Question

cos(90 - \theta) =
cos(31^{circ}) = sin()
cos(33^{circ}) = sin()
sin(28^{circ}) = cos()
90 - \theta
cos(\theta)
57^{circ}
59^{circ}
62^{circ}
sin(\theta)
90 + 33
90 + \theta

Explanation:

Step1: Use co - function identity

The co - function identity states that \(\cos\theta=\sin(90^{\circ}-\theta)\) and \(\sin\theta = \cos(90^{\circ}-\theta)\)

Step2: Solve \(\cos(90 - \theta)\)

By the co - function identity \(\cos(90-\theta)=\sin\theta\)

Step3: Solve \(\cos(31^{\circ})\)

Since \(\cos\alpha=\sin(90^{\circ}-\alpha)\), when \(\alpha = 31^{\circ}\), then \(\cos(31^{\circ})=\sin(90^{\circ}-31^{\circ})=\sin(59^{\circ})\)

Step4: Solve \(\cos(33^{\circ})\)

Using the co - function identity \(\cos\beta=\sin(90^{\circ}-\beta)\), for \(\beta = 33^{\circ}\), we have \(\cos(33^{\circ})=\sin(90^{\circ}-33^{\circ})=\sin(57^{\circ})\)

Step5: Solve \(\sin(28^{\circ})\)

Since \(\sin\gamma=\cos(90^{\circ}-\gamma)\), when \(\gamma=28^{\circ}\), then \(\sin(28^{\circ})=\cos(90^{\circ}-28^{\circ})=\cos(62^{\circ})\)

Answer:

\(\cos(90 - \theta)=\sin\theta\), \(\cos(31^{\circ})=\sin(59^{\circ})\), \(\cos(33^{\circ})=\sin(57^{\circ})\), \(\sin(28^{\circ})=\cos(62^{\circ})\)