मराठी

If Cos a = − 24 25 and Cos B = 3 5 , Where π < a < 3 π 2 and 3 π 2 < B < 2π, Find the Following: Cos (A + B) - Mathematics

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प्रश्न

If \[\cos A = - \frac{24}{25}\text{ and }\cos B = \frac{3}{5}\], where π < A < \[\frac{3\pi}{2}\text{ and }\frac{3\pi}{2}\]< B < 2π, find the following:
cos (A + B)

थोडक्यात उत्तर
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उत्तर

Given:
\[\cos A = - \frac{24}{25}\text{ and }\cos B = \frac{3}{5}\]
\[\text{ and }\pi < A < \frac{3\pi}{2}\text{ and }\frac{3\pi}{2} < B < 2\pi . \]
That is, A is in third quadrant and B is in fourth qudrant. 
We know that sine function is negative in third and fourth quadrants . 
Therefore, 
\[\sin A = - \sqrt{1 - \cos^2 A}\text{ and }\sin B = - \sqrt{1 - \cos^2 B}\]
\[ \Rightarrow \sin A = \sqrt{1 - \left( \frac{- 24}{25} \right)^2}\text{ and }\sin B = - \sqrt{1 - \left( \frac{3}{5} \right)^2}\]
\[ \Rightarrow \sin A = - \sqrt{1 - \frac{576}{625}}\text{ and }\sin B = - \sqrt{1 - \frac{9}{25}}\]
\[ \Rightarrow \sin A = - \sqrt{\frac{49}{625}}\text{ and }\sin B = - \sqrt{\frac{16}{25}}\]
\[ \Rightarrow \sin A = \frac{- 7}{25}\text{ and }\sin B = \frac{- 4}{5}\]
Now
\[ \cos\left( A + B \right) = \cos A \cos B - \sin A \sin B\]
\[ = \frac{- 24}{25} \times \frac{3}{5} - \frac{- 7}{25} \times \frac{- 4}{5}\]
\[ = \frac{- 72}{125} - \frac{28}{125}\]
\[ = \frac{- 100}{125}\]
\[ = \frac{- 4}{5}\]

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पाठ 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.1 [पृष्ठ १९]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 3.2 | पृष्ठ १९

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