Every formula you need for double and half angles — with derivations, the three forms of cos(2A), power-reducing and product-to-sum formulas, worked examples, and calculus applications.
Double-Angle
sin(2A) = 2 sin(A) cos(A)
cos(2A) = cos^2(A) - sin^2(A)
cos(2A) = 2 cos^2(A) - 1
cos(2A) = 1 - 2 sin^2(A)
tan(2A) = 2 tan(A) / (1 - tan^2(A))
Half-Angle
sin(A/2) = plus or minus sqrt((1 - cos A) / 2)
cos(A/2) = plus or minus sqrt((1 + cos A) / 2)
tan(A/2) = (1 - cos A) / sin A
tan(A/2) = sin A / (1 + cos A)
Power-Reducing
sin^2(A) = (1 - cos(2A)) / 2
cos^2(A) = (1 + cos(2A)) / 2
Every double-angle formula comes from the sum formulas with B set equal to A. You never need to memorize a separate set of formulas — just derive them from what you already know.
Start with the sine sum formula: sin(A + B) = sin(A) cos(B) + cos(A) sin(B). Set B equal to A:
sin(A + A) = sin(A) cos(A) + cos(A) sin(A)
sin(2A) = 2 sin(A) cos(A)
This is one of the most useful identities in calculus. It appears constantly in integration problems involving products of sin and cos.
Start with the cosine sum formula: cos(A + B) = cos(A) cos(B) - sin(A) sin(B). Set B equal to A:
cos(A + A) = cos(A) cos(A) - sin(A) sin(A)
Form 1: cos(2A) = cos^2(A) - sin^2(A)
Now substitute the Pythagorean identity sin^2(A) = 1 - cos^2(A) into Form 1:
cos(2A) = cos^2(A) - (1 - cos^2(A))
cos(2A) = cos^2(A) - 1 + cos^2(A)
Form 2: cos(2A) = 2 cos^2(A) - 1
Or substitute cos^2(A) = 1 - sin^2(A) into Form 1:
cos(2A) = (1 - sin^2(A)) - sin^2(A)
Form 3: cos(2A) = 1 - 2 sin^2(A)
When to use each form
Form 1 — use when you see both sin^2 and cos^2 and need to factor a difference of squares
Form 2 — use when you want to eliminate sin^2, or to derive the half-angle formula for cos
Form 3 — use when you want to eliminate cos^2, or to derive the half-angle formula for sin
Start with the tangent sum formula: tan(A + B) = (tan A + tan B) / (1 - tan A tan B). Set B equal to A:
tan(A + A) = (tan A + tan A) / (1 - tan A tan A)
tan(2A) = 2 tan(A) / (1 - tan^2(A))
Undefined when tan^2(A) = 1, i.e., when tan A = plus or minus 1, which is at A = 45 degrees, 135 degrees, etc.
Example 1: Find sin(2A) and cos(2A) given sin(A) = 3/5 with A in Quadrant II
Step 1: Find cos(A) using the Pythagorean identity.
sin^2(A) + cos^2(A) = 1
(3/5)^2 + cos^2(A) = 1
cos^2(A) = 1 - 9/25 = 16/25
cos(A) = -4/5 (negative because A is in Quadrant II)
Step 2: Apply the double-angle formulas.
sin(2A) = 2 sin(A) cos(A) = 2 (3/5)(-4/5) = -24/25
cos(2A) = cos^2(A) - sin^2(A) = 16/25 - 9/25 = 7/25
Answer: sin(2A) = -24/25 and cos(2A) = 7/25
Note: 2A is in Quadrant IV because A is between 90 and 180 degrees, so 2A is between 180 and 360 degrees. Sin is negative and cos is positive there — consistent.
Example 2: Find the exact value of sin(120 degrees) using a double-angle formula
Write 120 as 2 times 60:
sin(120) = sin(2 times 60)
= 2 sin(60) cos(60)
= 2 (sqrt(3)/2)(1/2)
= sqrt(3)/2 (matches the known unit-circle value)
Example 3: Simplify 2 sin(3x) cos(3x)
Recognize the pattern sin(2A) = 2 sin(A) cos(A) with A = 3x:
2 sin(3x) cos(3x) = sin(2 times 3x)
= sin(6x)
Example 4: Simplify cos^2(5x) - sin^2(5x)
Recognize Form 1 of cos(2A) = cos^2(A) - sin^2(A) with A = 5x:
cos^2(5x) - sin^2(5x) = cos(2 times 5x)
= cos(10x)
Half-angle formulas come from solving the double-angle formulas for sin^2 and cos^2, then replacing A with A/2. This means the derivation of half-angle formulas requires the double-angle formulas, so learn those first.
