Sin Cos Formulas — Full List, Identities, Examples

#Math Formula
TL;DR
The sin cos formulas are the family of trigonometric identities relating sine and cosine — the Pythagorean identity, the sum and difference formulas, the double-angle and half-angle identities, and the product-to-sum conversions. This article organises all of them into one map, proves the ones the others are built from, works six examples in degrees and radians, and shows which two formulas you actually have to memorise.
BT
Bhanzu TeamLast updated on June 22, 20268 min read

What Are the Sin Cos Formulas?

The sin cos formulas are the set of trigonometric identities that express relationships between the sine and cosine functions — and, through them, every other trig function. At the foundation is the right-triangle definition: for an acute angle θ\theta,

sin⁡θ=oppositehypotenuse,cos⁡θ=adjacenthypotenuse,tan⁡θ=sin⁡θcos⁡θ.\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \qquad \tan\theta = \frac{\sin\theta}{\cos\theta}.

On the unit circle, a point at angle θ\theta has coordinates exactly (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) — cosine is the xx-coordinate, sine is the yy-coordinate. These two anchors, the triangle and the circle, are the same fact in two costumes, and every formula below respects both.

The Pythagorean Identity — The One You Must Know Cold

Everything starts here:

;sin⁡2θ+cos⁡2θ=1;\boxed{;\sin^2\theta + \cos^2\theta = 1;}

It is the Pythagorean theorem written in trig: on the unit circle the point (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) sits at distance 11 from the origin, so x2+y2=1x^2 + y^2 = 1 becomes cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1. Dividing through by cos⁡2θ\cos^2\theta gives 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta; dividing by sin⁡2θ\sin^2\theta gives 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta. Our Pythagorean identities page proves all three in full. If you know this one identity cold, you can rebuild the cosine-from-sine relationship any time: cos⁡θ=±1−sin⁡2θ\cos\theta = \pm\sqrt{1 - \sin^2\theta}.

Sum and Difference Formulas

These give the sine and cosine of a combined angle:

;sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B;\boxed{;\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B;} ;cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B;\boxed{;\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B;}

Read the cosine sign carefully: the sign between the two products is the opposite of the sign between the angles (cos⁡(A+B)\cos(A+B) takes a minus). Sine keeps the same sign. These are the formulas the double- and half-angle identities are built from.

Double Angle Formulas

Set A=B=θA = B = \theta in the sum formulas and the double-angle identities drop out:

;sin⁡2θ=2sin⁡θcos⁡θ;\boxed{;\sin 2\theta = 2\sin\theta\cos\theta;} ;cos⁡2θ=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ;\boxed{;\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta;}

Cosine has three equivalent forms because sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 lets you trade one squared term for the other. The companion sin 2x derivation and cos2x formula pages each work one of these in depth.

Half Angle Formulas

Solve the cosine double-angle forms backwards (let θ→θ2\theta \to \frac{\theta}{2}) and you get the half-angle identities:

;sin⁡θ2=±1−cos⁡θ2,cos⁡θ2=±1+cos⁡θ2;\boxed{;\sin\frac{\theta}{2} = \pm\sqrt{\dfrac{1 - \cos\theta}{2}}, \qquad \cos\frac{\theta}{2} = \pm\sqrt{\dfrac{1 + \cos\theta}{2}};}

The ±\pm is fixed by the quadrant of θ2\frac{\theta}{2} — see the full half angle formula walk-through, including the sign chart.

Product-to-Sum and Sum-to-Product Formulas

Adding and subtracting the sum and difference formulas turns products into sums:

sin⁡Acos⁡B=12[sin⁡(A+B)+sin⁡(A−B)],cos⁡Acos⁡B=12[cos⁡(A−B)+cos⁡(A+B)].\sin A\cos B = \tfrac{1}{2}\big[\sin(A+B) + \sin(A-B)\big], \qquad \cos A\cos B = \tfrac{1}{2}\big[\cos(A-B) + \cos(A+B)\big].

Running the algebra the other direction gives the sum-to-product forms, e.g. sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}. These conversions are what make wave-on-wave physics — beats, interference, AM radio — tractable.

