Trigonometric Identities

A point on a circle of radius 1

Draw a circle of radius 1 with its center at the origin, and take any point on the circle. Join the point to the center, and let that radius make an angle $\theta$θ with the positive x-axis.

Drop a line from the point straight down to the x-axis. This makes a right triangle: the radius is the hypotenuse, of length 1, the horizontal leg runs along the x-axis, and the vertical leg rises to the point.

In that triangle, $\cos \theta$cos θ is the adjacent side divided by the hypotenuse. The hypotenuse is 1, so the horizontal leg is $\cos \theta$cos θ itself. In the same way the vertical leg is $\sin \theta$sin θ. So the point is at $\left(\cos \theta ,\sin \theta \right)$(cos θ, sin θ): its distance across from the center is $\cos \theta$cos θ, and its height is $\sin \theta$sin θ.

The point at 50° on a circle of radius 1. Its distance across is cos 50° = 0.64 and its height is sin 50° = 0.77, to 2 decimal places.

Pythagoras gives ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1

The legs of the triangle are $\sin \theta$sin θ and $\cos \theta$cos θ, and the hypotenuse is 1. By Pythagoras, ${\left(\sin \theta \right)}^2+{\left(\cos \theta \right)}^2=1^2=1$(sin θ)² + (cos θ)² = 1² = 1.

The square of $\sin \theta$sin θ is written ${\sin }^2\theta$sin²θ, with the 2 between the sin and the $\theta$θ. So the result is ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1. Note that ${\sin }^2\theta$sin²θ means ${\left(\sin \theta \right)}^2$(sin θ)², the sine squared; it does not mean the sine of ${\theta }^2$θ².

Check it at 30°. $\sin 30^{\circ }=\frac{1}{2}$sin 30° = 1/2 and $\cos 30^{\circ }=\frac{\sqrt{3}}{2}$cos 30° = √3/2, so ${\sin }^{2}30^{\circ }+{\cos }^{2}30^{\circ }=\frac{1}{4}+\frac{3}{4}=1$sin²30° + cos²30° = 1/4 + 3/4 = 1. At 50°, $0.7660^2+0.6428^2=0.5868+0.4132=1.0000$0.7660² + 0.6428² = 0.5868 + 0.4132 = 1.0000.

At $\theta =30^{\circ }$θ = 30° the gold square on the horizontal leg has area ${\cos }^2\theta =0.75$cos²θ = 0.75, and the green square on the vertical leg has area ${\sin }^2\theta =0.25$sin²θ = 0.25. The bar on the right stacks the two areas, and they fill it to exactly 1. Drag the point round the circle: one square grows as the other shrinks, and the total stays 1. At 45° the two squares are equal, each ½.

True for every angle

An equation such as $\sin x=\frac{1}{2}$sin x = 1/2 is true only for particular angles, such as 30° and 150°. An identity is an equation that is true for every angle. ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1 is an identity.

Past 90°, the point on the circle can be left of the center or below it, so $\cos \theta$cos θ or $\sin \theta$sin θ can be negative. The identity still holds, because a square is never negative. At 150°, $\sin 150^{\circ }=\frac{1}{2}$sin 150° = 1/2 and $\cos 150^{\circ }=-\frac{\sqrt{3}}{2}$cos 150° = −√3/2, and the squares are $\frac{1}{4}$1/4 and $\frac{3}{4}$3/4 again. At 240°, $\sin 240^{\circ }=-\frac{\sqrt{3}}{2}$sin 240° = −√3/2 and $\cos 240^{\circ }=-\frac{1}{2}$cos 240° = −1/2, and the squares are $\frac{3}{4}$3/4 and $\frac{1}{4}$1/4.

It holds for an angle in radians too. At 0.3 radians, sin 0.3 = 0.2955 and cos 0.3 = 0.9553, to 4 decimal places, and their squares are 0.0873 and 0.9127, which add to 1.0000. At 1.1 radians, sin 1.1 = 0.8912 and cos 1.1 = 0.4536, and the squares are 0.7943 and 0.2057, which add to 1.0000.

$y={\sin }^{2}x$y = sin²x in gold and $y={\cos }^{2}x$y = cos²x in plain chalk, over one turn, with one square across for every 90°. Each curve is the other upside down about the level y = ½: wherever one is above ½, the other is below it by the same amount, so at every x the two heights add to 1. They cross at 45°, 135°, 225° and 315°, where each is ½.

The tangent: $\tan \theta =\sin \theta /\cos \theta$tan θ = sin θ / cos θ

In the same triangle, $\tan \theta$tan θ is the opposite side divided by the adjacent side. The opposite side is $\sin \theta$sin θ and the adjacent side is $\cos \theta$cos θ, so $\tan \theta =\sin \theta /\cos \theta$tan θ = sin θ / cos θ. This is the second identity, and it is true for every angle where $\cos \theta$cos θ is not 0.

