HSF.TF.C.8 math practice. Learn by doing.

HSF.TF.C.8 practice covers prove the Pythagorean identity sin²(θ) + cos²(θ) = 1 and use it to find sin(θ), cos(θ), or tan(θ) given one of them and the quadrant of the angle. Work through 8 free questions with answers and explanations, then continue in the related math games.

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8 free questions · 0/0 correct

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Which identity equals cos⁡(2θ)?\cos(2\theta)\textsf{?}

One form of the double-angle identity is cos⁡(2θ)=cos⁡2θ−sin⁡2θ.\cos(2\theta) = \cos^{2}\theta - \sin^{2}\theta\textsf{.}

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Which is the Pythagorean identity?

The fundamental Pythagorean identity is sin⁡2θ+cos⁡2θ=1.\sin^{2}\theta + \cos^{2}\theta = 1\textsf{.}

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If sin⁡θ=35\sin\theta = \dfrac{3}{5} and cos⁡θ=45,\cos\theta = \dfrac{4}{5}\textsf{,} what is tan⁡θ?\tan\theta\textsf{?}

tan⁡θ=sin⁡θcos⁡θ=3/54/5=34.\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{4/5} = \dfrac{3}{4}\textsf{.}

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If cos⁡θ=23,\cos\theta = \dfrac{2}{3}\textsf{,} what is sec⁡θ?\sec\theta\textsf{?}

sec⁡θ=1cos⁡θ=12/3=32.\sec\theta = \dfrac{1}{\cos\theta} = \dfrac{1}{2/3} = \dfrac{3}{2}\textsf{.}

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If sin⁡θ=513\sin\theta = \dfrac{5}{13} and cos⁡θ=1213,\cos\theta = \dfrac{12}{13}\textsf{,} what is sin⁡(2θ)?\sin(2\theta)\textsf{?}

sin⁡(2θ)=2sin⁡θcos⁡θ=2⋅513⋅1213=120169.\sin(2\theta) = 2\sin\theta\cos\theta = 2 \cdot \dfrac{5}{13} \cdot \dfrac{12}{13} = \dfrac{120}{169}\textsf{.}

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Which identity equals 1+tan⁡2θ?1 + \tan^{2}\theta\textsf{?}

The Pythagorean identity is 1+tan⁡2θ=sec⁡2θ.1 + \tan^{2}\theta = \sec^{2}\theta\textsf{.}

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If sin⁡θ=−513\sin\theta = -\dfrac{5}{13} and cos⁡θ=1213,\cos\theta = \dfrac{12}{13}\textsf{,} what is tan⁡θ?\tan\theta\textsf{?}

tan⁡θ=sin⁡θcos⁡θ=−5/1312/13=−512.\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{-5/13}{12/13} = -\dfrac{5}{12}\textsf{.}

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If cos⁡θ=−47,\cos\theta = -\dfrac{4}{7}\textsf{,} what is sec⁡θ?\sec\theta\textsf{?}

sec⁡θ=1cos⁡θ=1−4/7=−74.\sec\theta = \dfrac{1}{\cos\theta} = \dfrac{1}{-4/7} = -\dfrac{7}{4}\textsf{.}

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