Trigonometric identities
Reciprocal and quotient identities
Suppose f\left(x\right) is a function. The reciprocal function of f\left(x\right) is the function \dfrac{1}{f\left(x\right)}. In particular, we can define three reciprocal functions from the three standard trigonometric functions.
The reciprocal functions of the three trigonometric functions \sin\left(\alpha\right), \cos\left(\alpha\right) and \tan\left(\alpha\right) are:
csc\left(\alpha\right)=\dfrac{1}{\sin\left(\alpha\right)}
sec\left(\alpha\right)=\dfrac{1}{\cos\left(\alpha\right)}
cot\left(\alpha\right)=\dfrac{1}{\tan\left(\alpha\right)}
Consider \alpha such that:
- \sin\left(\alpha\right)=\dfrac{1}{3}
- \cos\left(\alpha\right)=\dfrac{2\sqrt{2}}{3}
Then:
sec\left(\alpha\right)=\dfrac{1}{\cos\left(\alpha\right)}=\dfrac{1}{\dfrac{2\sqrt{2}}{3}}=\dfrac{3}{2\sqrt{2}}=\dfrac{3\sqrt{2}}{4}
csc\left(\alpha\right)=\dfrac{1}{\sin\left(\alpha\right)}=\dfrac{1}{\dfrac{1}{3}}=3
For any real \alpha such that \cos\left(\alpha\right)\neq0, we have:
\tan\left(\alpha\right)=\dfrac{\sin\left(\alpha\right)}{\cos\left(\alpha\right)}
Therefore:
\cot\left(\alpha\right)=\dfrac{\cos\left(\alpha\right)}{\sin\left(\alpha\right)}
Consider \alpha such that:
- \sin\left(\alpha\right)=\dfrac{2}{5}
- \cos\left(\alpha\right)=\dfrac{\sqrt{21}}{5}
Then:
\cot\left(\alpha\right)=\dfrac{\cos\left(\alpha\right)}{\sin\left(\alpha\right)}=\dfrac{\dfrac{\sqrt{21}}{5}}{\dfrac{2}{5}}=\dfrac{\sqrt{21}}{5}\cdot\dfrac{5}{2}=\dfrac{\sqrt{21}}{2}
Pythagorean identities
For any real number \alpha :
sin^{2}\left(\alpha\right)+cos^{2}\left(\alpha\right)=1
Consider a real number \alpha such that:
- \cos\left(\alpha\right)=-\dfrac{3}{5}
- \alpha is in Quadrant II.
We want to calculate \sin\left(\alpha\right). We know that:
sin^{2}\left(\alpha\right)+cos^{2}\left(\alpha\right)=1
So:
sin^{2}\left(\alpha\right)+\left(-\dfrac{3}{5}\right)^{2}=1
sin^{2}\left(\alpha\right)=1-\dfrac{9}{25}
sin^{2}\left(\alpha\right)=\dfrac{16}{25}
\sin\left(\alpha\right)=\pm \sqrt{\dfrac{16}{25}}
\alpha is in Quadrant II, therefore \sin\left(\alpha\right)\gt0. We get:
\sin\left(\alpha\right)=\dfrac{4}{5}
The Pythagorean identity for trigonometric functions is derived from the Pythagorean theorem. In the trigonometric circle with radius 1, we get:
It follows that for any real number \alpha :
1+tan^{2}\left(\alpha\right)=sec^{2}\left(\alpha\right)
1+cot^{2}\left(\alpha\right)=csc^{2}\left(\alpha\right)
Cofunction identities
For any real number \alpha :
\sin\left(\dfrac{\pi}{2}-\alpha\right)=\cos\left(\alpha\right)
\cos\left(\dfrac{\pi}{2}-\alpha\right)=\sin\left(\alpha\right)
Assume that:
\cos\left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt{2}+\sqrt{6}}{4}
Observe that:
\dfrac{\pi}{2}-\dfrac{\pi}{12}=\dfrac{5\pi}{12}
Therefore:
