Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 9th
Advertisement Remove all ads

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

Sum

Verify the following equalities:

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

Advertisement Remove all ads

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.

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Samacheer Kalvi Mathematics Class 9th Tamil Nadu State Board
Chapter 6 Trigonometry
Exercise 6.2 | Q 1. (iii) | Page 232
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×