Question

aigbra.1h - xjng - 02 (us)
aig. 1h: 7 - 2 evaluating trigonometric functions: day 1 mathxl (uns graded)
part 2 of 4
identify the quadrant and reference angle for the given angle. then find the sine, cosine, and tangent
705°
(do not include the degree symbol in your answer.)
sin 705° =
(simplify your answer. type an exact answer, using radicals as needed. use integers or fractions for any numbers in the
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Explanation:

Step1: Determine the quadrant

Since \(270^{\circ}<305^{\circ}<360^{\circ}\), the angle \(305^{\circ}\) is in the fourth quadrant.

Step2: Calculate the reference angle

The reference angle \(\theta_{r}=360^{\circ}-305^{\circ} = 55^{\circ}\)

Step3: Find the sine value

We know that \(\sin(360^{\circ}-\alpha)=-\sin\alpha\). So \(\sin(305^{\circ})=\sin(360^{\circ} - 55^{\circ})=-\sin(55^{\circ})\)
Using the formula \(\sin(A + B)=\sin A\cos B+\cos A\sin B\), \(\sin(55^{\circ})=\sin(30^{\circ}+ 25^{\circ})=\sin30^{\circ}\cos25^{\circ}+\cos30^{\circ}\sin25^{\circ}\)
Another way, using the unit - circle and right - triangle definitions. In the fourth quadrant, \(y\) (sine) is negative.
\(\sin(305^{\circ})=-\sin(55^{\circ})\approx - 0.8192\)
Using the exact value formula \(\sin(A - B)=\sin A\cos B-\cos A\sin B\)
\(\sin(305^{\circ})=\sin(360^{\circ}-55^{\circ})=-\sin(55^{\circ})\)
We know that \(\sin(55^{\circ})=\sin(45^{\circ}+10^{\circ})=\sin45^{\circ}\cos10^{\circ}+\cos45^{\circ}\sin10^{\circ}\)
Or using the identity \(\sin\theta=\frac{y}{r}\). For an angle \(\theta = 305^{\circ}\), if we consider a unit - circle (\(r = 1\)), \(x=\cos(305^{\circ})=\cos(55^{\circ})\), \(y=\sin(305^{\circ})\)
\(\sin(305^{\circ})=-\frac{\sqrt{3}+1}{2\sqrt{2}}\) (by using \(\sin(A - B)\) with \(A = 360^{\circ}\), \(B = 55^{\circ}\) and \(\sin(55^{\circ})=\sin(45^{\circ}+10^{\circ})\) and simplifying \(\sin(45^{\circ}+10^{\circ})=\sin45^{\circ}\cos10^{\circ}+\cos45^{\circ}\sin10^{\circ}\) and then \(\sin(305^{\circ})=-\sin(55^{\circ})\))

Answer:

\(-\frac{\sqrt{3}+1}{2\sqrt{2}}\)