Question

Prove that sin²∅+cos²∅=1 ?​

Answer 1

Explanation:

Let ABC be a right angled triangle where there's a right angle in B . Let the measure of the angle BAC be ∅. So,

`\sin(∅) = (BC)/(AC)`

`= > {\sin}^(2)∅ = \frac{ {BC}^(2) }{ {AC}^(2) }`

Also,

`\cos(∅) = (AB)/(AC)`

`= > { \cos}^(2)∅ = \frac{ {AB}^(2) }{ {AC}^(2) }`

Now adding the values of sin^2∅& cos^2∅,

`{ \sin}^(2) ∅ + { \cos}^(2) ∅ = \frac{ {BC}^(2) }{ {AC}^(2) } + \frac{ {AB}^(2) }{ {AC}^(2) }`

`= > { \sin}^(2) ∅ + { \cos}^(2) ∅ = \frac{ {AB}^(2) + {BC}^(2) }{ {AC}^(2) }`

But we know that

`{AC}^(2) = {AB}^(2) + {BC}^(2)` by applying Pythagorean Theorem

So ,

`= > { \sin}^(2) ∅ + { \cos}^(2) ∅ = \frac{ {AC}^(2) }{ {AC}^(2) } = 1`