Trigonometric Equivalences

In Trigonometry, and again in Calculus, you will run into numerous cases in which special conversion rules, or identities, which I call "equivalences" can be used.  To motivate you to read this page, here's an example of a problem that needs two trigonometric identities:

Solve for x:

cos(x/2) = -sin(x - p/2)

The first equivalence I need is

cos(x) = -sin(x-p/2)

This is one of a whole family of equivalences involving adding and subtracting a right angle (p/2).  The next one I need comes from the Law of Cosines.  It is

cos x = 2 cos²(x/2) - 1

Using these facts, you can solve the equation:

cos(x/2) = cos(x)
cos(x) - cos(x/2) = 0
2cos²(x/2)-1 - cos(x/2) = 0
2cos²(x/2) - cos(x/2) - 1 = 0

Using the quadratic formula,

cos(x/2) = 1/4 +/- sqrt(1+8)/4
cos(x/2) = 1/4 +/- 3/4
cos(x/2) = 1 or -1/2
x/2=0 or x/2=2p/3
x=0 or x=4p/3

And of course, there are infinitely many other answers, which you can get using the periodicity of the cos function.

Here are some of the many equivalences, with proofs or explanations where needed:

Adding p/2 (a right angle) in sin and cos

cos(x) = -sin(x-p/2)
cos(x) = sin(x+p/2)
sin(x) = -sin(x-p)
cos(x) = -cos(x-p)
sin(x) = sin(x+2p)
cos(x) = cos(x+2p)

Subtracting from p/2 (a right angle) in sin and cos

sin(x) = cos(p/2-x)
cos(x) = sin(p/2-x)
sin(x) = sin(p-x)
cos(x) = -cos(p-x)
sin(x) = -sin(-x) = -sin(2p-x)
cos(x) = cos(-x) = cos(2p-x)

Common Angles

I'll use the sin.  To get cos, use one of the equivalences, above.

sin(0) = 0
sin(p/6) = sqrt(1/4) = 1/2
sin(p/4) = sqrt(1/2)
sin(p/3) = sqrt(3/4)
sin(p/2) = 1
Here are a whole lot more Special Angles.

Half angle formulas

Half Angles and Sin² and Cos²

sin²x + cos²x = 1
sin(x) = sqrt(1 - cos²(x))
sin(x) = 2(sin x/2)(cos x/2)   --  delightfully simple proof
cos(x) = 1 - 2sin²(x/2) = 2cos²(x/2) - 1  --  from the Law of Cosines 
sin(x/2) = ± sqrt((1-cos x)/2) -- from the statement above,
cos(x/2) = ± sqrt((1+cos x)/2) -- from the statement above.

tan²x + 1 = sec²x -- you can see this by starting with sin²x + cos²x = 1 and then dividing every term by cos²x.
1 + cot²x = csc²x -- same method, except divide by sin²x

If you start from sec²a - tan²a = 1, then using the "difference of 2 squares",
(sec a + tan a)(sec a - tan a) = 1, so it follows that
ln Isec a + tan aI = -ln Isec a - tan aI, which is a very cute result!

Half Angle formula for tan x/2

tan(x/2) = (1-cos(x))/sin(x) = sin(x)/(1+cos(x)) -- proof follows:

by first using the sin(x/2) and cos(x/2) identities, above, and then rationalizing the denominator,
tan(x/2) = sin(x/2) / cos(x/2) = sqrt(1-cos x)/sqrt(1+cos x) = (1-cos x) / (sin x)

To see that (1-cos(x))/sin(x) = sin(x)/(1+cos(x)), you can either observe that sin²(x)=(1-cos(x))(1+cos(x)), or else you can start with the sin(x/2) and cos(x/2) identities, as above, and rationalize the numerator instead of the denominator.

Double Angles

cos 2x = cos²x - sin²x = 1 - 2sin²x = 2cos²x - 1
sin 2x = 2 sin x cos x
tan 2x = sin 2x / cos 2x = 2sin x cos x / (cos² x - sin² x) = 2tan x / (1 - tan² x)

Click here for larger multiples, like cos 3x, cos 4x, etc.

