Respuesta :

egoodrich1337
So this first wants you to find where sin is √3/2 when θ is between π and 3π/2. θ would therefore be located at 2π/3.

Now plug in the value of θ for cosine:

cos (2π/3) = -1/2

And tangent:

tan (2π/3) = -√3/3

Answer with explanation:

Let, A=Theta

→Used the identity

Sin²A+Cos²A=1

→Sine and cosine are negative in third Quadrant.

[tex]\sin A= \frac{-\sqrt{3}}{2}\\\\\pi < A<\frac{3\pi}{2}\\\\\sin^2A+\cos^2A=1\\\\[ \frac{-\sqrt{3}}{2}]^2+\cos^2A=1\\\\\cos^2A=1-\frac{3}{4}\\\\\cos^2A=\frac{1}{4}\\\\\cos A=\pm\ frac{1}{2}\\\\\text{As Angle lies in third Quadrant}\\\\ \cos A=\frac{-1}{2}\\\\\tanA=\frac{\sinA}{\cosA}\\\\=\frac{\frac{-\sqrt{3}}{2}}{\frac{-1}{2}}\\\\=\sqrt{3}[/tex]

Otras preguntas