How do you simplify ${(sin 60)}^{2} + {(cos 60)}^{2}$ $(sin 60)^2 + (cos 60)^2$ ?

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How do you simplify ${(sin 60)}^{2} + {(cos 60)}^{2}$ $(sin 60)^2 + (cos 60)^2$ ?

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Answer 1 · 2015-05-25T11:00:12

${({sin 60}^{o})}^{2} + {({cos 60}^{o})}^{2} = 1$ $(sin 60^o)^2 + (cos 60^o)^2 = 1$

In fact, for any angle $θ$ $theta$ we have ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta + cos^2theta = 1$

We can see this geometrically for positive angles less than $90^{o}$ $90^o$ (e.g. $60^{o}$ $60^o$ ) by considering a right angled triangle with angles $θ$ $theta$ , $90^{o} - θ$ $90^o - theta$ and $90^{o}$ $90^o$ and hypotenuse of length $1$ $1$ .

By the definitions of $sin θ$ $sin theta$ and $cos θ$ $cos theta$ , the side opposite the angle $θ$ $theta$ will be of length $sin θ$ $sin theta$ and the side adjacent to the angle $θ$ $theta$ , opposite the $90^{o} - θ$ $90^o - theta$ angle will be of length $cos θ$ $cos theta$ .

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How do you simplify ${(sin 60)}^{2} + {(cos 60)}^{2}$ $(sin 60)^2 + (cos 60)^2$ ?

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