Q182 / 269 · NDA 2025 (Phase 1) · Mathematics

Consider the following for the three (03) items that follow:
Let $p = \sin 35°$, $q = \sin 25°$ and $r = \sin(-95°)$.

What is $(p^2 + q^2 + r^2)$ equal to?

  • A
    $\dfrac{1}{2}$
  • B
    $1$
  • Correct answer
    C
    $\dfrac{3}{2}$
  • D
    $2$

Solution

Given $p = \sin 35^\circ$, $q = \sin 25^\circ$, and $r = \sin(-95^\circ)$.

Simplifying $r$:

$r = -\sin(90^\circ + 5^\circ) = -\cos 5^\circ$

We need to find the value of $p^2 + q^2 + r^2$:

$p^2 + q^2 + r^2 = \sin^2 35^\circ + \sin^2 25^\circ + \cos^2 5^\circ$

Using the identities $\sin^2 \theta = \dfrac{1 - \cos 2\theta}{2}$ and $\cos^2 \theta = \dfrac{1 + \cos 2\theta}{2}$:

$p^2 + q^2 + r^2 = \dfrac{1 - \cos 70^\circ}{2} + \dfrac{1 - \cos 50^\circ}{2} + \dfrac{1 + \cos 10^\circ}{2}$

$= \dfrac{3}{2} - \dfrac{1}{2}(\cos 70^\circ + \cos 50^\circ - \cos 10^\circ)$

Using the sum-to-product formula $\cos C + \cos D = 2 \cos\left(\dfrac{C+D}{2}\right) \cos\left(\dfrac{C-D}{2}\right)$:

$\cos 70^\circ + \cos 50^\circ = 2 \cos\left(\dfrac{70^\circ + 50^\circ}{2}\right) \cos\left(\dfrac{70^\circ - 50^\circ}{2}\right)$

$= 2 \cos 60^\circ \cos 10^\circ$

$= 2 \left(\dfrac{1}{2}\right) \cos 10^\circ = \cos 10^\circ$

Substituting this back into the expression:

$p^2 + q^2 + r^2 = \dfrac{3}{2} - \dfrac{1}{2}(\cos 10^\circ - \cos 10^\circ)$

$= \dfrac{3}{2} - 0 = \dfrac{3}{2}$

Answer: $\dfrac{3}{2}$

Hence, the correct answer is option: 'C'.

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