The fundamental trigonometric identities can be proved by definition using a
general point, (, ), in the coordinate plane.
For any coordinate point, P(a,b ), = \(\sqrt{x^2+ b^2}>0\)
The six trigonometric ratios are defined as:
Primary Trigonometric Ratio: sinθ = \(\frac{b}{r}\) cosθ = \(\frac{a}{r}\) tanθ = \(\frac{b}{a}\)
Secondary Trigonometric Ratio: cscθ = \(\frac{r}{b}\) ; secθ = \(\frac{r}{a}\) ; cotθ = \(\frac{a}{b}\)
Reciprocal identities: cscθ = \(\frac{1}{sinθ}\) secθ = \(\frac{1}{cosθ}\) cotθ = \(\frac{1}{tanθ}\)
cscθ = \(\frac{r}{b}\) =\(\frac{1}{b/r}\) = \(\frac{1}{sinθ}\)
secθ = \(\frac{r}{a}\) = \(\frac{1}{a/r}\) = \(\frac{1}{cosθ}\)
cotθ = \(\frac{a}{b}\) =\(\frac{1}{b/a}\) =\(\frac{1}{tanθ}\)
Quotient Identities: tan = \(\frac{sinθ}{cosθ}\) cot = \(\frac{cosθ}{sinθ}\)
\(\frac{sinθ}{cosθ}\) = \(\frac{b/r}{a/r}\) = \(\frac{b}{a}\) = tan
\(\frac{cosθ}{sinθ}\) = \(\frac{a/r}{b/r}\) = \(\frac{a}{b}\) = cot
Pythagorean Identities: \({sin^2θ+cos^2θ }\) =1 \({1+tan^2θ}\) = \({sec^2θ}\) \({1 +cot^2θ}\) = \({1 +csc^2θ}\)
(a) [latex]{b^2} +{a^2}[/latex] = \({r^2}\)
\(\frac{b^2}{r^2}+\frac{a^2}{r^2}\) = \(\frac{r^2}{r^2}\)
\({sin^2θ+cos^2θ }\) =1
(b) b2 + a 2= r2
\(\frac{a^2}{a^2}+\frac{b^2}{a^2}\) = \(\frac{r^2}{a^2}\)
\({1+tan^2θ}\) = \({sec^2θ}\)
(c) b2 + a 2= r2
\(\frac{b^2}{b^2}+\frac{a^2}{b^2}\) = \(\frac{r^2}{b^2}\)
\({1 +cot^2θ}\) = \({1 +csc^2θ}\)
Student’s Activity
Prove Complementary Angle or Cofunction Identity with the help of right triangle ABC shown below.
∠A + ∠C = 90° or ∠C = 90° − ∠A
Cofunction Identity
sin(90° − θ) = cosθ
cos(90° − θ) = sinθ
tan(90° − θ) = cotθ
Hint: sin(A) = \(\frac{CB}{AC}\)
cos C = cos(90° − A) = \(\frac{CB}{AC}\)
Solution to Student’s Activity
Provide a counter example to show that \(\sqrt{x^2}\) = is not an identity.
If = −1, then = \(\sqrt{x^2}\) = \(\sqrt{(-1)^2}\) = 1 and the = = −1
Since ≠ for = −1 the equation is a conditional equation and not an identity.
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