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 Math Help > Trigonometry > Trig Equivalences > Summary of Identities

This page contains a summary of all of the basic trigonometric identities that you will need to use in a high school trig class, organized by type.

### even/odd, co-function and phase shift identities

even/odd

 cos(x) = cos(-x) sec(x) = 1/cos(x) = 1/cos(-x) = sec(-x) tan(x) = sin(x)/cos(x) = -sin(-x)/cos(-x) = -tan(-x) sin(x) = -sin(-x) csc(x) = 1/sin(x) = 1/-sin(-x) = -csc(-x) cot(x) = cos(x)/sin(x) = cos(-x)/-sin(-x) = -cot(-x)

co-functions

 sin(x) = cos(π/2-x) tan(x) = cot(π/2-x) sec(x) = csc(π/2-x) cos(x) = sin(π/2-x) cot(x) = tan(π/2-x) csc(x) = sec(π/2-x)

phase shift

 sin(x) = cos(x-π/2) = -sin(x-π) = -cos(x+π/2) = sin(x+2π) csc(x) = sec(x-π/2) = -csc(x-π) = -sec(x+π/2) = csc(x+2π) tan(x) = -cot(x±π/2) = tan(x+π) tan(x+π/4) = (1+tan(x))/(1-tan(x))

### exact values of trig functions of common angles -- "Brain Dump"

If you're in a high school trig class, and you need to know the exact values of sin, cos, and tan for these "round number" angles, then I strongly recommend practicing this brain dump.  On a blank piece of paper, write down all the degrees that are multiples of 30 or 45, convert them into radians by multiplying by π/180, and then write down the sin and cos values  The repetitive nature of these two functions should help a lot.  You know tan=sin/cos, so just do the math, and write 'em in.

Once you can do this brain dump in two or three minutes, then you will be ready to do that during every test.  No need to bring a cheat sheet with you -- make a new one on scrap paper at the very first thing you do during a test.  (Some teachers may object at first, so it's best to clear this idea with your teacher first.)

 degrees radians sin cos tan 0 0 0 1 0 30 π/6 1/2 sqrt(3)/2 sqrt(3)/3 45 π/4 sqrt(2)/2 sqrt(2)/2 1 60 π/3 sqrt(3)/2 1/2 sqrt(3) 90 π/2 1 0 -- 120 2π/3 sqrt(3)/2 -1/2 -sqrt(3) 135 3π/4 sqrt(2)/2 -sqrt(2)/2 -1 150 5π/6 1/2 -sqrt(3)/2 -sqrt(3)/3 180 π 0 -1 0 210 7π/6 -1/2 -sqrt(3)/2 sqrt(3)/3 225 5π/4 -sqrt(2)/2 -sqrt(2)/2 1 240 4π/3 -sqrt(3)/2 -1/2 sqrt(3) 270 3π/2 -1 0 -- 300 5π/3 -sqrt(3)/2 1/2 -sqrt(3) 315 7π/4 -sqrt(2)/2 sqrt(2)/2 -1 330 11π/6 -1/2 sqrt(3)/2 -sqrt(3)/3 360 2π 0 1 0

This idea can be extended using some of the identities presented below.  Here are a whole lot more Special Angles.

### Pythagorean identity: sin2x + cos2x = 1

formula proof
sin2x + cos2x = 1 Apply the Pythagorean Theorem to this triangle:
 1 sin x cos x
tan2x + 1 = sec2 Start with the first formula, then divide every term by cos2x.
1 + cot2x = csc2 Start over, dividing by sin2x.

### half angle formulas

 formula proof sin(x/2) = ±sqrt((1-cos(x))/2) cos(x) = cos(x/2 + x/2) = cos2(x/2) - sin2(x/2) cos(x) = 1 - 2sin2(x/2), then solve for sin(x/2); cos(x) = 2cos2(x/2) - 1, then solve for cos(x/2). cos(x/2) = ±sqrt((1+cos(x))/2) tan(x/2) = (1-cos(x))/sin(x)      = sin(x)/(1+cos(x)) tan(x/2) = sin(x/2) / cos(x/2)    = sqrt(1-cos x)/sqrt(1+cos x)    = (1-cos x)/(sin x),  by rationalizing the denominator     = (sin x)/(1+cos x),  by rationalizing the numerator sin(x/2)-cos(x/2) = ±sqrt(1-sin(x)), cos(x/2)-sin(x/2) = ±sqrt(1-sin(x)) (sin(x/2)-cos(x/2))2 = 1-2sin(x/2)cos(x/2) = 1-sin(x); take the square root of both sides. sin(x/2)+cos(x/2) = �sqrt(1+sin(x)) (sin(x/2)+cos(x/2))2 = 1+2sin(x/2)cos(x/2) = 1+sin(x) tan(x/2+π/4) = (1+sin(x)) / cos(x)      = cos(x) / (1-sin(x)) tan(x/2+π/4) = sin(x/2+π/4)/cos(x/2+π/4)  = (sin(x/2)+cos(x/2))/(cos(x/2)-sin(x/2))  = sqrt(1+sin x)/sqrt(1-sin x), and then rationalize, as with tan(x/2), above.

