Apply Pythagorean identities Pythagorean identities express the fundamental relationships among real-valued trigonometric functions: sin 2 θ + cos 2 θ = 1 \sin^2\theta+\cos^2\theta=1 sin 2 θ + cos 2 θ = 1 , 1 + tan 2 θ = sec 2 θ 1+\tan^2\theta=\sec^2\theta 1 + tan 2 θ = sec 2 θ , and 1 + cot 2 θ = csc 2 θ 1+\cot^2\theta=\csc^2\theta 1 + cot 2 θ = csc 2 θ . The learner applies these identities to rewrite expressions, find missing trigonometric values, establish equivalent forms, and simplify equations while respecting signs, quadrants, and the domains where tan \tan tan , cot \cot cot , sec \sec sec , or csc \csc csc are defined; complex-valued, hyperbolic, and more advanced generalized identities are outside this scope.
Detailed Explanation: Apply Pythagorean identities Pythagorean identities help you replace one trigonometric expression with an equivalent one:
sin 2 θ + cos 2 θ = 1 \sin^2\theta+\cos^2\theta=1 sin 2 θ + cos 2 θ = 1 1 + tan 2 θ = sec 2 θ 1+\tan^2\theta=\sec^2\theta 1 + tan 2 θ = sec 2 θ 1 + cot 2 θ = csc 2 θ 1+\cot^2\theta=\csc^2\theta 1 + cot 2 θ = csc 2 θ Choose the identity that contains the function you know. Then use the quadrant to select the correct sign.
Worked example
Suppose θ \theta θ
tan θ = − 3 4 . \tan\theta=-\frac{3}{4}. tan θ = − 4 3 . Find sin θ \sin\theta sin θ cos θ \cos\theta cos θ
Because tangent is given, use
1 + tan 2 θ = sec 2 θ . 1+\tan^2\theta=\sec^2\theta. 1 + tan 2 θ = sec 2 θ . Substitute tan θ = − 3 4 \tan\theta=-\frac34 tan θ = − 4 3
1 + ( − 3 4 ) 2 = sec 2 θ 1+\left(-\frac34\right)^2=\sec^2\theta 1 + ( − 4 3 ) 2 = sec 2 θ 1 + 9 16 = sec 2 θ 1+\frac{9}{16}=\sec^2\theta 1 + 16 9 = sec 2 θ sec 2 θ = 25 16 . \sec^2\theta=\frac{25}{16}. sec 2 θ = 16 25 . Taking square roots gives
sec θ = ± 5 4 . \sec\theta=\pm\frac54. sec θ = ± 4 5 . In Quadrant II, cosine is negative. Since sec θ = 1 cos θ \sec\theta=\frac{1}{\cos\theta} sec θ = c o s θ 1
sec θ = − 5 4 . \sec\theta=-\frac54. sec θ = − 4 5 . Therefore,
cos θ = 1 sec θ = 1 − 5 4 = − 4 5 . \cos\theta=\frac{1}{\sec\theta}
=\frac{1}{-\frac54}
=-\frac45. cos θ = sec θ 1 = − 4 5 1 = − 5 4 . Now use tan θ = sin θ cos θ \tan\theta=\frac{\sin\theta}{\cos\theta} tan θ = c o s θ s i n θ
− 3 4 = sin θ − 4 5 . -\frac34=\frac{\sin\theta}{-\frac45}. − 4 3 = − 5 4 sin θ . Multiply by − 4 5 -\frac45 − 5 4
sin θ = ( − 3 4 ) ( − 4 5 ) = 3 5 . \sin\theta=\left(-\frac34\right)\left(-\frac45\right)=\frac35. sin θ = ( − 4 3 ) ( − 5 4 ) = 5 3 . Thus,
sin θ = 3 5 , cos θ = − 4 5 . \boxed{\sin\theta=\frac35,\qquad \cos\theta=-\frac45}. sin θ = 5 3 , cos θ = − 5 4 . The signs match Quadrant II: sine is positive and cosine is negative.