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Show that Sin 100° − Sin 10° is Positive. - Mathematics

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प्रश्न

Show that sin 100° − sin 10° is positive. 

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उत्तर

\[\text{ Let } f\left( \theta \right) = \sin100° -  \sin10°\] 

\[\text{ Multiplying and dividing by } \sqrt{1^2 + 1^2}, i . e . \text{ by } \sqrt{2}, \text{ we get }: \]

\[ \sqrt{2}\left( \frac{1}{\sqrt{2}}\sin100° - \frac{1}{\sqrt{2}}\sin10°\right)\]

\[ = \sqrt{2}\left( \cos45°\sin(90°+ 10°) - \sin45°\sin10°\right)\]

\[ = \sqrt{2}\left( \cos45°\cos10° - \sin45°\sin10° \right)\]

\[ = \sqrt{2}\cos(45°+ 10°) = \sqrt{2}co s55° , \text{ which is positive since \cos is positive in the first quadrant } .\]

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अध्याय 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.2 [पृष्ठ २६]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.2 | Q 3 | पृष्ठ २६

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