\sin\left(\dfrac{5\pi}{12}\right)=\sin\left(\dfrac{\pi}{2}-\dfrac{\pi}{12}\right)=\cos\left(\dfrac{\pi}{12}\right)
\sin\left(\dfrac{5\pi}{12}\right)=\dfrac{\sqrt{2}+\sqrt{6}}{4}
It follows that for any real number \alpha :
\tan\left(\dfrac{\pi}{2}-\alpha\right)=\cot\left(\alpha\right)
\cot\left(\dfrac{\pi}{2}-\alpha\right)=\tan\left(\alpha\right)
\sec\left(\dfrac{\pi}{2}-\alpha\right)=\csc\left(\alpha\right)
\csc\left(\dfrac{\pi}{2}-\alpha\right)=\sec\left(\alpha\right)
Odd-even identities
For any real number \alpha :
\sin\left(-\alpha\right)=-\sin\left(\alpha\right)
\cos\left(-\alpha\right)=\cos\left(\alpha\right)
We know that:
\sin\left(\dfrac{5\pi}{12}\right)=\dfrac{\sqrt{2}+\sqrt{6}}{4}
Therefore:
\sin\left(-\dfrac{5\pi}{12}\right)=-\sin\left(\dfrac{5\pi}{12}\right)
\sin\left(-\dfrac{5\pi}{12}\right)=-\dfrac{\sqrt{2}+\sqrt{6}}{4}
It follows that for any real number \alpha such that \cos\left(\alpha\right)\neq0 :
\tan\left(-\alpha\right)=-\tan\left(\alpha\right)
Using the previous formulas, we also get:
\cot\left(-\alpha\right)=-\cot\left(\alpha\right)
\sec\left(-\alpha\right)=\sec\left(\alpha\right)
\csc\left(-\alpha\right)=-\csc\left(\alpha\right)
Sum and difference formulas
If a and b are two real numbers, then:
\sin\left(a+b\right)=\sin\left(a\right)\cos\left(b\right)+\sin\left(b\right)\cos\left(a\right)
\sin\left(a-b\right)=\sin\left(a\right)\cos\left(b\right)-\sin\left(b\right)\cos\left(a\right)
\sin\left(15^\circ\right)=\sin\left(45^\circ-30^\circ\right)
Therefore:
\sin\left(15^\circ\right)=\sin\left(45^\circ\right)\cos\left(30^\circ\right)-\sin\left(30^\circ\right)\cos\left(45^\circ\right)
And:
\sin\left(15^\circ\right)=\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{6}}{4}-\dfrac{\sqrt{2}}{4}
If a and b are two real numbers, then:
\cos\left(a+b\right)=\cos\left(a\right)\cos\left(b\right)-\sin\left(a\right)\sin\left(b\right)
\cos\left(a-b\right)=\cos\left(a\right)\cos\left(b\right)+\sin\left(a\right)\sin\left(b\right)
We want to prove that:
\cos\left(\dfrac{\pi}{2}-\alpha\right)=\sin\left(\alpha\right)
We have:
\cos\left(\dfrac{\pi}{2}-\alpha\right)=\cos\left(\dfrac{\pi}{2}\right)\cos\left(\alpha\right)+\sin\left(\dfrac{\pi}{2}\right)\sin\left(\alpha\right)
Therefore:
\cos\left(\dfrac{\pi}{2}-\alpha\right)=0\cdot\cos\left(\alpha\right)+1\cdot\sin\left(\alpha\right)=\sin\left(\alpha\right)
If a and b are two real numbers, then:
\tan\left(a+b\right)=\dfrac{\tan\left(a\right)+\tan\left(b\right)}{1-\tan\left(a\right)\tan\left(b\right)}
\tan\left(a-b\right)=\dfrac{\tan\left(a\right)-\tan\left(b\right)}{1+\tan\left(a\right)\tan\left(b\right)}
If a and b are two real numbers, then:
\cot\left(a+b\right)=\dfrac{\cot\left(a\right)\cdot\cot\left(b\right)-1}{\cot\left(b\right)+\cot\left(a\right)}
\cot\left(a-b\right)=\dfrac{\cot\left(a\right)\cdot\cot\left(b\right)+1}{\cot\left(b\right)-\cot\left(a\right)}