(1 - sin 2x) / (1 + sin 2x) = (1 - tan x)²/(1 + tan)²

Proof: LHS = (1 - 2 sin x cos x) / (1 + 2 sin x cos x)
 = (cos²x - 2 sin x cos x + sin²x) / (cos²x + 2 sin x cos x + sin²x)
 = (cos x - sin x)² / (cos x + sin x)²
 = (1 - tan x)² / (1 + tan x)²

Sin x - Cos x, or Cos x - Sin x

sin 2x = 2 sin x cos x,    from the double-angle formula, above
sin 2x = 1 - (sin² x - 2 sin x cos x + cos² x),    because sin²x+cos²x=1
sin 2x = 1 - (sin x - cos x)²
(sin x - cos x)² = 1 - sin 2x
sin x - cos x = ± sqrt(1 - sin 2x), and
cos x - sin x = ± sqrt(1 - sin 2x)

Cos x+y and Sin x+y and Tan x+y

cos x+y = cos x cos y - sin x sin y   (proof)
sin x+y = 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(a+b) cos(a-b) = cos²b - sin²a

Proof:  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-y - cos x+y = 2 sin x sin y -- expand the cosine sums, and it's clear.
sin x+y - sin x-y = 2 cos x sin y

sin a + sin b,  cos a - cos b,  cos a + cos b

sin(a) + sin(b) = 2 cos((a-b)/2) sin((a+b)/2)

Proof: sin(a) + sin(b) = 
2 (cos a/2 sin a/2 +  sin b/2 cos b/2) =
2 (( cos² b/2 + sin² b/2) cos a/2 sin a/2 +  (cos² a/2 + sin² a/2) sin b/2 cos b/2) =
2 ( cos² b/2 cos a/2 sin a/2 + cos² a/2 cos b/2 sin b/2 + sin² a/2 sin b/2 cos b/2 + sin² b/2 sin a/2 cos a/2) =
2 (cos a/2 cos b/2 + sin a/2 sin b/2) (sin a/2 cos b/2 + cos a/2 sin b/2) =
2 cos((a-b)/2) sin((a+b)/2) 

Note: this works for sin(a) - sin(b) because this can be rewritten sin(a) + sin(-b).

cos(a) - cos(b) = -2 sin((a-b)/2) sin((a+b)/2)

Proof: cos(a) - cos(b) =
(cos² a/2 - sin² a/2) - (cos² b/2 - sin² b/2)
(sin² b/2 + cos² b/2)(cos² a/2 - sin² a/2) - (sin² a/2 + cos² a/2)(cos² b/2 - sin² b/2)
(sin² b/2 cos² a/2 - sin² b/2 sin² a/2 + cos² b/2 cos² a/2 - cos² b/2 sin² a/2)
    - (sin² a/2 cos² b/2 - sin² a/2 sin² b/2 + cos² a/2 cos² b/2 - cos² a/2 sin² b/2)
-2 (sin² a/2 cos² b/2 - cos² a/2 sin² b/2) =
-2 (sin a/2 cos b/2 - cos a/2 sin b/2) (sin a/2 cos b/2 + cos a/2 sin b/2) =
-2 sin((a-b)/2) sin((a+b)/2) 

cos(a) + cos(b) = 2 cos((a-b)/2) cos((a+b)/2)

Proof: cos(a) + cos(b) =
(cos² a/2 - sin² a/2) + (cos² b/2 - sin² b/2)
(sin² b/2 + cos² b/2)(cos² a/2 - sin² a/2) + (sin² a/2 + cos² a/2)(cos² b/2 - sin² b/2)
(sin² b/2 cos² a/2 - sin² b/2 sin² a/2 + cos² b/2 cos² a/2 - cos² b/2 sin² a/2)
    + (sin² a/2 cos² b/2 - sin² a/2 sin² b/2 + cos² a/2 cos² b/2 - cos² a/2 sin² b/2)
2 (cos² a/2 cos² b/2 - sin² a/2 sin² b/2) 
2 (cos a/2 cos b/2 + sin a/2 sin b/2) (cos a/2 cos b/2 - sin a/2 sin b/2) 
2 cos((a-b)/2) cos((a+b)/2)

