MCQ
Which is smaller, $\sin 64^{\circ}$ or $\cos 64^{\circ}$ ?
- A$\cos 64^{\circ}$
- B$\sin 64^{\circ}$
- Ccannot be compared
- Dboth are equal
✓
Answer
(a) $\cos 64^{\circ}$
Explanation: In quadrant $I , \sin \theta$ is increasing.
Now, $\cos 64^{\circ}=\cos \left(90^{\circ}-26^{\circ}\right)=\sin 26^{\circ}$.
Clearly, $\sin 26^{\circ}<\sin 64^{\circ} \Rightarrow \cos 64^{\circ}<\sin 64^{\circ}$
Explanation: In quadrant $I , \sin \theta$ is increasing.
Now, $\cos 64^{\circ}=\cos \left(90^{\circ}-26^{\circ}\right)=\sin 26^{\circ}$.
Clearly, $\sin 26^{\circ}<\sin 64^{\circ} \Rightarrow \cos 64^{\circ}<\sin 64^{\circ}$
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