Learning objectives
The Pythagorean theorem concerning right triangles can be expressed by means of trigonometric ratios.
The following Pythagorean identities can also be used to find trigonometric function values given certain information:\begin{aligned} \left(\sin \theta \right)^2 + \left(\cos \theta \right) ^2 = 1 \quad \text{ or } \quad \sin^2 \theta + \cos^2 \theta = 1 \\\ \left(\tan \theta \right) ^ 2 + 1 = \left(\sec \theta \right) ^ 2 \quad \text{ or } \quad \tan ^2 \theta + 1 = \sec^2 \theta \\\ 1 + \left( \cot \theta \right) ^ 2 = \left( \csc \theta \right) ^2 \quad \text{ or } \quad 1 + \cot^2 \theta = \csc^2 \theta \end{aligned}
Given the following diagram, prove the Pythagorean identity \sin ^2 \theta + \cos ^2 \theta = 1.
Use \sin^2 \theta + \cos^2 \theta = 1 to prove the Pythagorean identity 1 + \cot^2 \theta = \csc^2 \theta.
Find the acute value of x, in radians, for the equation \sin^{2} x + \cos^{2} x + \cos x = \dfrac{\sqrt{3}}{2}+1.
We can write equivalent expressions and find trigonometric function values using a combination of reciprocal identities and the Pythagorean identities given here:\begin{aligned} \sin^2 \theta + \cos^2 \theta = 1 \\\ \tan ^2 \theta + 1 = \sec^2 \theta \\\ 1 + \cot^2 \theta = \csc^2 \theta \end{aligned}
The sum and difference identities in trigonometry are formulas that give relationships between the sine, cosine, and tangent of sums and differences of angles. These identities are useful when simplifying complex trigonometric expressions, solving trigonometric equations, and proving other trigonometric identities.
Sum and difference identities for sine:
Sum and difference identities for cosine:
Sum and difference identities for tangent:
Double-angle identities are specific trigonometric identities that help us express trigonometric functions of double angles in terms of single angles. They are derived from the sum identities of trigonometric functions.
Sine Double-Angle Identity | Cosine Double-Angle Identity | Tangent Double-Angle Identity |
---|---|---|
\sin 2 \alpha=2\sin \alpha\cos \alpha | \cos 2\alpha=\cos ^{2}(\alpha)-\sin ^{2}(\alpha) | \tan 2\alpha=\dfrac{2\tan \alpha}{1-\tan ^{2}(\alpha)} |
By letting \beta = 2\alpha we obtain three corresponding half-angle identities. These identities are sometimes referred to in alternate forms which have been manipulated by applying the Pythagorean identities. The relationships remain the same
Sine Half-Angle Identities | Cosine Half-Angle Identities | Tangent Half-Angle Identities |
---|---|---|
\sin \beta=2\sin \left(\dfrac{\beta}{2}\right)\cos \left(\dfrac{\beta}{2}\right) | \cos \beta=\cos ^{2}\left(\dfrac{\beta}{2}\right)-\sin ^{2}\left(\dfrac{\beta}{2}\right) | \tan \beta=\dfrac{2\tan \left(\dfrac{\beta}{2}\right)}{1-\tan ^{2}\left(\dfrac{\beta}{2}\right)} |
\sin \left(\dfrac{\beta}{2}\right)=\pm \sqrt{\dfrac{1-\cos \beta}{2}} | \cos \left(\dfrac{\beta}{2}\right)=\pm \sqrt{\dfrac{1+\cos \beta}{2}} | \tan \left(\dfrac{\beta}{2}\right)=\pm \sqrt{\dfrac{1-\cos \beta}{1+\cos \beta}} |
Find the solutions in the interval \left[0,\,2\pi\right) for \sin \left(x+\dfrac{\pi}{4}\right) + \sin \left(x-\dfrac{\pi}{4}\right) = -1 in radians.
Find all solutions of 2\cos {x} + \sin {2x} = 0 in the interval [0,\,2\pi).
Sum and difference identities for sine:
Sum and difference identities for cosine:
Sum and difference identities for tangent:
Double-Angle Identities:
Sine | \sin 2 \alpha=2\sin \alpha\cos \alpha |
---|---|
Cosine | \cos 2\alpha=\cos ^{2}(\alpha)-\sin ^{2}(\alpha) |
Tangent | \tan 2\alpha=\dfrac{2\tan \alpha}{1-\tan ^{2}(\alpha)} |
Half-Angle Identities:
Sine | \sin \beta=2\sin \left(\dfrac{\beta}{2}\right)\cos \left(\dfrac{\beta}{2}\right) | \sin \left(\dfrac{\beta}{2}\right)=\pm \sqrt{\dfrac{1-\cos \beta}{2}} |
---|---|---|
Cosine | \cos \beta=\cos ^{2}\left(\dfrac{\beta}{2}\right)-\sin ^{2}\left(\dfrac{\beta}{2}\right) | \cos \left(\dfrac{\beta}{2}\right)=\pm \sqrt{\dfrac{1+\cos \beta}{2}} |
Tangent | \tan \beta=\dfrac{2\tan \left(\dfrac{\beta}{2}\right)}{1-\tan ^{2}\left(\dfrac{\beta}{2}\right)} | \tan \left(\dfrac{\beta}{2}\right)=\pm \sqrt{\dfrac{1-\cos \beta}{1+\cos \beta}} |
Recall that we can rewrite trig expressions by factoring them, or using substitution and different polynomial identities. We can apply these techniques and the zero product property to solve trig equations that mimic polynomial equations.
