What it shows
Why is sin(A + B) not just sin A + sin B? This page answers with a picture. Start with a line OP of length 1 at an angle A + B above the x-axis. Split the climb into two right triangles: OQ at angle A, and QP at right angles to it, so that the angle between OQ and OP is B. Then
- OQ = cos B and QP = sin B, because OP = 1;
- the height of P is the height of Q plus the extra climb from Q to P;
- and how far right P is equals how far right Q is, minus the distance P has stepped back.
Measure those pieces with sin A and cos A and you get the two compound angle formulae of Class 11 maths (trigonometric functions):
sin(A + B) = sin A cos B + cos A sin B and cos(A + B) = cos A cos B − sin A sin B.
The rest (sin(A − B), cos(A − B), tan(A ± B), sin 2A, cos 2A, tan 2A) follow by putting −B for B, dividing, or setting B = A, and the table below the picture checks every one of them at your angles.
Revise the ratios in a right triangle with trig ratio explorer first; the unit circle shows why the formulas still hold beyond 90°, and the radian ruler turns the degrees into radians.
How to use it
- Move the A and B sliders (0° to 80° each). The picture redraws with O, Q, P, the feet R and M on the x-axis, and N, the point level with Q directly below P.
- Tap sin(A + B) to see the heights: RQ in red and NP in green add up to MP, the height of P. Tap cos(A + B) to see the widths: OR in red minus NQ in green leaves OM.
- The numbered steps under the picture give the reason for each length with your numbers filled in, and the bold line adds them up.
- Make B = A turns the sum formulas into the double angle formulas.
- The table works out each formula twice: the left side straight from the angle, the right side from the formula.
Why the angle at P is A
Triangle QNP has a right angle at N. Its side QN is horizontal and NP vertical, while QP is at right angles to OQ. Turning a whole triangle through 90° keeps its angles, so the angle between PN and PQ equals the angle between the x-axis and OQ, which is A. That is why NP = QP cos A and NQ = QP sin A.
Worked example
With the starting angles A = 30° and B = 45°, the title reads sin(30° + 45°) = sin 75° = 0.9659. The steps show OQ = cos 45° = 0.7071 and QP = sin 45° = 0.7071, then RQ = sin 30° cos 45° = 0.3536 and NP = cos 30° sin 45° = 0.6124. The last line reads sin(A + B) = 0.3536 + 0.6124 = 0.9659, the same as sin 75°. The exact value is (√6 + √2)/4.
Switch to cos(A + B): OR = cos 30° cos 45° = 0.6124 and NQ = sin 30° sin 45° = 0.3536, so cos 75° = 0.6124 − 0.3536 = 0.2588.
FAQ
Does the proof only work for acute angles? The picture is drawn for A and B up to 80°, but A + B can pass 90° (P then sits left of the y-axis and cos(A + B) is negative). The formulas hold for all angles; the unit circle definitions extend the proof.
Why does tan(A + B) say “undefined” sometimes? When A + B = 90°, tan(A + B) has no value, and so 1 − tan A tan B = 0 on the right as well.
Is anything saved? No. Everything runs in your browser. fibub is not part of NCERT; the topics follow the syllabus but the words and pictures are our own.
Where the numbers come from
- The sum, difference and double angle formulae and their geometric proof by stacked right triangles are standard trigonometry. Wording, pictures and examples are fibub's own; no textbook text, figures or exercises are used.
- Checked against Python's math.sin and math.cos for every pair of angles from 0° to 80° in 5° steps, and against sympy for the exact values of sin 75° and cos 15°.
Standard scientific values, such as constants, atomic weights and melting points, are facts found in every textbook; we name where ours were checked. If you are an author or publisher and want a value, a credit or a table changed or removed, tell us and we will act quickly.
Last updated 2026-11-09.