How do you find $sin θ$ $sin theta$ if $cos θ = \frac{3}{5}$ $cos theta = 3/5$ and theta terminates in Q 1?

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Answer 1 · 2015-06-09T04:09:41

cos x = 3/5 --> sin x = 4/5

${sin}^{2} x = 1 - {cos}^{2} x = 1 - \frac{9}{25} = \frac{16}{25}$ $sin^2 x = 1 - cos^2 x = 1 - 9/25 = 16/25$ ->

--> $sin x = \pm \frac{4}{5} .$ $sin x = +- 4/5.$

Since x is in Quadrant I, the answer 4/5 is chosen.

How do you find sinθsin theta if cosθ=35cos theta = 3/5 and theta terminates in Q 1?