Question

Find sinθ.tan θ = -(√7)/2, sec θ \u003e 0

Accepted Answer

Identify the given information: $$ 	an 	heta = -\frac{\sqrt{7}}{2} $$ and $$ \sec 	heta \u003e 0 . $$Recall that $$ \sec 	heta = \frac{1}{\cos 	heta} $$, so $$ \sec 	heta \u003e 0 $$ means $$ \cos 	heta \u003e 0 . $$Determine the quadrant where $$ 	heta $$ lies. Since $$ 	an 	heta  is $$negative and $$ \cos 	heta  is $$positive, $$ 	heta $$ must be in the fourth quadrant (where cosine is positive and tangent is negative). Use the identity relating tangent and sine and cosine: $$ 	an 	heta = \frac{\sin 	heta}{\cos 	heta} . $$Let $$ \cos 	heta = x $$, then $$ \sin 	heta = 	an 	heta 	imes x = -\frac{\sqrt{7}}{2} x . $$Apply the Pythagorean identity: $$ \sin^2 	heta + \cos^2 	heta = 1 . $$Substitute $$ \sin 	heta = -\frac{\sqrt{7}}{2} x $$ and $$ \cos 	heta = x  to $$get $$ \left(-\frac{\sqrt{7}}{2} xight)^2 + x^2 = 1 . $$Solve the equation for $$ x  ($$which is $$ \cos 	heta $$), then use $$ \sin 	heta = -\frac{\sqrt{7}}{2} x  to $$find $$ \sin 	heta . $$Remember to choose the sign of $$ \sin 	heta $$ consistent with the quadrant (fourth quadrant means $$ \sin 	heta \u003c 0 ).$$

Find sinθ.
tan θ = -(√7)/2, sec θ > 0

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Key Concepts

Trigonometric Ratios and Their Relationships

Sign of Trigonometric Functions in Different Quadrants

Using Pythagorean Identities to Find Missing Ratios

Find sinθ.tan θ = -(√7)/2, sec θ > 0