### double angle formulas

 formula proof sin(2x) = 2 sin(x) cos(x) delightfully simple proof using an isosceles triangle cos(2x) = cos2(x) - sin2(x)   = 1 - 2sin2(x)   = 2cos2(x) - 1 From the cos sum formula, using cos(x+x) tan(2x) = sin(2x)/cos(2x)   = 2 sin(x) cos(x) / (cos2(x)-sin2(x))   = 2 tan(x) / (1-tan2(x)) Use the sin(2x) and cos(2x) formulas, then divide the numerator and denominator by cos2(x)

### triple, quadruple, etc. angle formulas

formula proof
cos 3x = -3cos+4cos3
cos 4x = 1-8cos2+8cos4
cos 5x = 5cos-20cos3+16cos5
cos 6x = -1+18cos2-48cos4+32cos6
cos 7x = -7cos+56cos3-112cos5+64cos7
Proof: Cos of multiples of x, in terms of cos x

("x" on the right hand side is omitted for clarity)

cos((n+1)x)=2cos(x)cos(nx)-cos((n-1)x)

sin 3x = (sin) (-1+4cos2)
sin 4x = (sin) (-4cos+8cos3)
sin 5x = (sin) (1-12cos2+16cos4)
sin 6x = (sin) (6cos-32cos3+32cos5)
sin 7x = (sin) (-1+24cos2-80cos4+64cos6)
Proof: Sine of multiples of x, in terms of cos x

. . . . . . use similar recurrence relation as above:  sin(nx)=2sin((n-1)x)cos(x)-sin((n-2)x)
In fact, sin(nx)/sinx=gn(cosx) for some polynomials gn (the Chebyshev polynomials of the second kind; polynomials for cos((n+1)x) above are called Chebyshev polynomials of the first kind). They satisfy the same recurrence relation. Easy way how to see that is just to differentiate the formula cos(nx)=fn(cosx).

continued fraction for tan nx:
 tan nx = n tan x 1- (n²-1²)tan²x 3- (n²-2²)tan²x 5- ...
If n is an integer, the continued fraction terminates.  Examples here.
tan((n+1)x/2) = (sin x + sin 2x + ... + sin nx) /
(cos x + cos 2x + ... + cos nx)
. . . . . . further reading: http://www.qbyte.org/puzzles/p111s.html

### double angle formulas expressed in terms of tan(x/2)

These are variously called the "Universal" substitutions and the Weierstrass t-substitutions, where t=tan(x/2).

 formula Weierstrass t-substitution proof tan(x) = 2 tan(x/2)/(1-tan2(x/2)) tan(x) = 2t/(1-t2) from the tan-of-sum formula, using tan(x/2 + x/2) cos(x) = (1-tan2(x/2))/(1+tan2(x/2)) cos(x) = (1-t2)/(1+t2) proof cos2(x/2) = (cos(x)+1)/2 (cos(x)+1)/2 = 1/(1+t2) . . . . . .  add a row for sin^2(x/2) sin(x) = cos(x)tan(x)      = 2 tan(x/2)/(1+tan2(x/2)) sin(x) = 2t/(1+t2) by combining the two formulas above. dx = 2 cos2(x/2) dt    = (cos(x)+1) dt dx = ((1-t2)/(1+t2) + 1) dt   = 2dt/(1+t2) t = tan(x/2), so dt/dx = (1/2) sec2(x/2), or x=2 arctan(t), so dx/dt = 2/(t2+1) sec(x) - tan(x)      = (1-tan(x/2))/(1+tan(x/2)) LHS = (1+tan2(x/2))/(1-tan2(x/2)) - 2 tan(x/2)/(1-tan2(x/2))   = (1-tan(x/2))2/(1-tan2(x/2))   = (1-tan(x/2)/(1+tan(x/2)) (1-sin(x))/(1+sin(x))      = (1-tan(x/2))2/(1+tan(x/2))2      = (sec(x)-tan(x))2 (1-sin(2x))/(1+sin(2x))   = (1-2sin(x)cos(x)) / (1+2sin(x)cos(x))    = (cos(x)-sin(x))2 / (cos(x)+sin(x))2    = (1-tan(x))2 / (1+tan(x))2