Start from Form 3 of cos(2A): cos(2A) = 1 - 2 sin^2(A). Solve for sin^2(A):
2 sin^2(A) = 1 - cos(2A)
sin^2(A) = (1 - cos(2A)) / 2
Now replace A with A/2 (so 2A becomes A):
sin^2(A/2) = (1 - cos A) / 2
sin(A/2) = plus or minus sqrt((1 - cos A) / 2)
Start from Form 2 of cos(2A): cos(2A) = 2 cos^2(A) - 1. Solve for cos^2(A):
2 cos^2(A) = 1 + cos(2A)
cos^2(A) = (1 + cos(2A)) / 2
Replace A with A/2:
cos^2(A/2) = (1 + cos A) / 2
cos(A/2) = plus or minus sqrt((1 + cos A) / 2)
Tangent has three equivalent half-angle forms. The ratio form avoids the plus-or-minus ambiguity:
tan(A/2) = plus or minus sqrt((1 - cos A) / (1 + cos A))
Derived from dividing the half-angle formulas for sin and cos.
tan(A/2) = sin(A) / (1 + cos A)
Multiply numerator and denominator of the sqrt form by sqrt(1 + cos A) / sqrt(1 + cos A).
tan(A/2) = (1 - cos A) / sin(A)
Multiply numerator and denominator of the sqrt form by sqrt(1 - cos A) / sqrt(1 - cos A).
Advantage of the ratio forms
The formulas tan(A/2) = sin(A) / (1 + cos A) and tan(A/2) = (1 - cos A) / sin(A) automatically give the correct sign without needing to check the quadrant of A/2. Use these whenever possible.
The sign is determined by the quadrant of A/2 — not the quadrant of A. Work through these steps every time:
A in Quadrant I (0 to 90 deg) → A/2 in 0 to 45 deg → Q1 → sin, cos, tan all positive
A in Quadrant II (90 to 180 deg) → A/2 in 45 to 90 deg → Q1 → all positive
A in Quadrant III (180 to 270 deg) → A/2 in 90 to 135 deg → Q2 → sin positive, cos negative
A in Quadrant IV (270 to 360 deg) → A/2 in 135 to 180 deg → Q2 → sin positive, cos negative
Example 1: Find the exact value of sin(22.5 degrees)
Write 22.5 as 45/2, so A = 45 degrees:
sin(22.5) = sin(45/2) = plus or minus sqrt((1 - cos 45) / 2)
A/2 = 22.5 degrees is in Quadrant I, so sin is positive:
= sqrt((1 - sqrt(2)/2) / 2)
= sqrt((2 - sqrt(2)) / 4)
= sqrt(2 - sqrt(2)) / 2
Example 2: Find the exact value of cos(112.5 degrees)
Write 112.5 as 225/2, so A = 225 degrees:
cos(112.5) = cos(225/2) = plus or minus sqrt((1 + cos 225) / 2)
A/2 = 112.5 degrees is in Quadrant II, so cos is negative:
cos(225) = -sqrt(2)/2
= -sqrt((1 + (-sqrt(2)/2)) / 2)
= -sqrt((2 - sqrt(2)) / 4)
= -sqrt(2 - sqrt(2)) / 2
Example 3: Given tan(A) = 4/3 with A in Quadrant III, find sin(A/2) and cos(A/2)
Step 1: Find sin(A) and cos(A). In Q3 both are negative.
sec^2(A) = 1 + tan^2(A) = 1 + 16/9 = 25/9
cos^2(A) = 9/25 → cos(A) = -3/5 (negative in Q3)
sin(A) = tan(A) cos(A) = (4/3)(-3/5) = -4/5
Step 2: Determine quadrant of A/2.
A in Q3 means 180 less than A less than 270, so 90 less than A/2 less than 135 → Q2
In Q2: sin is positive, cos is negative.
