\text{θ}\) = cos \(\text{θ}\) for all values of \(\text{θ}\). 
This is true when \(\text{θ}\) = 45°
As sin 45° =\(\frac{1}{\sqrt2}\) and cos 45° = \(\frac{1}{\sqrt2}\)
It is not true for other values of \(\text{θ}\)
sin 30° = \(\frac{1}{\sqrt2}\) and cos 30° = \(\frac{\sqrt3}{2}\)

sin 60° = \(\frac{\sqrt3}{2}\) and cos 60° = \(\frac{1}{\sqrt2}\)
sin 90° = 1 and cos 90° = 0
Hence, the given statement is false.

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(v) cot A is not defined for A = 0°
cot A = \(\frac{\text{cos A}}{\text{sin A}}\)

cot 0° = \(\frac{\text{cos 0°}}{\text{sin 0°}} = \frac{1}{0} =\) undefined
Hence, the given statement is true.