Question

How do you verify `(\sin \frac { t} { 6} + \cos \frac { t } { 6} ) ^ { 2} = 1+ \sin \frac { t} { 3}`?

Answer 1

Please see below.

Explanation:

As `sin^2A+cos^2A=1` and `2sinAcosA=sin2A`

we have

`(sin(t/6)+cos(t/6))^2`

= `sin^2(t/6)+cos^2(t/6)+2sin(t/6)cos(t/6)`

= `1+sin(2xxt/6)`

= `1+sin(t/3)`