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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

^2 76^ ^ ^ + ^2 16^ ^ ^ - 76^ ^ ^ 16^ ^ ^ is equal to

cos⁡276∘+cos⁡216∘−cos⁡76∘cos⁡16∘\cos ^2 76^{{}^{{}^{\circ}}}+\cos ^2 16^{{}^{{}^{\circ}}}-\cos 76^{{}^{{}^{\circ}}} \cos 16^{{}^{{}^{\circ}}} is equal to

Options

  1. A00
  2. B12\frac{1}{2}
  3. C−14-\frac{1}{4}
  4. D32\frac{3}{2}

Correct answer

D. 32\frac{3}{2}

Step-by-step solution

cos⁡276∘+cos⁡216∘−cos⁡76∘cos⁡16∘=cos⁡276∘+1−sin⁡216∘−12⋅2cos⁡76∘cos⁡16∘=1+cos⁡(76∘+16∘)cos⁡(76∘−16∘)−12(cos⁡92∘+cos⁡60∘)=1+12cos⁡92∘−12cos⁡92∘−14=1−14=34\begin{aligned} & \cos ^2 76^{{}^{{}^{\circ}}}+\cos ^2 16^{{}^{{}^{\circ}}}-\cos 76^{{}^{{}^{\circ}}} \cos 16^{{}^{{}^{\circ}}} \\ & =\cos ^2 76^{{}^{{}^{\circ}}}+1-\sin ^2 16^{{}^{{}^{\circ}}}-\frac{1}{2} \cdot 2 \cos 76^{{}^{{}^{\circ}}} \cos 16^{{}^{{}^{\circ}}} \\ & =1+\cos \left(76^{{}^{{}^{\circ}}}+16^{{}^{{}^{\circ}}}\right) \cos \left(76^{{}^{{}^{\circ}}}-16^{{}^{{}^{\circ}}}\right) \\ & \quad-\frac{1}{2}\left(\cos 92^{{}^{{}^{\circ}}}+\cos 60^{{}^{{}^{\circ}}}\right) \\ & =1+\frac{1}{2} \cos 92^{{}^{{}^{\circ}}}-\frac{1}{2} \cos 92^{{}^{{}^{\circ}}}-\frac{1}{4} \\ & =1-\frac{1}{4}=\frac{3}{4}\end{aligned}
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