### cos(x+y) etc. -- cos of sum, sin of sum, tan of sum formulas

 formula proof cos(x+y) = cos(x)cos(y) - sin(x)sin(y) geometrical proof sin(x+y) = sin(x)cos(y) + cos(x)sin(y) exponential proof of both identities at once: cos(x+y)+i sin(x+y) = ex+y = exey   =(cos(x)+i sin(x)) (cos(y)+i sin(y))   =(cos(x)cos(y)-sin(x)sin(y)) + i(sin(x)cos(y)+cos(x)sin(y)) tan(x+y) = (tan(x)+tan(y)) / (1-tan(x)tan(y)) proof and more info

### cos(x) cos(y) -- sum of cos, etc. -- converting between a sum and a product of trig functions

These Prosthaphaeresis Formulas are important for solving integrals of products of trig functions.  The trick is to convert these products into sums of functions that are easy to integrate.

 formula proof cos(a+b) cos(a-b) = cos2b - sin2a cos(x) cos(y) = cos2((x-y)/2)-sin2((x+y)/2),    where x=a+b; y=a-b;         a=(x+y)/2; b=(x-y)/2 cos(a+b) cos(a-b) = (cos a cos b - sin a sin b) (cos a cos b + sin a sin b) = cos2a cos2b - sin2a sin2b = (1-sin2a)cos2b - sin2a (1-cos2b) = cos2b - sin2a cos2b - sin2a + sin2a cos2b = cos2b - sin2a cos(x) cos(y) = cos(x+y)/2 + cos(x-y)/2 cos(x+y) = cos(x)cos(y) - sin(x)sin(y) cos(x-y) = cos(x)cos(y) + sin(x)sin(y) the proof is to add these equations together. cos(a) + cos(b) = 2 cos((a-b)/2) cos((a+b)/2) from above, where a=x+y; b=x-y; x=(a+b)/2; y=(a-b)/2; sin(x) sin(y) = cos(x-y)/2-cos(x+y)/2 cos(x-y) = cos(x)cos(y) + sin(x)sin(y) -cos(x+y) = -cos(x)cos(y) + sin(x)sin(y); the proof is to add these equations together. cos(b) - cos(a) = 2 sin((a-b)/2) sin((a+b)/2) from above, where a=x+y; b=x-y; x=(a+b)/2; y=(a-b)/2; cos(x) sin(y) = sin(x+y)/2 - sin(x-y)/2 sin(x+y) = sin(x)cos(y) + cos(x)sin(y)  -sin(x-y) = -sin(x)cos(y) + cos(x)sin(y); the proof is to add these equations together. sin(a) - sin(b) = 2 cos((a+b)/2) sin((a-b)/2) sin(a) + sin(b) = 2 cos((a-b)/2) sin((a+b)/2) from above, where a=x+y; b=x-y; x=(a+b)/2; y=(a-b)/2; 2nd equation replaces b with -b. (cos(b)-cos(a)) / (sin(a)+sin(b)) = tan((a-b)/2) (cos(b)-cos(a)) / (sin(a)-sin(b)) = tan((a+b)/2) divide the cos(b)-cos(a) and sin(a)+sin(b) identities, above. (sin(a)+sin(b)) / (cos(a)+cos(b)) = tan((a+b)/2) (sin(a)-sin(b)) / (cos(a)+cos(b)) = tan((a-b)/2) divide the sin(a)+sin(b) and cos(a)+cos(b) identities, above. cos(a−b) cos(t+u) − cos(a+b) cos(t−u) = sin(u+a) sin(b−t) − sin(u−a) sin(b+t) Proof: cosine and sine product.  This identity is used in the proof of Urquhart's Theorem.

### generalized phase shift

 formula proof a cos(x-A) + b cos(x-B) = r cos(x-θ), where      r = sqrt(a2 + b2 + 2ab cos(A-B)), and      θ = atan2(a cos A + b cos B, a sin A + b sin B) y = a cos(x-A) + b cos(x-B)   = cos x (a cos A + b cos B) + sin x (a sin A + b sin B)   = cos x (r cos θ) + sin x (r sin θ)   = r cos(x-θ) generalized phase shift explains this more completely.

