
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
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) cofunctions
sin(x) = cos(π/2x) tan(x) = cot(π/2x) sec(x) = csc(π/2x) cos(x) = sin(π/2x) cot(x) = tan(π/2x) csc(x) = sec(π/2x)
phase shift
(see also the generalized phase shift)
sin(x) = cos(xπ/2) = sin(xπ) = cos(x+π/2) = sin(x+2π) csc(x) = sec(xπ/2) = csc(xπ) = sec(x+π/2) = csc(x+2π) tan(x) = cot(x±π/2) = tan(x+π) tan(x+π/4) = (1+tan(x))/(1tan(x))
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.
formula proof sin^{2}x + cos^{2}x = 1 Apply the Pythagorean Theorem to this triangle:
1
sin xcos x
tan^{2}x + 1 = sec^{2}x Start with the first formula, then divide every term by cos^{2}x. 1 + cot^{2}x = csc^{2}x Start over, dividing by sin^{2}x.
formula proof sin(x/2) = ±sqrt((1cos(x))/2) cos(x) = cos(x/2 + x/2) = cos^{2}(x/2)  sin^{2}(x/2)
cos(x) = 1  2sin^{2}(x/2), then solve for sin(x/2);
cos(x) = 2cos^{2}(x/2)  1, then solve for cos(x/2).cos(x/2) = ±sqrt((1+cos(x))/2) tan(x/2) = (1cos(x))/sin(x)
= sin(x)/(1+cos(x))tan(x/2) = sin(x/2) / cos(x/2)
= sqrt(1cos x)/sqrt(1+cos x)
= (1cos x)/(sin x), by rationalizing the denominator
= (sin x)/(1+cos x), by rationalizing the numeratorsin(x/2)cos(x/2) = ±sqrt(1sin(x)),
cos(x/2)sin(x/2) = ±sqrt(1sin(x))(sin(x/2)cos(x/2))^{2} = 12sin(x/2)cos(x/2) = 1sin(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) / (1sin(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(1sin x),
and then rationalize, as with tan(x/2), above.
formula proof sin(2x) = 2 sin(x) cos(x) delightfully simple proof using an isosceles triangle cos(2x) = cos^{2}(x)  sin^{2}(x)
= 1  2sin^{2}(x)
= 2cos^{2}(x)  1From the cos sum formula, using cos(x+x) tan(2x) = sin(2x)/cos(2x)
= 2 sin(x) cos(x) / (cos^{2}(x)sin^{2}(x))
= 2 tan(x) / (1tan^{2}(x))Use the sin(2x) and cos(2x) formulas, then divide the numerator and denominator by cos^{2}(x)
formula proof cos 3x = 3cos+4cos^{3}
cos 4x = 18cos^{2}+8cos^{4}
cos 5x = 5cos20cos^{3}+16cos^{5}
cos 6x = 1+18cos^{2}48cos^{4}+32cos^{6}
cos 7x = 7cos+56cos^{3}112cos^{5}+64cos^{7}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((n1)x)
sin 3x = (sin) (1+4cos^{2})
sin 4x = (sin) (4cos+8cos^{3})
sin 5x = (sin) (112cos^{2}+16cos^{4})
sin 6x = (sin) (6cos32cos^{3}+32cos^{5})
sin 7x = (sin) (1+24cos^{2}80cos^{4}+64cos^{6})Proof: Sine of multiples of x, in terms of cos x . . . . . . use similar recurrence relation as above: sin(nx)=2sin((n1)x)cos(x)sin((n2)x)
In fact, sin(nx)/sinx=g_{n}(cosx) for some polynomials g_{n} (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)=f_{n}(cosx).continued fraction for tan nx:
tan nx = n tan x 1
(n²1²)tan²x 3
(n²2²)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
These are variously called the "Universal" substitutions and the Weierstrass tsubstitutions, where t=tan(x/2).
formula Weierstrass
tsubstitutionproof tan(x) = 2 tan(x/2)/(1tan^{2}(x/2)) tan(x) = 2t/(1t^{2}) from the tanofsum formula, using tan(x/2 + x/2) cos(x) = (1tan^{2}(x/2))/(1+tan^{2}(x/2)) cos(x) = (1t^{2})/(1+t^{2}) proof cos^{2}(x/2) = (cos(x)+1)/2 (cos(x)+1)/2 = 1/(1+t^{2}) . . . . . . add a row for sin^2(x/2) sin(x) = cos(x)tan(x)
= 2 tan(x/2)/(1+tan^{2}(x/2))sin(x) = 2t/(1+t^{2}) by combining the two formulas above. dx = 2 cos^{2}(x/2) dt
= (cos(x)+1) dtdx = ((1t^{2})/(1+t^{2}) + 1) dt
= 2dt/(1+t^{2})t = tan(x/2), so dt/dx = (1/2) sec^{2}(x/2), or
x=2 arctan(t), so dx/dt = 2/(t^{2}+1)sec(x)  tan(x)
= (1tan(x/2))/(1+tan(x/2))LHS = (1+tan^{2}(x/2))/(1tan^{2}(x/2))  2 tan(x/2)/(1tan^{2}(x/2))
= (1tan(x/2))^{2}/(1tan^{2}(x/2))
= (1tan(x/2)/(1+tan(x/2))(1sin(x))/(1+sin(x))
= (1tan(x/2))^{2}/(1+tan(x/2))^{2}
= (sec(x)tan(x))^{2}(1sin(2x))/(1+sin(2x))
= (12sin(x)cos(x)) / (1+2sin(x)cos(x))
= (cos(x)sin(x))^{2} / (cos(x)+sin(x))^{2}
= (1tan(x))^{2} / (1+tan(x))^{2}
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) = e^{x+y} = e^{x}e^{y}
=(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)) / (1tan(x)tan(y)) proof and more info
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(ab) = cos^{2}b  sin^{2}a
cos(x) cos(y) = cos^{2}((xy)/2)sin^{2}((x+y)/2),
where x=a+b; y=ab;
a=(x+y)/2; b=(xy)/2cos(a+b) cos(ab) =
(cos a cos b  sin a sin b) (cos a cos b + sin a sin b) =
