If \( \sin 27^{\circ}=q \), express each of the following in tems of \( q \) \( \begin{array}{ll}\text { a) } \sin 117^{\circ} & \text { b) } \cos \left(-27^{\circ}\right)\end{array} \)

To express the given trigonometric functions in terms of \( q = \sin 27^{\circ} \), we'll use fundamental trigonometric identities. ### a) Expressing \( \sin 117^{\circ} \) in terms of \( q \) First, recognize that: \[ 117^{\circ} = 90^{\circ} + 27^{\circ} \] Using the **sine addition formula**: \[ \sin(90^{\circ} + \theta) = \cos \theta \] Letting \( \theta = 27^{\circ} \): \[ \sin 117^{\circ} = \cos 27^{\circ} \] Now, using the **Pythagorean identity**: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Solving for \( \cos \theta \): \[ \cos \theta = \sqrt{1 - \sin^2 \theta} \] Since \( \sin 27^{\circ} = q \) and \( 27^{\circ} \) is in the first quadrant (where cosine is positive): \[ \cos 27^{\circ} = \sqrt{1 - q^2} \] Therefore: \[ \sin 117^{\circ} = \sqrt{1 - q^2} \] ### b) Expressing \( \cos(-27^{\circ}) \) in terms of \( q \) The **cosine function** is even, meaning: \[ \cos(-\theta) = \cos \theta \] Thus: \[ \cos(-27^{\circ}) = \cos 27^{\circ} \] Using the same logic as in part (a): \[ \cos 27^{\circ} = \sqrt{1 - q^2} \] Therefore: \[ \cos(-27^{\circ}) = \sqrt{1 - q^2} \] ### Summary Both expressions in terms of \( q \) are: \[ \boxed{ \begin{aligned} \text{a) } \sin 117^{\circ} &= \sqrt{1 - q^2} \\ \text{b) } \cos(-27^{\circ}) &= \sqrt{1 - q^2} \end{aligned} } \]