(cos(a)-cos(b)) / (sin(a)+sin(b)) = tan((b-a)/2)

Proof: Using two of the identities, above, the LHS is
- 2 sin((a-b)/2) sin((a+b)/2) / (2 cos((a-b)/2) sin((a+b)/2)), which equals the RHS.

(sin(a)+sin(b)) / (cos(a)+cos(b)) = tan((a+b)/2)

Proof: Using two of the identities, above, the LHS is
2 cos((a-b)/2) sin((a+b)/2) / (2 cos((a-b)/2) cos((a+b)/2)), which equals the RHS.

Half Angle Formulas Expressed in Terms of Tan

tan(x) = 2 tan(x/2)/(1-tan²(x/2)) -- from the tan-of-sum formula

cos(x) = (1-tan²(x/2))/(1+tan²(x/2)) -- proof

sin(x) = cos(x)tan(x) = 2 tan(x/2)/(1+tan²(x/2)) -- by combining the two formulas above.

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

sec(x) - tan(x) = (1-tan(x/2))/(1+tan(x/2))

Proof: 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))

tan(x) = 2sin²x / (2 sin x cos x) = (1-cos²x+sin²x)/sin(2x) = (1-cos(2x))/sin(2x)

tan(x/2) = (1-cos(x))/sin(x) = sin(x)/(1+cos(x)) -- from the result above and sin²(x) = (1-cos(x))(1+cos(x))

Continued Fractions for Tan of multiples of x

If n is an integer, the continued fraction terminates.  Examples here.

Sin 2x - Sin 2y = 2 cos(x+y) sin(x-y)

Proof:
sin 2x - sin 2y =
2 cos x sin x - 2 cos y sin y =
2 (cos x sin x - cos y sin y) =
2 (cos x sin x (cos² y + sin² y)  -  cos y sin y (cos² x + sin² x) ) =
2 (cos x sin x cos² y - cos y sin y cos² x - cos y sin y sin² x + cos x sin x sin² y) =
2 (cos x cos y - sin x sin y) (sin x cos y - cos x sin y) =
2 cos(x+y) sin(x-y)

cos(a−b) cos(t + u) − cos(a +b) cos(t−u) = sin(u +a) sin(b−t) − sin(u−a) sin(b + t)

Just expand the following expression to show that it is zero:

cos(a-b) cos(t + u) - cos(a +b) cos(t-u) - sin(u +a) sin(b-t) + sin(u-a) sin(b + t)

=(cos a cos b + sin a sin b)(cos t cos u - sin t sin u)
-(cos a cos b - sin a sin b)(cos t cos u + sin t sin u)
-(sin u cos a + cos u sin a)(sin b cos t - cos b sin t)
+(sin u cos a - cos u sin a)(sin b cos t + cos b sin t)

= cos a cos b cos t cos u + sin a sin b cos t cos u - cos a cos b sin t sin u - sin a sin b sin t sin u
- cos a cos b cos t cos u + sin a sin b cos t cos u - cos a cos b sin t sin u + sin a sin b sin t sin u
- cos a sin b cos t sin u - sin a sin b cos t cos u + cos a cos b sin t sin u + sin a cos b sin t cos u
+ cos a sin b cos t sin u - sin a sin b cos t cos u + cos a cos b sin t sin u - sin a cos b sin t cos u

= 0

This identity is used in the proof of Urquhart's Theorem.

Tan^-1(1/2) + Tan^-1(1/5) + Tan^-1(1/8) = p/4

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

Gregory's Formula for Arctan, Machin's formula for p/4

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

p/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 p.)

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

p/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).)

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).

Mathworld article: Leibniz Series.

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

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.

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

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