A Quick-Reference Table

Family

Sine form

Cosine form

Pythagorean

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

—

Sum / difference

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A\pm B) = \sin A\cos B \pm \cos A\sin B

cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A\pm B) = \cos A\cos B \mp \sin A\sin B

Double angle

sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta

cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2\theta

Half angle

sin⁡θ2=±1−cos⁡θ2\sin\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{2}}

cos⁡θ2=±1+cos⁡θ2\cos\frac{\theta}{2} = \pm\sqrt{\frac{1+\cos\theta}{2}}

Negative angle

sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta

cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta

The full identity catalogue, including reciprocal and cofunction relations, lives in our trigonometric identities reference.

Examples of Sin Cos Formulas

Example 1

Given sin⁡θ=35\sin\theta = \dfrac{3}{5} with θ\theta in Quadrant I, find cos⁡θ\cos\theta.

Use the Pythagorean identity, solved for cosine. In Quadrant I cosine is positive:

cos⁡θ=1−sin⁡2θ=1−925=1625=45.\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \tfrac{9}{25}} = \sqrt{\tfrac{16}{25}} = \frac{4}{5}.

Final answer: cos⁡θ=45\cos\theta = \dfrac{4}{5}.

Example 2

Given sin⁡θ=35\sin\theta = \dfrac{3}{5} with θ\theta in Quadrant II, find cos⁡θ\cos\theta.

A natural reflex is to repeat Example 1 exactly and report cos⁡θ=45\cos\theta = \frac{4}{5}.

Wrong path. Writing cos⁡θ=+1−925=+45\cos\theta = +\sqrt{1 - \frac{9}{25}} = +\frac{4}{5}. The magnitude is right, but the sign is wrong: in Quadrant II, sine is positive while cosine is negative. A positive cosine puts the angle in Quadrant I, contradicting the given.

The break. The Pythagorean identity only ever fixes cos⁡2θ\cos^2\theta, never the sign of cos⁡θ\cos\theta. The square root's ±\pm is settled by the quadrant, not by the algebra. Skip that step and the formula can only ever return a positive cosine.

The rescue. Quadrant II forces the negative root:

cos⁡θ=−1−925=−45.\cos\theta = -\sqrt{1 - \tfrac{9}{25}} = -\frac{4}{5}.

Final answer: cos⁡θ=−45\cos\theta = -\dfrac{4}{5}.

Example 3

Compute sin⁡75°\sin 75° exactly using a sum formula.

Write 75°=45°+30°75° = 45° + 30° and apply sin⁡(A+B)\sin(A+B):

sin⁡75°=sin⁡45°cos⁡30°+cos⁡45°sin⁡30°=22⋅32+22⋅12=6+24.\sin 75° = \sin 45°\cos 30° + \cos 45°\sin 30° = \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} + \frac{\sqrt2}{2}\cdot\frac12 = \frac{\sqrt6 + \sqrt2}{4}.

In radians, 75°=5π1275° = \frac{5\pi}{12}. Final answer: sin⁡75°=6+24≈0.966\sin 75° = \dfrac{\sqrt6 + \sqrt2}{4} \approx 0.966.

Example 4

Given sin⁡θ=1213\sin\theta = \dfrac{12}{13} with θ\theta acute, find sin⁡2θ\sin 2\theta and cos⁡2θ\cos 2\theta.

First cos⁡θ=1−144169=513\cos\theta = \sqrt{1 - \frac{144}{169}} = \frac{5}{13}. Then apply the double-angle identities:

sin⁡2θ=2sin⁡θcos⁡θ=2⋅1213⋅513=120169,\sin 2\theta = 2\sin\theta\cos\theta = 2\cdot\frac{12}{13}\cdot\frac{5}{13} = \frac{120}{169}, cos⁡2θ=1−2sin⁡2θ=1−2⋅144169=1−288169=−119169.\cos 2\theta = 1 - 2\sin^2\theta = 1 - 2\cdot\frac{144}{169} = 1 - \frac{288}{169} = -\frac{119}{169}.

