KCET · Maths · Determinants
The value of \( \cos ^{2} 45^{\circ}-\sin ^{2} 15^{\circ} \) is
- A \( \frac{\sqrt{3}}{2} \)
- B \( \frac{\sqrt{3}}{4} \)
- C \( \frac{\sqrt{3}+1}{2 \sqrt{2}} \)
- D \( \frac{\sqrt{3}-1}{2 \sqrt{2}} \)
Answer & Solution
Correct Answer
(B) \( \frac{\sqrt{3}}{4} \)
Step-by-step Solution
Detailed explanation
Given that, \(\cos ^{2} 45^{\circ}-\sin ^{2} 45^{\circ}\)
We know that,
\(\cos (A+B) \cdot \cos (A-B)=\cos ^{2} A-\sin ^{2} B\)
So,
\(\cos ^{2} 45^{\circ}-\sin ^{2} 15^{\circ}=\cos (45+15) \cdot \cos (45-15)\)
\(\cos 60^{\circ} \cdot \cos 30^{\circ}=\frac{1}{2} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}\)
We know that,
\(\cos (A+B) \cdot \cos (A-B)=\cos ^{2} A-\sin ^{2} B\)
So,
\(\cos ^{2} 45^{\circ}-\sin ^{2} 15^{\circ}=\cos (45+15) \cdot \cos (45-15)\)
\(\cos 60^{\circ} \cdot \cos 30^{\circ}=\frac{1}{2} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}\)
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