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

Verify the following equalities: cos 90° = 1 – 2sin2 45° = 2cos2 45° – 1 - Mathematics

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Question

Verify the following equalities:

cos 90° = 1 – 2sin2 45° = 2cos2 45° – 1

Sum

Solution

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

cos 90° = 0  ...(1)

1 – 2 sin2 45° = `1 - 2(1/sqrt(2))^2`

= `1 - 2 xx 1/2`

= 1 – 1 = 0 → (2)

2 cos2 45° – 1 = `2(1/sqrt(2))^2 - 1`

= `2/2 - 1`

= `(2 - 2)/2`

= 0 → (3)

From (1), (2) and (3) we get

cos 90° = 1 – 2 sin2 45° = 2 cos2 45° – 1

Hence it is proved.

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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 1. (iii) | Page 232
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