Step 3: Apply half-angle formulas.
sin(A/2) = +sqrt((1 - cos A) / 2) = sqrt((1 - (-3/5)) / 2) = sqrt((8/5) / 2) = sqrt(4/5) = 2/sqrt(5)
cos(A/2) = -sqrt((1 + cos A) / 2) = -sqrt((1 + (-3/5)) / 2) = -sqrt((2/5) / 2) = -sqrt(1/5) = -1/sqrt(5)
Answer: sin(A/2) = 2 sqrt(5) / 5 and cos(A/2) = -sqrt(5) / 5
Power-reducing formulas rewrite squared trig functions as first-power expressions involving double angles. They are derived directly from the double-angle formulas and are essential in calculus integration.
sin squared
sin^2(A) = (1 - cos(2A)) / 2
From cos(2A) = 1 - 2 sin^2(A), solve for sin^2(A).
cos squared
cos^2(A) = (1 + cos(2A)) / 2
From cos(2A) = 2 cos^2(A) - 1, solve for cos^2(A).
tan squared
tan^2(A) = (1 - cos(2A)) / (1 + cos(2A))
Divide the sin^2 formula by the cos^2 formula.
Calculus Application: Integrate sin^2(x) dx
Direct integration of sin^2(x) is not possible. Use the power-reducing formula:
sin^2(x) = (1 - cos(2x)) / 2
Now integrate term by term:
integral of sin^2(x) dx = integral of (1/2 - cos(2x)/2) dx
= x/2 - sin(2x)/4 + C
Answer: x/2 - (1/4) sin(2x) + C
Calculus Application: Integrate cos^2(x) dx
cos^2(x) = (1 + cos(2x)) / 2
integral of cos^2(x) dx = integral of (1/2 + cos(2x)/2) dx
= x/2 + (1/4) sin(2x) + C
Calculus Application: Integrate sin^4(x) dx using power-reducing twice
First reduce the power:
sin^4(x) = (sin^2(x))^2 = ((1 - cos(2x)) / 2)^2
= (1 - 2 cos(2x) + cos^2(2x)) / 4
Apply power-reducing to cos^2(2x) = (1 + cos(4x)) / 2:
= (1 - 2 cos(2x) + (1 + cos(4x))/2) / 4
= 3/8 - (1/2) cos(2x) + (1/8) cos(4x)
Integrate term by term:
= 3x/8 - (1/4) sin(2x) + (1/32) sin(4x) + C
These formulas convert between products and sums of trig functions. They are derived by adding or subtracting pairs of sum/difference formulas.
Add sin(A + B) = sin A cos B + cos A sin B and sin(A - B) = sin A cos B - cos A sin B, then divide by 2:
sin(A) cos(B) = (1/2) [sin(A + B) + sin(A - B)]
sin(A) sin(B) = (1/2) [cos(A - B) - cos(A + B)]
cos(A) cos(B) = (1/2) [cos(A - B) + cos(A + B)]
cos(A) sin(B) = (1/2) [sin(A + B) - sin(A - B)]
Example: Convert sin(3x) cos(x) to a sum
sin(3x) cos(x) = (1/2) [sin(3x + x) + sin(3x - x)]
= (1/2) [sin(4x) + sin(2x)]
Let A = (u + v)/2 and B = (u - v)/2 in the product-to-sum formulas and simplify. These convert sums into products, which is useful for solving equations and factoring.
sin(u) + sin(v) = 2 sin((u + v)/2) cos((u - v)/2)
sin(u) - sin(v) = 2 cos((u + v)/2) sin((u - v)/2)
cos(u) + cos(v) = 2 cos((u + v)/2) cos((u - v)/2)
cos(u) - cos(v) = -2 sin((u + v)/2) sin((u - v)/2)
Example: Convert sin(5x) + sin(3x) to a product
sin(5x) + sin(3x) = 2 sin((5x + 3x)/2) cos((5x - 3x)/2)
= 2 sin(4x) cos(x)
Example: Simplify cos(5x) - cos(3x)
cos(5x) - cos(3x) = -2 sin((5x + 3x)/2) sin((5x - 3x)/2)
= -2 sin(4x) sin(x)
Work on one side only. Transform that side using algebra and identities until it matches the other side. Never move terms across the equal sign.
