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.