Skip to content
NCERT Exemplar · Q52

Q.If sin⁡θ=−45\sin\theta = \dfrac{-4}{5} and θ\theta lies in third quadrant then the value of cos⁡θ2\cos\dfrac{\theta}{2} is
(A) 15\dfrac{1}{5}
(B) −110-\dfrac{1}{\sqrt{10}}
(C) −15-\dfrac{1}{\sqrt{5}}
(D) 110\dfrac{1}{\sqrt{10}}

Chhattisgarh CgbseMCQ· 1mImportance★★★★★est
84% · 126/150 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

In Q3, cos⁡θ=−35\cos\theta=-\frac{3}{5}; since θ2\frac{\theta}{2} lands in Q2 the cosine is negative, giving cos⁡θ2=−15\cos\frac{\theta}{2}=-\frac{1}{\sqrt{5}} — option (C).

Step 1 — Find cos⁡θ\cos\theta. With sin⁡θ=−45\sin\theta=-\dfrac{4}{5} and θ\theta in the third quadrant (cosine negative):

cos⁡θ=−1−sin⁡2θ=−1−1625=−35\cos\theta=-\sqrt{1-\sin^2\theta}=-\sqrt{1-\tfrac{16}{25}}=-\frac{3}{5}

Step 2 — Quadrant of θ2\dfrac{\theta}{2}. Since 180∘<θ<270∘180^\circ<\theta<270^\circ, we have 90∘<θ2<135∘90^\circ<\dfrac{\theta}{2}<135^\circ, so θ2\dfrac{\theta}{2} lies in the second quadrant, where cosine is negative. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.