Prove: sin(2A) / (1 + cos(2A)) = tan(A)
Work on the left side. Apply double-angle formulas:
sin(2A) = 2 sin(A) cos(A)
cos(2A) = 2 cos^2(A) - 1
Left side = 2 sin(A) cos(A) / (1 + 2 cos^2(A) - 1)
= 2 sin(A) cos(A) / (2 cos^2(A))
= sin(A) / cos(A)
= tan(A) ✓
Prove: cos(2A) = (1 - tan^2(A)) / (1 + tan^2(A))
Work on the right side. Substitute tan(A) = sin(A)/cos(A):
(1 - sin^2(A)/cos^2(A)) / (1 + sin^2(A)/cos^2(A))
Multiply numerator and denominator by cos^2(A):
(cos^2(A) - sin^2(A)) / (cos^2(A) + sin^2(A))
Apply Pythagorean identity cos^2 + sin^2 = 1 in denominator:
(cos^2(A) - sin^2(A)) / 1
= cos(2A) ✓
Prove: sin(3A) = 3 sin(A) - 4 sin^3(A)
Write sin(3A) = sin(2A + A) and expand using sum formula:
sin(2A + A) = sin(2A) cos(A) + cos(2A) sin(A)
Substitute double-angle formulas:
= 2 sin(A) cos(A) cos(A) + (1 - 2 sin^2(A)) sin(A)
= 2 sin(A) cos^2(A) + sin(A) - 2 sin^3(A)
Substitute cos^2(A) = 1 - sin^2(A):
= 2 sin(A)(1 - sin^2(A)) + sin(A) - 2 sin^3(A)
= 2 sin(A) - 2 sin^3(A) + sin(A) - 2 sin^3(A)
= 3 sin(A) - 4 sin^3(A) ✓
Prove: cos^4(A) - sin^4(A) = cos(2A)
Factor the left side as a difference of squares:
cos^4(A) - sin^4(A) = (cos^2(A) + sin^2(A))(cos^2(A) - sin^2(A))
Apply Pythagorean identity to the first factor:
= (1)(cos^2(A) - sin^2(A))
= cos(2A) ✓
When a trig equation contains double angles, use the double-angle formula to rewrite everything in terms of a single angle. Factor, then solve each factor separately.
Solve: sin(2x) = sin(x) for x in [0, 2 pi)
Replace sin(2x) with the double-angle formula:
2 sin(x) cos(x) = sin(x)
Move everything to one side and factor — never divide by sin(x):
2 sin(x) cos(x) - sin(x) = 0
sin(x)(2 cos(x) - 1) = 0
Set each factor equal to zero:
sin(x) = 0 → x = 0, pi
2 cos(x) - 1 = 0 → cos(x) = 1/2 → x = pi/3, 5pi/3
Solutions: x = 0, pi/3, pi, 5pi/3
Solve: cos(2x) + cos(x) = 0 for x in [0, 2 pi)
Use Form 3 of cos(2x): cos(2x) = 1 - 2 sin^2(x) — wait, the variable is cos(x), so use Form 2:
Use cos(2x) = 2 cos^2(x) - 1 so everything is in terms of cos(x):
2 cos^2(x) - 1 + cos(x) = 0
2 cos^2(x) + cos(x) - 1 = 0
Factor the quadratic in cos(x):
(2 cos(x) - 1)(cos(x) + 1) = 0
Solve each factor:
cos(x) = 1/2 → x = pi/3, 5pi/3
cos(x) = -1 → x = pi
Solutions: x = pi/3, pi, 5pi/3
Solve: cos(2x) = 1 - sin(x) for x in [0, 2 pi)
The right side has sin(x), so use Form 3: cos(2x) = 1 - 2 sin^2(x):
1 - 2 sin^2(x) = 1 - sin(x)
-2 sin^2(x) = -sin(x)
2 sin^2(x) - sin(x) = 0
sin(x)(2 sin(x) - 1) = 0
sin(x) = 0 → x = 0, pi
2 sin(x) - 1 = 0 → sin(x) = 1/2 → x = pi/6, 5pi/6
Solutions: x = 0, pi/6, 5pi/6, pi
Solve: sin(2x) + cos(x) = 0 for x in [0, 2 pi)
2 sin(x) cos(x) + cos(x) = 0
cos(x)(2 sin(x) + 1) = 0
cos(x) = 0 → x = pi/2, 3pi/2
2 sin(x) + 1 = 0 → sin(x) = -1/2 → x = 7pi/6, 11pi/6
Solutions: x = pi/2, 7pi/6, 3pi/2, 11pi/6
Double-angle and power-reducing formulas appear throughout integral calculus. Knowing them cold saves significant time on exams.