Final answer: sin⁡2θ=120169\sin 2\theta = \dfrac{120}{169}, cos⁡2θ=−119169\cos 2\theta = -\dfrac{119}{169}. (Check: (120169)2+(119169)2=14400+141611692=1\left(\frac{120}{169}\right)^2 + \left(\frac{119}{169}\right)^2 = \frac{14400 + 14161}{169^2} = 1.)

Example 5

Verify cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1 at θ=60°\theta = 60° (π3\frac{\pi}{3}).

Left side: cos⁡120°=−12\cos 120° = -\frac12. Right side: 2cos⁡260°−1=2(12)2−1=2⋅14−1=−122\cos^2 60° - 1 = 2\left(\frac12\right)^2 - 1 = 2\cdot\frac14 - 1 = -\frac12. Both equal −12-\frac12. Final answer: identity verified, −12=−12-\frac12 = -\frac12.

Example 6

Rewrite 2sin⁡3θcos⁡θ2\sin 3\theta\cos\theta as a sum.

Use the product-to-sum formula 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B = \sin(A+B) + \sin(A-B) with A=3θA = 3\theta, B=θB = \theta:

2sin⁡3θcos⁡θ=sin⁡(3θ+θ)+sin⁡(3θ−θ)=sin⁡4θ+sin⁡2θ.2\sin 3\theta\cos\theta = \sin(3\theta + \theta) + \sin(3\theta - \theta) = \sin 4\theta + \sin 2\theta.

Final answer: 2sin⁡3θcos⁡θ=sin⁡4θ+sin⁡2θ2\sin 3\theta\cos\theta = \sin 4\theta + \sin 2\theta — a product turned into a sum, the move that linearises wave interference.

Where Sin and Cos Formulas Carry Real Weight

Sine and cosine are how every periodic thing in the universe gets described, and these identities are the algebra that makes the description usable.

  • Sound and music — beats. When two notes of nearly equal frequency play together, the sum-to-product formula rewrites sin⁡(ω1t)+sin⁡(ω2t)\sin(\omega_1 t) + \sin(\omega_2 t) as a single tone amplitude-modulated at the difference frequency. That throb you hear when tuning a guitar is a sin-cos identity made audible.

  • Electrical engineering — AC power. Alternating current is V0cos⁡(ωt)V_0\cos(\omega t); computing power means multiplying two cosines, and the product-to-sum identity is what separates the average (real) power from the oscillating part.

  • Optics and interferometry. Overlapping light waves combine through the sum formulas — the foundation of everything from anti-reflective coatings to LIGO's gravitational-wave detection.

  • Computer graphics. Rotating a point by angle θ\theta multiplies it by a matrix of sin⁡θ\sin\theta and cos⁡θ\cos\theta; composing two rotations uses the sum formulas directly.

For Class 11 students, the immediate payoff is exact-value computation and identity proofs — but the same five families reappear in calculus (the derivative of sine needs the sum formula), physics, and signal processing.

Tripping Points to Avoid

Mistake 1: Writing (sin θ)² as sin θ²

Where it slips in: Reading or writing sin⁡2θ\sin^2\theta — which means (sin⁡θ)2(\sin\theta)^2 — as sin⁡(θ2)\sin(\theta^2).

Don't do this: Square the angle. sin⁡2θ\sin^2\theta squares the output of sine, not the input.

The correct way: sin⁡2θ=(sin⁡θ)(sin⁡θ)\sin^2\theta = (\sin\theta)(\sin\theta). The exponent sits on the function value. In the McKinney TX Grade 11 cohort, this notation slip surfaces on close to half of first attempts when the Pythagorean identity is first introduced — the second-guesser archetype writes it correctly, then "fixes" it into sin⁡θ2\sin\theta^2 because the spacing looks ambiguous.

Mistake 2: Confusing sin²θ + cos²θ = 1 with (sin θ + cos θ)² = 1

Where it slips in: Squaring the sum of sine and cosine and expecting 11.