### "geometric progression" identities

I call these "geometric progression" identities, because if a, b, c are in geometric progression then a/b = b/c...

 formula proof (1+sin(x)) / cos(x) = cos(x) / (1-sin(x)) 1-sin2(x)=cos2(x) (1+sin(x))(1-sin(x))=cos2(x) (1+sin(x))/cos(x)=cos(x)/(1-sin(x)) Note: this equals tan(x/2+π/4) -- see the half angle formula section of this page. (1+sin(x)) / cos(x) = cos(x) / (1-sin(x))  = (1+cos(x)+sin(x)) / (1+cos(x)-sin(x)) Linear Combination of Fractions: If A/B = C/D, then (rA+sC)/(rB+sD) = C/D (proof) (1-cos(x)) / sin(x) = sin(x) / (1+cos(x))  = (1+sin(x)-cos(x)) / (1+sin(x)+cos(x)) Note: this equals tan(x/2) -- half angle. and Linear Combination of Fractions. (sec(x)-1) / tan(x) = tan(x) / (sec(x)+1))  = (sec(x)+tan(x)-1) / (sec(x)+tan(x)+1) similar to the others; and Linear Combination of Fractions. (sec(x)-tan(x)) = 1 / (sec(x)+tan(x))  = (1+sec(x)-tan(x))/(1+sec(x)+tan(x)) ln Isec(x)-tan(x)I = -ln Isec(x)+tan(x)I similar to the others; and Linear Combination of Fractions; Watch out for equivalent solutions of integrals! (csc(x)-1) / cot(x) = cot(x) / (csc(x)+1)) Any difference of squares can be turned into a geometric progression identity! A2-B2=C2 —> (A-B)/C = C/(A+B) (csc(x)-cot(x)) = 1/ (csc(x)+cot(x))

### Triangle identities -- sin(arccos(x)), etc.

I've noticed some students have trouble understanding the meaning of sin(arccos(x)), so I'll dissect it for you in this paragraph.  Let an angle, A, represented by the red line in the triangles, below, have a cosine of x.  That is cos(A)=x, or in other words, A=arccos(x).  So then sin(arccos(x))=sin(A).  To rephrase the whole thing, sin(arccos(x)) is saying "what is the sine of the angle whose cosine is x?"

formula proof
sin(arccos(x)) = ±sqrt(1-x2
 1 sqrt(1-x2) x
tan(arccos(x)) = ±sqrt(1-x2)/x
cos(arcsin(x)) = ±sqrt(1-x2
 1 x sqrt(1-x2)
tan(arcsin(x)) = ±x/sqrt(1-x2
cos(arctan(x)) = ±1/sqrt(1+x2)
 sqrt(1+x2) x 1
sin(arctan(x)) = ±x/sqrt(1+x2)
sin(arcsec(x)) = ±sqrt(x2-1)/x
 x sqrt(x2-1) 1
tan(arcsec(x)) = ±sqrt(x2-1)
cos(arccsc(x)) = ±sqrt(x2-1)/x
 x 1 sqrt(x2-1)
tan(arccsc(x)) = ±1/sqrt(x2-1)
cos(arccot(x)) = ±x/sqrt(1+x2)
 sqrt(1+x2) 1 x
sin(arccot(x)) = ±1/sqrt(1+x2)

### Euler identities and hyperbolic functions

 formula proof ex = 1 + x/1! + x2/2! + x3/3! + ...  cos(x) = 1  - x2/2! + x4/4! - x6/6! + ...  sin(x) = x/1! - x3/3! + x5/5! - ...  cosh(x) = 1  + x2/2! + x4/4! + x6/6! + ...  sin(x) = x/1! + x3/3! + x5/5! + ... Euler's formula for ex see Hyperbolic Functions cosh(x) = (ex+e-x)/2 sinh(x) = (ex-e-x)/2 tanh(x) = (ex-e-x)/(ex+e-x) eix = cos(x) + i sin(x) = cosh(ix) + sinh(ix) ex = cos(ix) - i sin(ix) = cosh(x) + sinh(x) Euler's formula for ex, then replace x with -ix cosh(ix) = cos(x) sinh(ix) = i sin(x) cos(ix) = cosh(x) sin(ix) = i sinh(x) arcsinh(x) = ln(x + sqrt(x² + 1)) arccosh(x) = ln(x + sqrt(x² - 1)) arctanh(x) = (1/2) (ln(1+x) - ln(1-x)) Hyperbolic Functions cosh²(x) - sinh²(x) = 1

If additional trig identities come to light (send me an email if yours is missing from this page!) then I'll add them here, until I can find a "home" for them in the other sections of this page.

 formula proof tan(x-y) + tan(y-z) + tan(z-x) = tan(x-y) *tan(y-z)*tan(z-x) (proof)

### Tan-1(1/2) + Tan-1(1/5) + Tan-1(1/8) = π/4

tan-1(1/2) + tan-1(1/5) + tan-1(1/8)=π/4.
See the diagram to the right.

