Further symmetry properties and the Pythagorean identity
Theory
The symmetry properties of the circular functions relate an angle to its negative: \(\sin\) and \(\tan\) are odd while \(\cos\) is even, so \(\sin(-x)=-\sin x\), \(\cos(-x)=\cos x\) and \(\tan(-x)=-\tan x\). Complementary and supplementary angles give further relationships, and the Pythagorean identity \(\sin^2 x+\cos^2 x=1\) links sine and cosine so that, from one ratio and the quadrant, the others can be found. All angles are measured in radians.
A function is odd when \(f(-x)=-f(x)\) and even when \(f(-x)=f(x)\). On the unit circle the point at angle \(x\) is \((\cos x,\sin x)\); turning through \(-x\) instead reflects that point in the \(x\)-axis, so the \(x\)-coordinate is unchanged and the \(y\)-coordinate is negated. Hence \(\cos(-x)=\cos x\) (even), \(\sin(-x)=-\sin x\) (odd) and, since \(\tan x=\dfrac{\sin x}{\cos x}\), also \(\tan(-x)=-\tan x\) (odd).
Two angles are complementary if they add to \(\dfrac{\pi}{2}\) and supplementary if they add to \(\pi\). Complementary angles swap sine and cosine: \(\cos\!\left(\dfrac{\pi}{2}-x\right)=\sin x\) and \(\sin\!\left(\dfrac{\pi}{2}-x\right)=\cos x\). Supplementary angles keep the sine but reverse the cosine: \(\sin(\pi-x)=\sin x\), \(\cos(\pi-x)=-\cos x\) and \(\tan(\pi-x)=-\tan x\).
Because \((\cos x,\sin x)\) lies on the circle of radius \(1\), its coordinates satisfy \(x^2+y^2=1\), giving the Pythagorean identity \(\sin^2 x+\cos^2 x=1\). This holds for every angle and rearranges to \(\sin^2 x=1-\cos^2 x\) and \(\cos^2 x=1-\sin^2 x\). Given one ratio and the quadrant, the identity finds the size of another ratio while the quadrant fixes its sign.
Symmetry under a change of sign — \(\sin\) and \(\tan\) are odd, \(\cos\) is even:
Complementary angles (adding to \(\tfrac{\pi}{2}\)) swap sine and cosine; supplementary angles (adding to \(\pi\)) keep the sine:
The Pythagorean identity and its rearrangements:
Finding the other ratios from one ratio and the quadrant
- Note the quadrant. It fixes the signs: sine is \(+\) in quadrants 1 and 2; cosine is \(+\) in quadrants 1 and 4; tangent is \(+\) in quadrants 1 and 3.
- Apply the Pythagorean identity. Use \(\sin^2 x=1-\cos^2 x\) or \(\cos^2 x=1-\sin^2 x\) to find the square of the unknown ratio.
- Square-root and choose the sign. Take \(\pm\sqrt{\;}\), then keep the sign that the quadrant requires.
- Build the last ratio. Use \(\tan x=\dfrac{\sin x}{\cos x}\) to obtain the remaining value.
- Simplify first with symmetry. For a negated, complementary or supplementary angle, replace it using the symmetry rules before doing any arithmetic.
| \(\sin\!\left(-\dfrac{\pi}{6}\right)\) | \(=\) | \(-\sin\dfrac{\pi}{6}=-\dfrac{1}{2}\) |
| \(\cos\!\left(-\dfrac{\pi}{4}\right)\) | \(=\) | \(\cos\dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}\) |
| \(\tan\!\left(-\dfrac{\pi}{3}\right)\) | \(=\) | \(-\tan\dfrac{\pi}{3}=-\sqrt{3}\) |
| \(\sin(\pi-x)\) | \(=\) | \(\sin x\) |
| \(\cos\!\left(\dfrac{\pi}{2}-x\right)\) | \(=\) | \(\sin x\) |
| \(-\sin(-x)\) | \(=\) | \(-(-\sin x)=\sin x\) |
| expression | \(=\) | \(\sin x+\sin x+\sin x\) |
| \(=\) | \(3\sin x\) |
| \(\sin^2 x\) | \(=\) | \(1-\left(\dfrac{3}{5}\right)^2=1-\dfrac{9}{25}\) |
| \(\sin^2 x\) | \(=\) | \(\dfrac{16}{25}\) |
| \(\sin x\) | \(=\) | \(+\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}\) |
| \(\tan x\) | \(=\) | \(\dfrac{4/5}{3/5}=\dfrac{4}{3}\) |
| \(\sin x\) | \(=\) | \(-\dfrac{3}{4}\cos x\) |
| \(\dfrac{9}{16}\cos^2 x+\cos^2 x\) | \(=\) | \(1\) |
| \(\dfrac{25}{16}\cos^2 x\) | \(=\) | \(1\) |
| \(\cos^2 x\) | \(=\) | \(\dfrac{16}{25}\) |
| \(\cos x\) | \(=\) | \(-\dfrac{4}{5}\) |
| \(\sin x\) | \(=\) | \(-\dfrac{3}{4}\times\left(-\dfrac{4}{5}\right)=\dfrac{3}{5}\) |
Common pitfalls
Frequently asked questions
Is sine an odd or even function?
Sine is odd, so \(\sin(-x)=-\sin x\), and tangent is odd, so \(\tan(-x)=-\tan x\). Cosine is even, so \(\cos(-x)=\cos x\). Replacing \(x\) by \(-x\) reflects the unit-circle point in the \(x\)-axis, keeping the cosine and negating the sine.
What does cos(π/2 minus x) equal?
Complementary angles swap sine and cosine: \(\cos\!\left(\dfrac{\pi}{2}-x\right)=\sin x\) and \(\sin\!\left(\dfrac{\pi}{2}-x\right)=\cos x\), because \(\dfrac{\pi}{2}-x\) and \(x\) add to \(\dfrac{\pi}{2}\).
What is sin(π minus x)?
For supplementary angles, \(\sin(\pi-x)=\sin x\), \(\cos(\pi-x)=-\cos x\) and \(\tan(\pi-x)=-\tan x\). An angle and its supplement share the same sine but have opposite cosine.
What is the Pythagorean identity?
It is \(\sin^2 x+\cos^2 x=1\), true for all \(x\), because \((\cos x,\sin x)\) lies on the unit circle. Rearranging gives \(\sin^2 x=1-\cos^2 x\) and \(\cos^2 x=1-\sin^2 x\).
How do you find sin x given cos x and the quadrant?
Use \(\sin^2 x=1-\cos^2 x\), take the square root, then choose the sign from the quadrant. Sine is positive in quadrants 1 and 2, negative in 3 and 4. If \(\cos x=\dfrac{3}{5}\) in quadrant 1, then \(\sin x=\dfrac{4}{5}\).
Why does cos(minus x) equal cos x?
Because cosine is even. Turning through \(-x\) reflects the angle \(x\) point in the \(x\)-axis, which leaves the \(x\)-coordinate unchanged, and cosine is that \(x\)-coordinate, so \(\cos(-x)=\cos x\).