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

# 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

#### 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?

#### APPEARS IN

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