Case 1: Even powers of both sin and cos
Use power-reducing formulas repeatedly to reduce all even powers to first powers.
integral of sin^2(x) cos^2(x) dx
= integral of ((1 - cos(2x))/2)((1 + cos(2x))/2) dx
= (1/4) integral of (1 - cos^2(2x)) dx
= (1/4) integral of sin^2(2x) dx
= (1/4) integral of (1 - cos(4x))/2 dx
= x/8 - sin(4x)/32 + C
Case 2: Odd power of sin or cos
Save one factor for the differential, convert the rest using the Pythagorean identity, then use substitution.
integral of sin^3(x) dx = integral of sin^2(x) sin(x) dx
= integral of (1 - cos^2(x)) sin(x) dx
Let u = cos(x), du = -sin(x) dx:
= integral of (1 - u^2)(-du)
= -(u - u^3/3) + C
= -cos(x) + cos^3(x)/3 + C
Case 3: Using sin(2x) = 2 sin(x) cos(x) in reverse
integral of sin(x) cos(x) dx
Rewrite as (1/2) sin(2x) and integrate:
= -(1/4) cos(2x) + C
(Alternatively: let u = sin(x) to get sin^2(x)/2 + C — both answers are equivalent)
Double-angle identities are also used to simplify expressions that arise after trig substitution. When evaluating definite or indefinite integrals involving sqrt(a^2 - x^2), you substitute x = a sin(theta), and terms like cos^2(theta) and sin^2(theta) arise — handled with power-reducing formulas.
Integral of sqrt(1 - x^2) dx using x = sin(theta):
sqrt(1 - sin^2(theta)) = sqrt(cos^2(theta)) = cos(theta)
dx = cos(theta) d theta
integral becomes integral of cos^2(theta) d theta = (1 + cos(2 theta))/2 d theta
= theta/2 + sin(2 theta)/4 + C → back-substitute for final answer
Selected Answers
1 (double): sin(A) = 12/13, so sin(2A) = 2(12/13)(-5/13) = -120/169; cos(2A) = (25-144)/169 = -119/169
1 (half): cos(15) = cos(30/2) = sqrt((1 + cos 30)/2) = sqrt((1 + sqrt(3)/2)/2) = sqrt(2 + sqrt(3))/2
1 (power): sin^2(x)cos^2(x) = sin^2(2x)/4 = (1 - cos(4x))/8
3 (solve): cos(2x) = 3 cos(x) + 1 becomes 2cos^2(x) - 3cos(x) - 2 = 0, so (2cos(x)+1)(cos(x)-2)=0; cos(x)=-1/2; x = 2pi/3, 4pi/3
4 (solve): 2sin(x)cos(x) - cos(x) = 0; cos(x)(2sin(x)-1)=0; x = pi/2 + npi or x = pi/6 + 2npi or x = 5pi/6 + 2npi
Mistake: Dividing both sides by sin(x) when solving equations
Always move all terms to one side and factor. Dividing by sin(x) loses all solutions where sin(x) = 0.
Mistake: Applying the wrong form of cos(2A)
Identify what other functions appear in the equation. If the equation has only cos(x), use Form 2: cos(2A) = 2cos^2(A) - 1. If it has only sin(x), use Form 3: cos(2A) = 1 - 2sin^2(A).
Mistake: Using the quadrant of A instead of A/2 for the plus-or-minus sign
Always find the quadrant of A/2 explicitly. Write out the inequality: if 90 less than A less than 180, then 45 less than A/2 less than 90, so A/2 is in Q1.
Mistake: Writing sin(2A) = 2 sin(A) or sin(A + A) = 2 sin(A)
sin(2A) does NOT equal 2 sin(A). This is a frequent algebra error. Always use the formula: sin(2A) = 2 sin(A) cos(A).
Mistake: Forgetting that the tan(A/2) ratio forms do not need a plus-or-minus
tan(A/2) = sin(A)/(1 + cos A) and tan(A/2) = (1 - cos A)/sin(A) are exact — the sign is automatic. Only the sqrt form needs the plus-or-minus determination.
