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\( \left. \begin{array} { l } { \frac { 3 ) \cos ^ { 4 } 90 ^ { \circ } - \sin ^ { 4 } ( - 10 ^ { \circ } ) } { \cos 55 ^ { \circ } \cos 325 ^ { \circ } } } \\ { \frac { \cos ( 40 ^ { \circ } - x ) \cos x - \sin ( 40 ^ { \circ } - x ) \sin x } { \sin 2 \cos ^ { \circ } \cos 155 ^ { \circ } } = \pi } \end{array} \right. \)

Ask by Gough Zimmerman. in South Africa
Mar 15,2025

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Answer

The first expression evaluates to approximately \(-445.080726\), and the second equation has no real solutions.

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Extra Insights

Firstly, let’s unravel the mysteries behind the angles mentioned! The angle \( 90^\circ \) for cosine gives us \( 0 \) since \( \cos 90^\circ = 0 \). Therefore, the first part of the expression simplifies dramatically! As for the sine, note that \( \sin(-10^\circ) = -\sin(10^\circ) \), so you’ll be squaring it. Feel free to mix in all those trigonometric identities for maximum flavor! Now, when \( \cos(40^\circ - x) \cos(x) - \sin(40^\circ - x) \sin(x) \) is present, that’s just the cosine of a sum formula sneaking into the mix, turning into \( \cos(40^\circ) \), which makes solving for \( x \) a fun puzzle. Who doesn’t love a good trigonometric treasure hunt? Just make sure to keep track of your identities and signs!

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