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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 9

Find the value of the following: (sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°) - Mathematics

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Question

Find the value of the following:

(sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)

Sum

Solution

sin 90° = 1, cos 60° = `1/2`, cos 45° = `1/sqrt(2)`, sin 30° = `1/2`, cos 0° = 1

(sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)

= `(1 + 1/2 +  1/sqrt(2)) xx (1/1 + 1 - 1/sqrt(2))`

= `((2sqrt(2) + sqrt(2) + 2)/(2sqrt(2))) xx ((sqrt(2) + 2sqrt(2) - 2)/(2sqrt(2)))`

= `((3sqrt(2) + 2)/(2sqrt(2))) xx ((3sqrt(2) - 2)/(2sqrt(2)))`

= `((3sqrt(2))^2 - (2)^2)/8`

= `(9(2) - 4)/8`

= `(18 - 4)/8`

= `14/8`

= `7/4`

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Chapter 6: Trigonometry - Exercise 6.2 [Page 232]

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Samacheer Kalvi Mathematics [English] Class 9 TN Board
Chapter 6 Trigonometry
Exercise 6.2 | Q 2. (ii) | Page 232
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