Double-angle and half-angle formulas
For any real number \alpha :
\sin\left(2\alpha\right)=2\sin\left(\alpha\right)\cos\left(\alpha\right)
\sin\left(120^\circ\right)=\sin\left(2\cdot 60^\circ\right)=2\sin\left(60^\circ\right)\cos\left(60^\circ\right)=2\cdot\dfrac{\sqrt{3}}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt{3}}{2}
For any real number \alpha :
\cos\left(2\alpha\right)=cos^{2}\left(\alpha\right)-sin^{2}\left(\alpha\right)
Therefore:
\cos\left(2\alpha\right)=2cos^{2}\left(\alpha\right)-1
And:
\cos\left(2\alpha\right)=1-2sin^{2}\left(\alpha\right)
It follows that:
\tan\left(2\alpha\right)=\dfrac{2\tan\left(\alpha\right)}{1-tan^{2}\left(\alpha\right)}
For any real number \alpha :
\sin^2\left(\dfrac{a}{2}\right)=\dfrac{1-\cos\left(a\right)}{2}
\cos^2\left(\dfrac{a}{2}\right)=\dfrac{1+\cos\left(a\right)}{2}
The half angle identity allows us to find \sin\left(\dfrac{\pi}{12}\right).
Observe that:
\dfrac{\pi}{12}=\dfrac{1}{2}\cdot \dfrac{\pi}{6}
By the half-angle identity for sine:
\sin^2\left(\dfrac{\pi}{12}\right)=\dfrac{1-\cos\left(\dfrac{\pi}{6}\right)}{2}=\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}=\dfrac{2-\sqrt{3}}{4}
Therefore:
\sin\left(\dfrac{\pi}{12}\right)=\sqrt{\dfrac{2-\sqrt{3}}{4}}=\dfrac{\sqrt{2-\sqrt{3}}}{2}
Product-to-sum and sum-to-product formulas
For any real numbers \alpha and \beta :
\sin\left(\alpha\right)\sin\left(\beta\right)=\dfrac{1}{2}\left[ \cos\left(\alpha-\beta\right)-\cos\left(\alpha+\beta\right) \right]
\cos\left(\alpha\right)\cos\left(\beta\right)=\dfrac{1}{2}\left[ \cos\left(\alpha-\beta\right)+\cos\left(\alpha+\beta\right) \right]
\sin\left(\alpha\right)\cos\left(\beta\right)=\dfrac{1}{2}\left[ \sin\left(\alpha+\beta\right)+\sin\left(\alpha-\beta\right) \right]
\sin\left(75^\circ\right)\cdot\sin\left(15^\circ\right)=\dfrac{1}{2}\left[ \cos\left(75^\circ-15^\circ\right)-\cos\left(75^\circ+15^\circ\right) \right]
\sin\left(75^\circ\right)\cdot\sin\left(15^\circ\right)=\dfrac{1}{2}\left[ \cos\left(60^\circ\right)-\cos\left(90^\circ\right) \right]
\sin\left(75^\circ\right)\cdot\sin\left(15^\circ\right)=\dfrac{1}{2}\left[ \dfrac{1}{2}-0 \right]=\dfrac{1}{4}
It follows that for any real numbers \alpha and \beta :
\sin\left(\alpha\right)+\sin\left(\beta\right)=2\cdot\sin\left(\dfrac{\alpha+\beta}{2}\right)\cdot\cos\left(\dfrac{\alpha-\beta}{2}\right)
\sin\left(\alpha\right)-\sin\left(\beta\right)=2\cdot\sin\left(\dfrac{\alpha-\beta}{2}\right)\cdot\cos\left(\dfrac{\alpha+\beta}{2}\right)
It also follows that for any real numbers \alpha and \beta :
\cos\left(\alpha\right)+\cos\left(\beta\right)=2\cdot\cos\left(\dfrac{\alpha+\beta}{2}\right)\cdot\cos\left(\dfrac{\alpha-\beta}{2}\right)
\cos\left(\alpha\right)-\cos\left(\beta\right)=-2\cdot\sin\left(\dfrac{\alpha+\beta}{2}\right)\cdot\sin\left(\dfrac{\alpha-\beta}{2}\right)
