determine cos theta to three decimal places where sin theta=6/square root 61 and theta is an angle in the second quadrant
Sin θ=661Sin2 θ+Cos2 θ=1Cos2 θ=1−Sin2 θCos θ=1−Sin2 θCos θ=1−Sin2 θCos θ=1−(661)2Cos θ=1−3661Cos θ=61−3661=2561Cos θ=±561Sin\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}}Sin θ=616Sin2 θ+Cos2 θ=1Cos2 θ=1−Sin2 θCos θ=1−Sin2 θCos θ=1−Sin2 θCos θ=1−(616)2Cos θ=1−6136Cos θ=6161−36=6125Cos θ=±615
second quadrant Cos θCos\space θCos θ is negative
=−57.810Cos θ=−0.640=-\frac{5}{7.810}\\ Cos\space θ =-0.640=−7.8105Cos θ=−0.640
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