Answer to Question #247259 in Trigonometry for kai

Question

determine cos theta to three decimal places where sin theta=6/square root 61 and theta is an angle in the second quadrant

Answer 1

$Sin\space θ =\frac{6}{\sqrt{61}}\\Sin^2\space θ + Cos^2\space θ =1\\Cos^2\space θ =1-Sin^2\space θ \\Cos\space θ =\sqrt{1-Sin^2\space θ }\\\\Cos\space θ =\sqrt{1-Sin^2 \space θ }\\ Cos\space θ =\sqrt{1-(\frac{6}{\sqrt{61}})^2}\\ Cos\space θ =\sqrt{1-\frac{36}{61}}\\ Cos\space θ =\sqrt{\frac{61-36}{61}}=\frac{\sqrt{25}}{\sqrt{61}}\\ Cos\space θ =\pm \frac{5}{\sqrt{61}}$

second quadrant $Cos\space θ$ is negative

$=-\frac{5}{7.810}\\ Cos\space θ =-0.640$