Mistake: Applying power-reducing formulas with the wrong sign
sin^2 uses MINUS: (1 - cos(2A))/2. cos^2 uses PLUS: (1 + cos(2A))/2. The minus/plus matches the minus/plus in Form 3 and Form 2 of cos(2A) respectively.
| Category | Formula | Key Use |
|---|---|---|
| Double-Angle Sin | sin(2A) = 2 sin(A) cos(A) | Simplify products; integration |
| Double-Angle Cos (Form 1) | cos(2A) = cos^2(A) - sin^2(A) | Factoring; both functions present |
| Double-Angle Cos (Form 2) | cos(2A) = 2 cos^2(A) - 1 | Eliminate sin; derive cos half-angle |
| Double-Angle Cos (Form 3) | cos(2A) = 1 - 2 sin^2(A) | Eliminate cos; derive sin half-angle |
| Double-Angle Tan | tan(2A) = 2 tan(A) / (1 - tan^2(A)) | When working in tan only |
| Half-Angle Sin | sin(A/2) = +/- sqrt((1 - cos A) / 2) | Exact values; sign from A/2 quadrant |
| Half-Angle Cos | cos(A/2) = +/- sqrt((1 + cos A) / 2) | Exact values; sign from A/2 quadrant |
| Half-Angle Tan (ratio form) | tan(A/2) = sin(A) / (1 + cos A) | No sign ambiguity; preferred |
| Half-Angle Tan (ratio form 2) | tan(A/2) = (1 - cos A) / sin(A) | Alternate ratio form; also no +/- |
| Power-Reducing Sin | sin^2(A) = (1 - cos(2A)) / 2 | Integrate even powers of sin |
| Power-Reducing Cos | cos^2(A) = (1 + cos(2A)) / 2 | Integrate even powers of cos |
| Product-to-Sum | sin A cos B = (1/2)[sin(A+B) + sin(A-B)] | Convert products to sums |
| Sum-to-Product | sin u + sin v = 2 sin((u+v)/2) cos((u-v)/2) | Factor sums for solving equations |
The double-angle formulas are: sin(2A) = 2 sin(A) cos(A). Cosine has three equivalent forms: cos(2A) = cos^2(A) - sin^2(A) = 2 cos^2(A) - 1 = 1 - 2 sin^2(A). Tangent: tan(2A) = 2 tan(A) / (1 - tan^2(A)). All three are derived by setting B = A in the sum formulas sin(A+B) and cos(A+B).
Starting from cos(2A) = cos^2(A) - sin^2(A), substitute the Pythagorean identity sin^2(A) = 1 - cos^2(A) to get cos(2A) = 2 cos^2(A) - 1. Alternatively substitute cos^2(A) = 1 - sin^2(A) to get cos(2A) = 1 - 2 sin^2(A). Each form is useful in a different context: the first for factoring differences of squares, the second for deriving the half-angle formula for cos, and the third for deriving the half-angle formula for sin.
The half-angle formulas use a plus-or-minus sign: sin(A/2) = plus-or-minus sqrt((1 - cos A) / 2) and cos(A/2) = plus-or-minus sqrt((1 + cos A) / 2). The sign depends on the quadrant of A/2, not A. First determine what quadrant A/2 falls in. If A/2 is in Quadrant I or II, sin(A/2) is positive. If A/2 is in Quadrant I or IV, cos(A/2) is positive. Always determine the quadrant of A/2 before choosing the sign.
Power-reducing formulas rewrite squared trig functions without exponents: sin^2(A) = (1 - cos(2A)) / 2 and cos^2(A) = (1 + cos(2A)) / 2 and tan^2(A) = (1 - cos(2A)) / (1 + cos(2A)). These come directly from solving the double-angle formulas for cos(2A) for sin^2 and cos^2. They are essential in calculus for integrating sin^2(x) and cos^2(x), and in proving identities involving even powers of trig functions.
Product-to-sum formulas convert products of trig functions into sums: sin(A) cos(B) = (1/2)[sin(A+B) + sin(A-B)]. sin(A) sin(B) = (1/2)[cos(A-B) - cos(A+B)]. cos(A) cos(B) = (1/2)[cos(A-B) + cos(A+B)]. These are derived by adding or subtracting the sum and difference formulas. They are used in calculus integration and in signal processing.
When a trig equation involves sin(2x) or cos(2x), replace it using a double-angle formula to express everything in terms of a single angle. For example, sin(2x) = sin(x) becomes 2 sin(x) cos(x) = sin(x), then 2 sin(x) cos(x) - sin(x) = 0, then sin(x)(2 cos(x) - 1) = 0. This gives sin(x) = 0 or cos(x) = 1/2, both of which you can solve with the unit circle. Never divide both sides by sin(x) since that would lose solutions where sin(x) = 0.
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