Which identity equals cos ( 2 θ ) \cos(2\theta) cos ( 2 θ ) ?
cos 2 θ + sin 2 θ \cos^2\theta + \sin^2\theta cos 2 θ + sin 2 θ cos 2 θ − sin 2 θ \cos^2\theta - \sin^2\theta cos 2 θ − sin 2 θ sin 2 θ − cos 2 θ \sin^2\theta - \cos^2\theta sin 2 θ − cos 2 θ 2 sin θ cos θ 2\sin\theta\cos\theta 2 sin θ cos θ One form of the double-angle identity is cos ( 2 θ ) = cos 2 θ − sin 2 θ \cos(2\theta) = \cos^2\theta - \sin^2\theta cos ( 2 θ ) = cos 2 θ − sin 2 θ .
Show the answer Which is the Pythagorean identity?
sin 2 θ + cos 2 θ = 1 \sin^2\theta + \cos^2\theta = 1 sin 2 θ + cos 2 θ = 1 tan 2 θ − sec 2 θ = 1 \tan^2\theta - \sec^2\theta = 1 tan 2 θ − sec 2 θ = 1 sin 2 θ − cos 2 θ = 1 \sin^2\theta - \cos^2\theta = 1 sin 2 θ − cos 2 θ = 1 sin θ + cos θ = 1 \sin\theta + \cos\theta = 1 sin θ + cos θ = 1 The fundamental Pythagorean identity is sin 2 θ + cos 2 θ = 1 \sin^2\theta + \cos^2\theta = 1 sin 2 θ + cos 2 θ = 1 .
Show the answer If sin θ = 3 5 \sin\theta = \dfrac{3}{5} sin θ = 5 3 and cos θ = 4 5 \cos\theta = \dfrac{4}{5} cos θ = 5 4 , what is tan θ \tan\theta tan θ ?
3 5 \dfrac{3}{5} 5 3 12 25 \dfrac{12}{25} 25 12 3 4 \dfrac{3}{4} 4 3 4 3 \dfrac{4}{3} 3 4 tan θ = sin θ cos θ = 3 / 5 4 / 5 = 3 4 \tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{4/5} = \dfrac{3}{4} tan θ = cos θ sin θ = 4/5 3/5 = 4 3 .
Show the answer If cos θ = 2 3 \cos\theta = \dfrac{2}{3} cos θ = 3 2 , what is sec θ \sec\theta sec θ ?
2 3 \dfrac{2}{3} 3 2 3 2 \dfrac{3}{2} 2 3 1 3 \dfrac{1}{3} 3 1 5 2 \dfrac{5}{2} 2 5 sec θ = 1 cos θ = 1 2 / 3 = 3 2 \sec\theta = \dfrac{1}{\cos\theta} = \dfrac{1}{2/3} = \dfrac{3}{2} sec θ = cos θ 1 = 2/3 1 = 2 3 .
Show the answer If sin θ = 5 13 \sin\theta = \dfrac{5}{13} sin θ = 13 5 and cos θ = 12 13 \cos\theta = \dfrac{12}{13} cos θ = 13 12 , what is sin ( 2 θ ) \sin(2\theta) sin ( 2 θ ) ?
10 13 \dfrac{10}{13} 13 10 119 169 \dfrac{119}{169} 169 119 60 169 \dfrac{60}{169} 169 60 120 169 \dfrac{120}{169} 169 120 sin ( 2 θ ) = 2 sin θ cos θ = 2 ⋅ 5 13 ⋅ 12 13 = 120 169 \sin(2\theta) = 2\sin\theta\cos\theta = 2 \cdot \dfrac{5}{13} \cdot \dfrac{12}{13} = \dfrac{120}{169} sin ( 2 θ ) = 2 sin θ cos θ = 2 ⋅ 13 5 ⋅ 13 12 = 169 120 .
Show the answer Which identity equals 1 + tan 2 θ 1 + \tan^2\theta 1 + tan 2 θ ?
sec 2 θ \sec^2\theta sec 2 θ cot 2 θ \cot^2\theta cot 2 θ cos 2 θ \cos^2\theta cos 2 θ csc 2 θ \csc^2\theta csc 2 θ The Pythagorean identity 1 + tan 2 θ = sec 2 θ 1 + \tan^2\theta = \sec^2\theta 1 + tan 2 θ = sec 2 θ .
Show the answer If sin θ = − 5 13 \sin\theta = -\dfrac{5}{13} sin θ = − 13 5 and cos θ = 12 13 \cos\theta = \dfrac{12}{13} cos θ = 13 12 , what is tan θ \tan\theta tan θ ?
− 5 12 \dfrac{-5}{12} 12 − 5 5 12 \dfrac{5}{12} 12 5 − 12 5 -\dfrac{12}{5} − 5 12 − 60 169 -\dfrac{60}{169} − 169 60 tan θ = sin θ cos θ = − 5 / 13 12 / 13 = − 5 12 \tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{-5/13}{12/13} = -\dfrac{5}{12} tan θ = cos θ sin θ = 12/13 − 5/13 = − 12 5 .
Show the answer If cos θ = − 4 7 \cos\theta = -\dfrac{4}{7} cos θ = − 7 4 , what is sec θ \sec\theta sec θ ?
7 4 \dfrac{7}{4} 4 7 − 7 4 \dfrac{-7}{4} 4 − 7 − 4 7 -\dfrac{4}{7} − 7 4 − 5 4 \dfrac{-5}{4} 4 − 5 sec θ = 1 cos θ = 1 − 4 / 7 = − 7 4 \sec\theta = \dfrac{1}{\cos\theta} = \dfrac{1}{-4/7} = -\dfrac{7}{4} sec θ = cos θ 1 = − 4/7 1 = − 4 7 .
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