The following polynomial identities also apply to trigonometric expressions of the same form:
Polynomial Identity | |
---|---|
\text{Distributive property /} \\ \text{Greatest common factor (GCF)} | AB+AC+\ldots =A(B+C+\ldots) |
\text{Perfect square trinomial} | A^2+2AB+B^2=(A+B)^2 \\A^2-2AB+B^2=(A-B)^2 |
\text{Difference of squares} | A^2-B^2=(A+B)(A-B) |
\text{Sum of cubes} | A^3+B^3=(A+B)(A^2-AB+B^2) |
\text{Difference of cubes} | A^3-B^3=(A-B)(A^2+AB+B^2) |
\text{Factoring quadratics (leading coefficient of 1)} | x^2+(A+B)x+AB=(x+A)(x+B) |
\text{Distributive property / Grouping in pairs} | (A+B)(C+D)=AC+AD+BC+BD |
\text{Quadratic formula} | \text{If }Ax^2+Bx+C=0 \text{ then } \\x=\dfrac{-B\pm \sqrt{B^2-4AC}}{2A} |
\text{Zero product property} | \text{The equation }AB=0 \text{ is true if and only if }\\A=0 \text{ or }B=0. |
Recall the basic trigonometric identities:
\begin{aligned} \sin \theta = \dfrac{1}{\csc \theta} \qquad \cos \theta = \dfrac{1}{\sec \theta} \qquad \tan \theta = \dfrac{1}{\cot \theta} \qquad \tan \theta = \dfrac{ \sin \theta}{\cos \theta} \\\\ \csc \theta= \dfrac{1}{\sin \theta} \qquad \sec \theta = \dfrac{1}{\cos \theta} \qquad \cot \theta = \dfrac{1}{\tan \theta} \qquad \cot \theta = \dfrac{\cos \theta}{\sin \theta} \end{aligned}
Cofunction identities are relationships between functions of complementary angles. These relationships allow us to express one trigonometric function in terms of another function of the complement of the angle. Here are the basic cofunction identities for sine, cosine, tangent, cotangent, secant, and cosecant:
\begin{aligned} \sin \alpha = \cos(90\degree - \alpha) \qquad \tan \alpha = \cot(90\degree - \alpha) \qquad \sec \alpha = \csc(90\degree - \alpha) \\ \qquad \cos \alpha = \sin(90\degree - \alpha) \qquad \cot \alpha = \tan (90\degree - \alpha) \qquad \csc \alpha = \sec(90\degree - \alpha) \end{aligned}
The key when solving problems involving these identities is to figure out which to use and when.
Solve \cot x\cos ^2x=2\cot x.
Consider \sin \theta = -\dfrac{2}{3}, where 0 < \theta < 2 \pi.
State any quadrant angle \theta lies in.
Find \cos \theta when \pi < \theta < \dfrac{3 \pi}{2}.
Find the value of \tan \theta.
Consider the equation \left(\sin \theta + \dfrac{\sqrt{3}}{2}\right) \left(\cos \theta + \dfrac{\sqrt{3}}{2}\right) = 0 for 0 \degree \lt \theta \lt 90 \degree.
How many solutions for \theta does the equation have?
Find one possible solution, in degrees, for the equation \sec \alpha = \csc\left(\dfrac{\alpha}{2} + 20 \degree\right).
The following polynomial identities also apply to trigonometric expressions of the same form:
Polynomial Identity | |
---|---|
\text{Distributive property /} \\ \text{Greatest common factor (GCF)} | AB+AC+\ldots =A(B+C+\ldots) |
\text{Perfect square trinomial} | A^2+2AB+B^2=(A+B)^2 \\A^2-2AB+B^2=(A-B)^2 |
\text{Difference of squares} | A^2-B^2=(A+B)(A-B) |
\text{Sum of cubes} | A^3+B^3=(A+B)(A^2-AB+B^2) |
\text{Difference of cubes} | A^3-B^3=(A-B)(A^2+AB+B^2) |
\text{Factoring quadratics (leading coefficient of 1)} | x^2+(A+B)x+AB=(x+A)(x+B) |
\text{Distributive property / Grouping in pairs} | (A+B)(C+D)=AC+AD+BC+BD |
\text{Quadratic formula} | \text{If }Ax^2+Bx+C=0 \text{ then } \\x=\dfrac{-B\pm \sqrt{B^2-4AC}}{2A} |
\text{Zero product property} | \text{The equation }AB=0 \text{ is true if and only if }\\A=0 \text{ or }B=0. |
Basic trigonometric identities:
\begin{aligned} \sin \theta = \dfrac{1}{\csc \theta} \qquad \cos \theta = \dfrac{1}{\sec \theta} \qquad \tan \theta = \dfrac{1}{\cot \theta} \qquad \tan \theta = \dfrac{ \sin \theta}{\cos \theta} \\\\ \csc \theta= \dfrac{1}{\sin \theta} \qquad \sec \theta = \dfrac{1}{\cos \theta} \qquad \cot \theta = \dfrac{1}{\tan \theta} \qquad \cot \theta = \dfrac{\cos \theta}{\sin \theta} \end{aligned}
Cofunction identities:
\begin{aligned} \sin \alpha = \cos(90\degree - \alpha) \qquad \tan \alpha = \cot(90\degree - \alpha) \qquad \sec \alpha = \csc(90\degree - \alpha) \\ \qquad \cos \alpha = \sin(90\degree - \alpha) \qquad \cot \alpha = \tan (90\degree - \alpha) \qquad \csc \alpha = \sec(90\degree - \alpha) \end{aligned}