The vertical leg divided by the horizontal leg is the gradient of the radius, so $\tan \theta$tan θ is the slope of the radius at angle $\theta$θ. Check it at 1.1 radians: $\sin 1.1/\cos 1.1=0.8912/0.4536=1.9648$sin 1.1 / cos 1.1 = 0.8912 / 0.4536 = 1.9648, and tan 1.1 is 1.9648. At 45°, the two legs are equal, so tan 45° = 1.

At 90° the point is at the top of the circle, (0, 1). There cos 90° = 0, and dividing by 0 has no value, so tan 90° is undefined: the radius is vertical and has no gradient.

Finding one ratio from another

The identity turns a known sine into the cosine of the same angle. Suppose $\theta$θ is acute and $\sin \theta =\frac{5}{13}$sin θ = 5/13. Then ${\cos }^2\theta =1-{\sin }^2\theta =1-\frac{25}{169}=\frac{144}{169}$cos²θ = 1 − sin²θ = 1 − 25/169 = 144/169.

Take the square root: $\cos \theta =\frac{12}{13}$cos θ = 12/13 or $-\frac{12}{13}$−12/13. An acute angle has a positive cosine, so $\cos \theta =\frac{12}{13}$cos θ = 12/13. Then $\tan \theta =\sin \theta /\cos \theta =\frac{5}{13}÷\frac{12}{13}=\frac{5}{12}$tan θ = sin θ / cos θ = 5/13 ÷ 12/13 = 5/12.

Check with a triangle. Sides of 5, 12 and 13 make a right triangle, because $5^2+12^2=25+144=169=13^2$5² + 12² = 25 + 144 = 169 = 13². With 5 opposite $\theta$θ and 13 the hypotenuse, the adjacent side is 12, and $\cos \theta =\frac{12}{13}$cos θ = 12/13.

If $\theta$θ were obtuse instead, between 90° and 180°, the point on the circle would be left of the center, and $\cos \theta$cos θ would be $-\frac{12}{13}$−12/13. The identity gives the size; the angle decides the sign.

A right triangle with sides 5, 12 and 13. The side of 5 is opposite $\theta$θ, so $\sin \theta =\frac{5}{13}$sin θ = 5/13, $\cos \theta =\frac{12}{13}$cos θ = 12/13 and $\tan \theta =\frac{5}{12}$tan θ = 5/12.

Rewriting an expression with the identity

The identity can be turned round: ${\sin }^2\theta =1-{\cos }^2\theta$sin²θ = 1 − cos²θ, and ${\cos }^2\theta =1-{\sin }^2\theta$cos²θ = 1 − sin²θ. When an expression mixes ${\sin }^2\theta$sin²θ and ${\cos }^2\theta$cos²θ, one of these turns it into a single function.

Take ${\cos }^2\theta +2{\sin }^2\theta$cos²θ + 2 sin²θ. Write it as $\left({\cos }^2\theta +{\sin }^2\theta \right)+{\sin }^2\theta$(cos²θ + sin²θ) + sin²θ. The bracket is 1, so ${\cos }^2\theta +2{\sin }^2\theta =1+{\sin }^2\theta$cos²θ + 2 sin²θ = 1 + sin²θ.

Check at two angles. At 0.3 radians, ${\cos }^2\theta +2{\sin }^2\theta =0.9127+2\times 0.0873=1.0873$cos²θ + 2 sin²θ = 0.9127 + 2 × 0.0873 = 1.0873, and $1+{\sin }^2\theta =1.0873$1 + sin²θ = 1.0873. At 1.1 radians, 0.2057 + 2 × 0.7943 = 1.7943, and 1 + 0.7943 = 1.7943. A check at one or two angles cannot prove an identity, but it catches a mistake quickly.

The usual mistakes

Reading ${\sin }^2\theta$sin²θ as the sine of ${\theta }^2$θ². At 0.3 radians, ${\sin }^2\theta$sin²θ is $0.2955^2=0.0873$0.2955² = 0.0873, but sin(0.09) is 0.0899.

Writing $\sin \theta +\cos \theta =1$sin θ + cos θ = 1. It is the squares that add to 1. At 45°, $\sin 45^{\circ }+\cos 45^{\circ }=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$sin 45° + cos 45° = 1/√2 + 1/√2 = √2, about 1.41.

Giving $\sin \theta$sin θ again when asked for $\cos \theta$cos θ. If $\sin \theta =\frac{5}{13}$sin θ = 5/13, the cosine is found from ${\cos }^2\theta =1-\frac{25}{169}$cos²θ = 1 − 25/169, and it is $\frac{12}{13}$12/13.

Dividing the wrong way for the tangent. $\tan \theta$tan θ is $\sin \theta /\cos \theta$sin θ / cos θ, opposite over adjacent; $\cos \theta /\sin \theta$cos θ / sin θ is its reciprocal.

Forgetting the sign. The square root gives a positive and a negative answer, and the quadrant of $\theta$θ decides which one is the cosine.