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Evaluate :

(sin 77°cos 13°)2\Big(\dfrac{\text{sin 77°}}{\text{cos 13°}}\Big)^2 + (cos 77°sin 13°)2\Big(\dfrac{\text{cos 77°}}{\text{sin 13°}}\Big)^2 - 2 cos2 45°

Trigonometric Identities

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Answer

(sin 77°cos 13°)2+(cos 77°sin 13°)2−2 cos245°=(sin (90° - 13°)cos 13°)2+(cos (90° - 13°)sin 13°)2−2 cos245°=(cos 13°cos 13°)2+(sin 13°sin 13°)2−2 cos245°=(cos13°cos13°)2+(sin13°sin13°)2−2 cos245°=12+12−2×(12)2=1+1−2×(12)=2−1=1\Big(\dfrac{\text{sin 77°}}{\text{cos 13°}}\Big)^2 + \Big(\dfrac{\text{cos 77°}}{\text{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = \Big(\dfrac{\text{sin (90° - 13°)}}{\text{cos 13°}}\Big)^2 + \Big(\dfrac{\text{cos (90° - 13°)}}{\text{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = \Big(\dfrac{\text{cos 13°}}{\text{cos 13°}}\Big)^2 + \Big(\dfrac{\text{sin 13°}}{\text{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = \Big(\dfrac{\cancel{cos 13°}}{\cancel{cos 13°}}\Big)^2 + \Big(\dfrac{\cancel{sin 13°}}{\cancel{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = 1^2 + 1^2 - 2 \times \Big(\dfrac{1}{\sqrt2}\Big)^2\\[1em] = 1 + 1 - 2 \times \Big(\dfrac{1}{2}\Big)\\[1em] = 2 - 1\\[1em] = 1

Hence, (sin 77°cos 13°)2+(cos 77°sin 13°)−2 cos245°=1\Big(\dfrac{\text{sin 77°}}{\text{cos 13°}}\Big)^2 + \Big(\dfrac{\text{cos 77°}}{\text{sin 13°}}\Big) - \text{2 cos}^2 45° = 1.

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