State whether the following are true or false. Justify your answer.
(i) False
(ii) True
(iii) False
(iv) False
(v) True
We check each statement using trigonometric identities and basic behaviour of \(\sin\theta\) and \(\cos\theta\).
(i) \(\sin(A+B)=\sin A+\sin B\)
False. The correct identity is:
\(\sin(A+B)=\sin A\cos B+\cos A\sin B\).
Justification by example: Take \(A=30^\circ\), \(B=30^\circ\).
LHS: \(\sin(30^\circ+30^\circ)=\sin 60^\circ=\dfrac{\sqrt{3}}{2}\).
RHS: \(\sin 30^\circ+\sin 30^\circ=\dfrac{1}{2}+\dfrac{1}{2}=1\).
Since \(\dfrac{\sqrt{3}}{2} \ne 1\), the statement is false.
(ii) “The value of \(\sin\theta\) increases as \(\theta\) increases.”
True (for \(0^\circ \le \theta \le 90^\circ\)).
In the first quadrant, as \(\theta\) increases, the opposite side (compared to hypotenuse) effectively increases, so \(\sin\theta\) increases.
Quick table check: \(\sin 0^\circ=0\), \(\sin 30^\circ=\dfrac{1}{2}\), \(\sin 45^\circ=\dfrac{\sqrt{2}}{2}\), \(\sin 60^\circ=\dfrac{\sqrt{3}}{2}\), \(\sin 90^\circ=1\). These values clearly increase.
Student Note: This “increasing” behaviour is commonly discussed for \(0^\circ\) to \(90^\circ\). Outside this range, \(\sin\theta\) does not keep increasing forever.
(iii) “The value of \(\cos\theta\) increases as \(\theta\) increases.”
False (for \(0^\circ \le \theta \le 90^\circ\)).
In the first quadrant, \(\cos\theta\) actually decreases as \(\theta\) increases.
Quick table check: \(\cos 0^\circ=1\), \(\cos 30^\circ=\dfrac{\sqrt{3}}{2}\), \(\cos 45^\circ=\dfrac{\sqrt{2}}{2}\), \(\cos 60^\circ=\dfrac{1}{2}\), \(\cos 90^\circ=0\). These values go down.
(iv) \(\sin\theta=\cos\theta\) for all values of \(\theta\)
False. \(\sin\theta\) equals \(\cos\theta\) only for some specific angles, not for all.
Justification by example: At \(\theta=45^\circ\), \(\sin 45^\circ=\cos 45^\circ=\dfrac{\sqrt{2}}{2}\) (so it is true here).
But at \(\theta=30^\circ\): \(\sin 30^\circ=\dfrac{1}{2}\) and \(\cos 30^\circ=\dfrac{\sqrt{3}}{2}\), which are not equal. Hence it is not true for all \(\theta\).
(v) “\(\cot A\) is not defined for \(A=0^\circ\)”
True. We know:
\(\cot A = \dfrac{\cos A}{\sin A}\).
At \(A=0^\circ\), \(\sin 0^\circ = 0\) and \(\cos 0^\circ = 1\).
So, \(\cot 0^\circ = \dfrac{1}{0}\), which is not defined.
Final Answers: (i) False, (ii) True, (iii) False, (iv) False, (v) True.