Sum

Verify the following equalities:

cos 90° = 1 – 2sin^{2} 45° = 2cos^{2} 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 sin^{2} 45° = `1 - 2(1/sqrt(2))^2`

= `1 - 2 xx 1/2`

= 1 – 1 = 0 → (2)

2 cos^{2} 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 sin^{2} 45° = 2 cos^{2} 45° – 1

Hence it is proved.

Is there an error in this question or solution?

Advertisement Remove all ads

#### APPEARS IN

Advertisement Remove all ads

Advertisement Remove all ads