Question

sin theta barabar root 1 minus cos square

Answer 1

First finding the value of sin²θ + cos²θ : -

sin²θ + cos²θ

⇒ $\dfrac{\mathbf{height^{2}}}{\mathbf{hypotenuse^{2}}} + \dfrac{\mathbf{base^{2}}}{\mathbf{hypotenuse^{2}}}$

⇒ $\dfrac{\mathbf{height^2 + base^2}}{\mathbf{hypotenuse^2}}$

From pythagoras theorem, height^2 + base^2 = hypotenuse^2.Now, substituting the value of height^2 + base^2.

⇒ $\dfrac{hypotenuse^2}{hypotenuse^2}$

⇒ 1

∴ sin²θ + cos²θ = 1

Adding - cos²θ on both sides,

⇒ sin²θ + cos²θ - cos²θ = 1 - cos²θ

⇒ sin²θ = 1 - cos²θ

 

Square root on both sides

⇒ √sin² = √1 - cos²θ

∴ sinθ = $\sqrt{1-cos \theta}$

Hence, proved that sinθ = √1 - cos²θ