use an identity to write the expression as a single trigonometric function value or as a single number.\n1/3…

use an identity to write the expression as a single trigonometric function value or as a single number.\n1/3 sin²25.7° - 1/6\n1/3 sin²25.7° - 1/6 = \n(simplify your answer. type an exact answer, using radicals as needed. do not include the degree symbol in your answer.)

use an identity to write the expression as a single trigonometric function value or as a single number.\n1/3 sin²25.7° - 1/6\n1/3 sin²25.7° - 1/6 = \n(simplify your answer. type an exact answer, using radicals as needed. do not include the degree symbol in your answer.)

Answer

Explanation:

Step1: Use the double - angle identity

The double - angle identity for cosine is (\cos(2\theta)=1 - 2\sin^{2}\theta), which can be rewritten as (\sin^{2}\theta=\frac{1-\cos(2\theta)}{2}). For the expression (\frac{1}{3}-\frac{1}{6}\sin^{2}(25.7^{\circ})), substitute (\sin^{2}(25.7^{\circ})=\frac{1 - \cos(51.4^{\circ})}{2}). [ \begin{align*} \frac{1}{3}-\frac{1}{6}\sin^{2}(25.7^{\circ})&=\frac{1}{3}-\frac{1}{6}\times\frac{1 - \cos(51.4^{\circ})}{2}\ &=\frac{1}{3}-\frac{1}{12}+\frac{\cos(51.4^{\circ})}{12}\ &=\frac{4 - 1+\cos(51.4^{\circ})}{12}\ &=\frac{3+\cos(51.4^{\circ})}{12} \end{align*} ] Another way: We know that (\cos(2\theta)=1 - 2\sin^{2}\theta\Rightarrow\frac{1}{2}-\sin^{2}\theta=\frac{\cos(2\theta)}{2}) Multiply the given expression (\frac{1}{3}-\frac{1}{6}\sin^{2}(25.7^{\circ})) by (6) to get (2-\sin^{2}(25.7^{\circ})) [ \begin{align*} \frac{1}{3}-\frac{1}{6}\sin^{2}(25.7^{\circ})&=\frac{1}{6}(2-\sin^{2}(25.7^{\circ}))\ \end{align*} ] Using the identity (\cos(2\theta) = 1-2\sin^{2}\theta\Rightarrow\sin^{2}\theta=\frac{1 - \cos(2\theta)}{2}) [ \begin{align*} \frac{1}{3}-\frac{1}{6}\sin^{2}(25.7^{\circ})&=\frac{1}{6}\left(2-\frac{1-\cos(51.4^{\circ})}{2}\right)\ &=\frac{1}{6}\left(\frac{4-(1 - \cos(51.4^{\circ}))}{2}\right)\ &=\frac{1}{12}(3+\cos(51.4^{\circ})) \end{align*} ] Or use the identity (\frac{1}{2}-\sin^{2}\alpha=\frac{\cos(2\alpha)}{2}) [ \begin{align*} \frac{1}{3}-\frac{1}{6}\sin^{2}(25.7^{\circ})&=\frac{1}{6}(2-\sin^{2}(25.7^{\circ}))\ &=\frac{1}{6}\left(\frac{1}{2}+\frac{3}{2}-\sin^{2}(25.7^{\circ})\right)\ &=\frac{1}{6}\left(\frac{1}{2}+\frac{3 - 2\sin^{2}(25.7^{\circ})}{2}\right)\ \end{align*} ] Since (3-2\sin^{2}\alpha=2 + 1-2\sin^{2}\alpha=2+\cos(2\alpha)) [ \begin{align*} \frac{1}{3}-\frac{1}{6}\sin^{2}(25.7^{\circ})&=\frac{1}{12}+\frac{1}{12}\cos(51.4^{\circ}) \end{align*} ]

Step2: Calculate the numerical value

We know that (\cos(51.4^{\circ})\approx0.6249) [ \begin{align*} \frac{1}{3}-\frac{1}{6}\sin^{2}(25.7^{\circ})&=\frac{1}{12}(3 + 0.6249)\ &=\frac{3.6249}{12}\ &= 0.3021 \end{align*} ]

Answer:

(0.3021)