Question

If cosine theta = -2/5 and tangent theta > 0, what is the value of sine theta?

Answer 1

Answer:

It is given that

cosθ = -2/5

We know that if tanθ > 0 cosθ < 0 then sinθ < 0 when θ is in third quadrant

We can use the formula

sinθ = -√(1 - cos2θ)

Substituting the values

sinθ = -√(1 - (-2/5)2)

sinθ = -√(1 - (4/25))

Taking LCM

sinθ = -√[25 - 4]/25

sinθ = -√(21/25)

sinθ = -√21/5

Therefore, the value of sinθ is -√21/5.

Answer 2

$\text{Given that,}\\\\\cos \theta = -\dfrac 25 \\\\\implies \cos^2 \theta = \dfrac 4{25}\\\\\\\implies 1- \sin^2 \theta = \dfrac{4}{25}\\\\\\\implies \sin^2 \theta = 1-\dfrac{4}{25}\\\\\\\implies \sin^2 \theta =\dfrac{21}{25}\\\\\\\implies \sin \theta = \pm \sqrt{\dfrac {21}{25}} = \pm \dfrac{\sqrt{21}} 5\\\\\text{Since}~ \tan \theta>0,~ ~\text{the value of} ~\sin \theta ~\text{wii be:} \\\\\sin \theta = -\dfrac{\sqrt{21}}5$