cos^{2}a cos^{2}b  sin^{2}a sin^{2}b =
(1sin^{2}a)cos^{2}b  sin^{2}a (1cos^{2}b) =
cos^{2}b  sin^{2}a cos^{2}b  sin^{2}a + sin^{2}a cos^{2}b =
cos^{2}b  sin^{2}acos(x) cos(y) = cos(x+y)/2 + cos(xy)/2 cos(x+y) = cos(x)cos(y)  sin(x)sin(y)
cos(xy) = cos(x)cos(y) + sin(x)sin(y)
the proof is to add these equations together.cos(a) + cos(b) = 2 cos((ab)/2) cos((a+b)/2) from above, where a=x+y; b=xy; x=(a+b)/2; y=(ab)/2; sin(x) sin(y) = cos(xy)/2cos(x+y)/2 cos(xy) = 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((ab)/2) sin((a+b)/2) from above, where a=x+y; b=xy; x=(a+b)/2; y=(ab)/2; cos(x) sin(y) = sin(x+y)/2  sin(xy)/2 sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
sin(xy) = 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((ab)/2)
sin(a) + sin(b) = 2 cos((ab)/2) sin((a+b)/2)from above, where a=x+y; b=xy; x=(a+b)/2; y=(ab)/2;
2nd equation replaces b with b.(cos(b)cos(a)) / (sin(a)+sin(b)) = tan((ab)/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((ab)/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.
formula proof a cos(xA) + b cos(xB) = r cos(xθ), where
r = sqrt(a^{2} + b^{2} + 2ab cos(AB)), and
θ = atan2(a cos A + b cos B, a sin A + b sin B)y = a cos(xA) + b cos(xB)
= 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(xθ)
generalized phase shift explains this more completely.
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) / (1sin(x)) 1sin^{2}(x)=cos^{2}(x)
(1+sin(x))(1sin(x))=cos^{2}(x)
(1+sin(x))/cos(x)=cos(x)/(1sin(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) / (1sin(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)(1cos(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)Isimilar 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!
A^{2}B^{2}=C^{2} —> (AB)/C = C/(A+B)(csc(x)cot(x)) = 1/ (csc(x)+cot(x))
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(1x^{2})
1
sqrt(1x^{2})x
tan(arccos(x)) = ±sqrt(1x^{2})/x cos(arcsin(x)) = ±sqrt(1x^{2})
1
xsqrt(1x^{2})
tan(arcsin(x)) = ±x/sqrt(1x^{2}) cos(arctan(x)) = ±1/sqrt(1+x^{2})
sqrt(1+x^{2})
x1
sin(arctan(x)) = ±x/sqrt(1+x^{2}) sin(arcsec(x)) = ±sqrt(x^{2}1)/x
x
sqrt(x^{2}1)1
tan(arcsec(x)) = ±sqrt(x^{2}1) cos(arccsc(x)) = ±sqrt(x^{2}1)/x
x
1sqrt(x^{2}1)
tan(arccsc(x)) = ±1/sqrt(x^{2}1) cos(arccot(x)) = ±x/sqrt(1+x^{2})
sqrt(1+x^{2})
1x
sin(arccot(x)) = ±1/sqrt(1+x^{2})
See also,
Hyperbolic Functions
formula proof e^{x} = 1 + x/1! + x^{2}/2! + x^{3}/3! + ...
cos(x) = 1  x^{2}/2! + x^{4}/4!  x^{6}/6! + ...
sin(x) = x/1!  x^{3}/3! + x^{5}/5!  ...
cosh(x) = 1 + x^{2}/2! + x^{4}/4! + x^{6}/6! + ...
sin(x) = x/1! + x^{3}/3! + x^{5}/5! + ...Euler's formula for e^{x}
see Hyperbolic Functions cosh(x) = (e^{x}+e^{x})/2
sinh(x) = (e^{x}e^{x})/2
tanh(x) = (e^{x}e^{x})/(e^{x}+e^{x})e^{ix} = cos(x) + i sin(x) = cosh(ix) + sinh(ix)
e^{x} = cos(ix)  i sin(ix) = cosh(x) + sinh(x)Euler's formula for e^{x},
then replace x with ixcosh(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(1x))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(xy) + tan(yz) + tan(zx) = tan(xy) *tan(yz)*tan(zx) (proof)
tan^{1}(1/2) + tan^{1}(1/5) + tan^{1}(1/8)=π/4.
See the diagram to the right.
This was discovered in 1672 by James Gregory (16381675). 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 isarctan x = x  x^{3}/3 + x^{5}/5  x^{7}/7 + x^{9}/9  ...
where 1 < x < 1
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 (16461716), 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 (16801752).)
Derivation of the formula:
Remember that tan x+y = (tan x + tan y) / (1  tan x tan y) let z be arctan 1/5. Now let w be arctan 5/12 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 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 tan x+y = (tan x + tan y) / (1  tan x tan y) 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 = (1a) / (1 + a) Here's another one that adds up to π/4: 3 arctan 1/4 + arctan 5/99 
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 DoubleAngle, HalfAngle, and MultipleAngle,
Mathworld: Prosthaphaeresis and Werner formulas, which convert sums of sines and cosines into products of sines and cosines, and viceversa.
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 tsubstitution 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.