Functions
Tangents from external point to sin curve
MJAT_TS4_P1
Grade 12
Question:
Tangents are drawn from $\left(-\dfrac{\pi}{2}, 0\right)$ to $f:[0,\infty)\to\mathbb{R}$, $f(x)=\sin x$. Let $(a_i, \sin a_i)$ be the points of contact for $i=1,2,3,\ldots$ with $a_{i+1}>a_i$. Choose the correct option(s):
A) $a_5 + \dfrac{\pi}{2} = \tan a_5$
B) $a_{10} - a_9 < \pi$
C) $a_{11} - a_9 > 2\pi$
D) $\sin a_{10} - \sin a_9 < 0$
Step-by-Step Solution
Key Concept: A tangent from $(-\pi/2, 0)$ to $(a,\sin a)$: slope $= \frac{\sin a - 0}{a-(-\pi/2)} = \cos a$. So $\sin a = \cos a(a+\pi/2)$, i.e., $a+\pi/2 = \tan a$ (A ✓ — satisfied by all contact points).
A ✓, B ✓, D ✓. Answer: A, B, D.
Correct Answer: ABD