# How would calculate (sin 150^o)^2+(cos30^o)^2?

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You should notice that `150^o = 180^o - 30^o` , hence, you may evaluate `sin 150^o` , such that:

`sin 150^o = sin(180^o - 30^o) => sin 150^o = cos 30^o`

Replacing `cos 30^o` for `sin 150^o` yields:

`(sin 150^o)^2+(cos30^o)^2 = cos^2 30^o + cos^2 3o^o`

`(sin 150^o)^2+(cos30^o)^2 = 2cos^2 30^o`

Since `cos 30^o = sqrt3/2,` yields:

`(sin 150^o)^2+(cos30^o)^2 = 2 *(sqrt3/2)^2`

`(sin 150^o)^2+(cos30^o)^2 = 2 * (3/4)`

Reducing duplicate factors, yields:

`(sin 150^o)^2+(cos30^o)^2 = 3/2`

**Hence, evaluating the given summation, yields **`(sin 150^o)^2+(cos30^o)^2 = 3/2.`

**Sources:**

`sin(150^0)=sin(180^0-30^0)`

`=sin(30^0)`

`sin^2(150^0)+cos^(30^0)=sin^2(30^0)+cos^2(30^0)`

`=1`

Thus

`sin^2(150^0)+cos^2(30^0)=1`