$\sin^{2} 45^{\circ} + \cos^{2} 45^{\circ}$ equals:
A$0$
B$1$
C$\dfrac{1}{2}$
D$2$
Answer & Solution
Correct answer: B. $1$
Pythagorean identity: $\sin^{2}\theta + \cos^{2}\theta = 1$ for any $\theta$.
Direct check: $(1/\sqrt{2})^{2} + (1/\sqrt{2})^{2} = 1/2 + 1/2 = 1$ ✓.
Related questions
Penalty for wrong answers : THERE WILL BE PENALTY FOR WRONG ANSWERS MARKED BY A CANDIDATE Penalty for wrong answers : THERE WILL BE PENALTY FOR WRONG ANSWERS MARKED BY A CANDIDATE For [[1,2],[2,4]], the determinant equals:If A is invertible, |A^{−1}| equals:For a 3 × 3 non-singular A with |A| = 5, the value of |adj A| is:For a non-singular square matrix A of order n, |adj A| equals:For non-singular A, B of the same order, (AB)^{−1} equals:For a non-singular A, (A^{−1})^{−1} equals: