4. Calculati: \( \begin{array}{ll}\text { a) } \sin ^{2} 45^{\circ}+\cos ^{2} 135^{\circ} ; & \text { b) } \cos ^{2} 130^{\circ}+\sin ^{2} 50^{\circ} \\ \text { c) } \sin ^{2} 50^{\circ}+\sin ^{2} 40^{\circ}+2 \cos ^{2} 120^{\circ} ; & \text { d) } \cos 14^{\circ}+\cos 166^{\circ}\end{array} \)

Let's calculate each part step by step. ### a) \( \sin^2 45^\circ + \cos^2 135^\circ \) First, we find the values of \( \sin 45^\circ \) and \( \cos 135^\circ \): - \( \sin 45^\circ = \frac{\sqrt{2}}{2} \) - \( \cos 135^\circ = -\sin 45^\circ = -\frac{\sqrt{2}}{2} \) Now we calculate: \[ \sin^2 45^\circ = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2} \] \[ \cos^2 135^\circ = \left(-\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2} \] Adding these together: \[ \sin^2 45^\circ + \cos^2 135^\circ = \frac{1}{2} + \frac{1}{2} = 1 \] ### b) \( \cos^2 130^\circ + \sin^2 50^\circ \) Next, we find the values of \( \cos 130^\circ \) and \( \sin 50^\circ \): - \( \cos 130^\circ = -\cos 50^\circ \) - \( \sin 50^\circ = \sin 50^\circ \) Using the identity \( \sin^2 x + \cos^2 x = 1 \): \[ \cos^2 130^\circ = \sin^2 50^\circ \] Thus: \[ \cos^2 130^\circ + \sin^2 50^\circ = \sin^2 50^\circ + \sin^2 50^\circ = 1 \] ### c) \( \sin^2 50^\circ + \sin^2 40^\circ + 2 \cos^2 120^\circ \) First, we find \( \cos 120^\circ \): - \( \cos 120^\circ = -\frac{1}{2} \) Now calculate \( \cos^2 120^\circ \): \[ \cos^2 120^\circ = \left(-\frac{1}{2}\right)^2 = \frac{1}{4} \] Now we can substitute this into the expression: \[ 2 \cos^2 120^\circ = 2 \cdot \frac{1}{4} = \frac{1}{2} \] Now we need to find \( \sin^2 50^\circ + \sin^2 40^\circ \). Using the identity \( \sin^2 x + \cos^2 x = 1 \): \[ \sin^2 50^\circ + \sin^2 40^\circ = 1 - \cos^2 50^\circ + 1 - \cos^2 40^\circ \] However, we can also directly calculate: \[ \sin^2 50^\circ + \sin^2 40^\circ + 2 \cos^2 120^\circ = \sin^2 50^\circ + \sin^2 40^\circ + \frac{1}{2} \] Using the identity \( \sin^2 x + \cos^2 x = 1 \): \[ \sin^2 50^\circ + \sin^2 40^\circ + \frac{1}{2} = 1 \] ### d) \( \cos 14^\circ + \cos 166^\circ \) Using the cosine angle addition formula: \[ \cos 166^\circ = -\cos 14^\circ \] Thus: \[ \cos 14^\circ + \cos 166^\circ = \cos 14^\circ - \cos 14^\circ = 0 \] ### Summary of Results - a) \( 1 \) - b) \( 1 \) - c) \( 1 \) - d) \( 0 \)