Question

Prove that sin²∅+cos²∅=1 ?​

Answer 1

Step-by-step 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,

$\sin(∅) = \frac{BC}{AC}$

$= > {\sin}^{2}∅ = \frac{ {BC}^{2} }{ {AC}^{2} }$

Also,

$\cos(∅) = \frac{AB}{AC}$

$= > { \cos}^{2}∅ = \frac{ {AB}^{2} }{ {AC}^{2} }$

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 ,

$= > { \sin}^{2} ∅ + { \cos}^{2} ∅ = \frac{ {AC}^{2} }{ {AC}^{2} } = 1$