### Gregory's Formula for Arctan, Machin's formula for π/4

arctan x = x - x3/3 + x5/5 - x7/7 + x9/9 - ...
where -1 < x < 1

This was discovered in 1672 by James Gregory (1638-1675).  It's a useful formula because it converges quickly when x is small.  π/4 is arctan 1, so it provides a formula for π/4, which is

π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - ...

(Gregory never explicitly wrote down this formula but another famous mathematician of the time, Gottfried Leibniz (1646-1716), mentioned it in print first in 1682, and so this special case of Gregory's series is usually called Leibnitz Formula for π.)

This formula converges very slowly.  It is better to note that

π/4 = arctan(1) = 4 arctan(1/5) - arctan(1/239)

(This formula is called Machin's formula, discovered In 1706 by John Machin (1680-1752).)

 Derivation of the formula: Remember that tan x+y = (tan x + tan y) / (1 - tan x tan y) so tan 2x = (2 tan x) / (1 - tan2 x) let z be arctan 1/5. tan 2z = (2 tan z) / (1 - tan2 z) = (2/5)/(1-1/25) = (2/5)/(24/25) = 50/120 = 5/12 Now let w be arctan 5/12 tan 2w = (2 tan 2) / (1 - tan2 w) = (5/6)/(1-25/144) = (5/6)/(119/144) = 720/714 = 120/119 So arctan 120/119 is an angle 4 times as large as arctan 1/5. Now let x be arctan 120/119 and y be arctan -1/239 tan x+y = (120/119 - 1/239) / (1 + (120/119)(1/239)) = (120*239-119) / (28441+120) = 28561/28561 = 1 In other words, the sum of 4 arctan 1/5 and arctan -1/239 is equal to the arctan of 1, i.e. π/4. How can you discover your own group of small angles with rational tangents whose sum is π/4? In other words, if you have an angle arctan a, what is the value of b for which arctan a + arctan b = π/4? Let x = arctan a, and let y = arctan b, and let x + y = π/4 tan x+y = (tan x + tan y) / (1 - tan x tan y) 1 = (a + b) / (1 - ab) a + b = 1 - ab b + ab = 1 - a b(1+a) = (1-a) b = (1-a) / (1+a) After John Machin calculated that 4 arctan 1/5 is equal to arctan 120/119, then he presumably set a=120/119 and used this formula for b b = (1-a) / (1 + a) = (1 - 120/119) / (1 + 120/119) = (-1/119) / (239/119) = -1/239 Here's another one that adds up to π/4: 3 arctan 1/4 + arctan 5/99

### Internet references

SOS Math: Table of Trigonometric Identities

Gottfried Wilhelm Leibniz (b. 1646, d. 1716) was a German philosopher, mathematician, and logician who is probably most well known for having invented the differential and integral calculus (independently of Sir Isaac Newton).

Wikipedia: List of trigonometric identities

Mathworld article: Leibniz Series.

Mathworld: Trigonometric Addition Formulas, including formulas for Double-Angle, Half-Angle, and Multiple-Angle

Mathworld: Prosthaphaeresis and Werner formulas, which convert sums of sines and cosines into products of sines and cosines, and vice-versa.

### Related Pages in this website

Special Angles

Sin or Cos 3x, 4x, etc. -- trig functions of any multiple of an angle.

d/dx (sin x) = cos x, in the calculus section of this website

Weierstrass t-substitution is a clever trig substitution that lets you solve the kind of integrals that naturally come up in the polar form of certain functions. The substitution is t=tan(x/2).

Hyperbolic Functions -- sinh(x) and cosh(x), which, together with exp(x) and the circular functions sin(x) and cos(x) form a family of functions.

Phase shift by an arbitrary angle

Table of Integrals -- derivations of various special integrals requires extensive use of the trig identities on this page.

The webmaster and author of this Math Help site is Graeme McRae.