Solve trigonometric equations
Equations of the form \cos\left(x\right)=\cos\left(a\right) and \cos\left(x\right)=a
Consider the equation :
\cos\left(x\right)=\cos\left(a\right)
Since the cosine function is periodic with period of 2\pi, the solutions are:
\left\{ a+2k\pi, k\in\mathbb{Z} \right\} and \left\{ -a+2k\pi, k\in\mathbb{Z} \right\}
Consider the equation:
\cos\left(2x\right)=\cos\left(4x\right)
The solutions are all real numbers x such that:
\begin{cases} 2x=4x+2k\pi \cr \cr \text{or} \cr \cr 2x=-4x+2k\pi \end{cases}
We have:
2x=4x+2k\pi
\Leftrightarrow -2x=2k\pi
\Leftrightarrow x=-k\pi, k\in\mathbb{Z}
And:
2x=-4x+2k\pi
\Leftrightarrow 6x=2k\pi
\Leftrightarrow x=\dfrac{k\pi}{3}, k\in\mathbb{Z}
Consider the equation:
\cos\left(x\right)=a
It follows that the solutions are:
\left\{\arccos\left(a\right)+2k\pi, k\in\mathbb{Z}\right\} or \left\{-\arccos\left(a\right)+2k\pi, k\in\mathbb{Z}\right\}
We want to determine all the solutions to the equation:
\cos\left(x\right)=\dfrac{\sqrt{3}}{2}
The solutions are:
\begin{cases} x=\arccos\left(\dfrac{\sqrt{3}}{2}\right)+2k\pi \cr \cr\text{or} \cr \cr x=-\arccos\left(\dfrac{\sqrt{3}}{2}\right)+2k\pi \end{cases}
So:
\begin{cases} x=\dfrac{\pi}{6}+2k\pi, k\in\mathbb{Z} \cr \cr\text{or} \cr \cr x=-\dfrac{\pi}{6}+2k\pi, k\in\mathbb{Z}\end{cases}
Equations of the form \sin\left(x\right)=\sin\left(a\right) and \sin\left(x\right)=a
Consider the equation:
\sin\left(x\right)=\sin\left(a\right)
Since the sine function is periodic with period of 2\pi, the solutions are:
\left\{ a+2k\pi, k\in\mathbb{Z} \right\} or \left\{ \pi-a+2k\pi, k\in\mathbb{Z} \right\}
Consider the equation:
\sin\left(2x\right)=\sin\left(4x\right)
The solutions are all real numbers x such that:
\begin{cases} 2x=4x+2k\pi \cr \cr \text{or} \cr \cr 2x=\pi-4x+2k\pi \end{cases}
We have:
2x=4x+2k\pi
\Leftrightarrow -2x=2k\pi
\Leftrightarrow x=-k\pi, k\in\mathbb{Z}
And:
2x=\pi-4x+2k\pi
\Leftrightarrow 6x=\left(2k+1\right)\pi
\Leftrightarrow x=\dfrac{\left(2k+1\right)\pi}{6}, k\in\mathbb{Z}
Consider the equation:
\sin\left(x\right)=a
It follows that the solutions are:
\left\{\arcsin\left(a\right)+2k\pi, k\in\mathbb{Z}\right\} or \left\{\pi-\arcsin\left(a\right)+2k\pi, k\in\mathbb{Z}\right\}
We want to determine all the solutions to the equation:
\sin\left(x\right)=\dfrac{\sqrt{3}}{2}
The solutions are:
\begin{cases} x=\arcsin\left(\dfrac{\sqrt{3}}{2}\right)+2k\pi \cr \cr\text{or} \cr \cr x=\pi-\arcsin\left(\dfrac{\sqrt{3}}{2}\right)+2k\pi \end{cases}
So:
\begin{cases} x=\dfrac{\pi}{6}+2k\pi, k\in\mathbb{Z} \cr \cr\text{or} \cr \cr x=\dfrac{5\pi}{6}+2k\pi, k\in\mathbb{Z}\end{cases}
Equations using trigonometric identities
The properties above can be used to solve more complicated yet related equations.