Don't do this: Assume (sin⁡θ+cos⁡θ)2=sin⁡2θ+cos⁡2θ(\sin\theta + \cos\theta)^2 = \sin^2\theta + \cos^2\theta. The cross term is missing.

The correct way: (sin⁡θ+cos⁡θ)2=sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ=1+sin⁡2θ(\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 1 + \sin 2\theta. The 2sin⁡θcos⁡θ2\sin\theta\cos\theta middle term is exactly sin⁡2θ\sin 2\theta — so the squared sum equals 1+sin⁡2θ1 + \sin 2\theta, not 11.

Mistake 3: Copying the cosine sum sign from the angle operation

Where it slips in: Writing cos⁡(A+B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A + B) = \cos A\cos B + \sin A\sin B — matching the inner ++ to the expansion.

Don't do this: Let the sign inside the cosine dictate the sign in the expansion. For cosine they are opposite.

The correct way: cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A\cos B - \sin A\sin B — a minus for a sum. The memorizer archetype, fluent at sine's same-sign rule, carries that habit into cosine and reverses the one sign that flips.

Conclusion

  • The sin cos formulas are five families: Pythagorean, sum and difference, double angle, half angle, and product-to-sum.

  • The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 is the one to know cold — the cosine-from-sine relationship rebuilds from it.

  • The sum and difference formulas generate the double- and half-angle identities by substitution.

  • The most common errors are notation (sin⁡2θ\sin^2\theta vs sin⁡θ2\sin\theta^2), the missing cross term, and copying the cosine sum sign.

  • These identities power beats in music, AC power analysis, interferometry, and the rotation math in computer graphics.

Practice These Before Moving On

  1. Given cos⁡θ=817\cos\theta = \frac{8}{17} with θ\theta acute, find sin⁡θ\sin\theta, sin⁡2θ\sin 2\theta, and cos⁡2θ\cos 2\theta.

  2. Compute cos⁡15°\cos 15° exactly using a difference formula.

  3. Rewrite 2cos⁡5θcos⁡3θ2\cos 5\theta\cos 3\theta as a sum.

If Problem 2 returned a minus sign between the products, return to Mistake 3 — the cosine difference formula takes a plus.

Want a live Bhanzu trainer to walk your child through the trigonometric identities and the Class 11 chapter? Book a free demo class — online globally.

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Frequently Asked Questions

What are the sin cos formulas?
They are the trigonometric identities relating sine and cosine: the Pythagorean identity, the sum and difference formulas, the double- and half-angle identities, and the product-to-sum conversions.
What is sin θ / cos θ equal to?
It equals tan⁡θ\tan\theta. The quotient of sine over cosine is the definition of the tangent function.
How do you find cos from sin?
Use the Pythagorean identity: cos⁡θ=±1−sin⁡2θ\cos\theta = \pm\sqrt{1 - \sin^2\theta}. Pick the sign from the quadrant θ\theta lives in.
How many sin cos formulas do I actually need to memorise?
Two, really — the Pythagorean identity and the cosine difference formula. Everything else (double angle, half angle, the rest of sum/difference) can be rebuilt from those two in under a minute.
Are the sin cos formulas the same in degrees and radians?
Yes. sin⁡π3\sin\frac{\pi}{3} and sin⁡60°\sin 60° are the same number. The identities never change form; only your calculator's mode does.
What is the difference between sin²θ and sin 2θ?
sin⁡2θ=(sin⁡θ)2\sin^2\theta = (\sin\theta)^2 — sine squared. sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta — the sine of a doubled angle. Different quantities that look similar in print.
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Bhanzu Team
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Bhanzu’s editorial team, known as Team Bhanzu, is made up of experienced educators, curriculum experts, content strategists, and fact-checkers dedicated to making math simple and engaging for learners worldwide. Every article and resource is carefully researched, thoughtfully structured, and rigorously reviewed to ensure accuracy, clarity, and real-world relevance. We understand that building strong math foundations can raise questions for students and parents alike. That’s why Team Bhanzu focuses on delivering practical insights, concept-driven explanations, and trustworthy guidance-empowering learners to develop confidence, speed, and a lifelong love for mathematics.
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