We want to determine all of the solutions to the equation:
\cos^2\left(\dfrac{x}{2}\right)=\dfrac{\cos\left(3\right)+1}{2}
By the half-angle identity for cosine:
\cos^2\left(\dfrac{x}{2}\right)=\dfrac{1+\cos\left(x\right)}{2}
Therefore, the equation is equivalent to:
\dfrac{1+\cos\left(x\right)}{2}=\dfrac{\cos\left(3\right)+1}{2}
And:
1+\cos\left(x\right)=\cos\left(3\right)+1
\cos\left(x\right)=\cos\left(3\right)
Therefore, the solutions to the equation are:
x=\begin{cases} 3 + 2k\pi & k\in \mathbb{Z} \cr \cr \mbox{and} \cr \cr -3+2k\pi & k\in \mathbb{Z} \end{cases}
Quadratic equations that involve trigonometric functions
The properties above can be used to solve quadratic equations involving trigonometric functions.
Find all the solutions of the following equation at the interval \left[0{,}2\pi\right) :
\cos\left(2x\right)=\sin\left(x\right)
We have:
\cos\left(2x\right)=1-2sin^{2}\left(x\right)
The given equation is equivalent to:
1-2sin^{2}\left(x\right)=\sin\left(x\right)
If we substitute \sin\left(x\right)=u, then we get the following quadratic equation:
1-2u^{2}=u
Which is equivalent to:
2u^{2}+u-1=0
Which is equivalent to:
\left(u+1\right)\left(2u-1\right)=0
A product is zero if one of the terms of the product is zero. So:
u=-1 or u=\dfrac{1}{2}
Going back to the substitution, we have to solve the equations:
\sin\left(x\right)=-1 and \sin\left(x\right)=\dfrac{1}{2}
The equation \sin\left(x\right)=-1 has the solution:
x=\dfrac{3\pi}{2}
The equation \sin\left(x\right)=\dfrac{1}{2} has the solutions:
x=\dfrac{\pi}{6} and x=\dfrac{7\pi}{6}
In conclusion, the solutions to the equation at the interval \left[0{,}2\pi\right) are:
x=\dfrac{3\pi}{2}, x=\dfrac{\pi}{6} and x=\dfrac{7\pi}{6}.
Law of Sines and Cosines
Law of sines
In any triangle \Delta ABC with angles A, B and C and opposite sides a, b, and c, the following equation is true:
\dfrac{\sin\left(A\right)}{a}=\dfrac{\sin\left(B\right)}{b}=\dfrac{\sin\left(C\right)}{c}
Law of cosines
In any triangle \Delta ABC with angles A, B and C and opposite sides a, b, and c, the following equations are true:
a^{2}=b^{2}+c^{2}-2 b c\cos\left(A\right)
b^{2}=a^{2}+c^{2}-2 a c\cos\left(B\right)
c^{2}=a^{2}+b^{2}-2 a b\cos\left(C\right)
The law of sines and the law of cosines can be used to solve triangles.
Consider the following triangle:
To solve the triangle, find the missing angle \alpha and the two missing sides x and y.
Solve for \alpha
To solve for \alpha, use the fact that the sum of the three angles of a triangle is 180^\circ :
30^\circ+15^\circ+\alpha=180^\circ\\\alpha=135^\circ
Solve for x
To solve for the missing side x, use the law of sines:
\dfrac{x}{\sin\left(135^\circ\right)}=\dfrac{7}{\sin\left(30^\circ\right)}\\\dfrac{x}{\dfrac{\sqrt{2}}{2}}=\dfrac{7}{\dfrac{1}{2}}\\x=\dfrac{\sqrt{2}}{2}\cdot 14=7\sqrt{2}
Solve for y
To solve for the missing side y, use the law of cosines:
y^2=\left(7\sqrt{2}\right)^2+7^2-2\left(7\sqrt{2}\right)\left(7\right)\dfrac{1+\sqrt{3}}{2\sqrt{2}}\\=49\left(2-\sqrt{3}\right)\\y=7\sqrt{2-\sqrt{3}}