The Master Identity and Its Family

Everything in the rest of this chapter grows from one seed:

cos⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡y\cos(x + y) = \cos x \cos y - \sin x \sin y

The proof is pure geometry. Place four points on the unit circle: P4(1,0)P_4(1, 0), P1(cos⁡x,sin⁡x)P_1(\cos x, \sin x), P2(cos⁡(x+y),sin⁡(x+y))P_2(\cos(x+y), \sin(x+y)) and P3(cos⁡y,−sin⁡y)P_3(\cos y, -\sin y) (angle −y-y). Rotating the circle by −x-x carries P2P_2 to P1P_1 and P4P_4 to P3P_3, so the chords P2P4P_2P_4 and P1P3P_1P_3 are equal. Computing both by the distance formula and equating gives exactly the identity. ∎

Unit circle chord proof of the cosine addition formula

Harvesting the family

Replace y by −y-y (using even-odd): cos⁡(x−y)=cos⁡xcos⁡y+sin⁡xsin⁡y\cos(x - y) = \cos x \cos y + \sin x \sin y

Put x=π2x = \frac{\pi}{2} in the difference formula: cos⁡(π2−y)=sin⁡y\cos\left(\frac{\pi}{2} - y\right) = \sin y, and swapping roles, sin⁡(π2−y)=cos⁡y\sin\left(\frac{\pi}{2} - y\right) = \cos y — the co-function pair. Then

sin⁡(x+y)=cos⁡(π2−x−y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin(x + y) = \cos\left(\frac{\pi}{2} - x - y\right) = \sin x \cos y + \cos x \sin y sin⁡(x−y)=sin⁡xcos⁡y−cos⁡xsin⁡y\sin(x - y) = \sin x \cos y - \cos x \sin y

Key Point (sign pattern): cos mixes cos·cos with sin·sin and FLIPS the middle sign; sin mixes sin·cos with cos·sin and KEEPS the sign. "cos is a contrarian, sin is loyal."

[Board Important] These four formulas plus the two co-function rules generate every identity in this chapter — derivations are examinable, especially the chord proof of cos(x + y).

Allied Angles — Reducing Any Angle

Feeding special values into the sum-difference formulas produces the allied angle (reduction) table:

  • cos⁡(π2+x)=−sin⁡x\cos\left(\frac{\pi}{2} + x\right) = -\sin x, sin⁡(π2+x)=cos⁡x\sin\left(\frac{\pi}{2} + x\right) = \cos x
  • cos⁡(π−x)=−cos⁡x\cos(\pi - x) = -\cos x, sin⁡(π−x)=sin⁡x\sin(\pi - x) = \sin x
  • cos⁡(π+x)=−cos⁡x\cos(\pi + x) = -\cos x, sin⁡(π+x)=−sin⁡x\sin(\pi + x) = -\sin x
  • cos⁡(2π−x)=cos⁡x\cos(2\pi - x) = \cos x, sin⁡(2π−x)=−sin⁡x\sin(2\pi - x) = -\sin x

Allied angle reduction table for sine and cosine

Two-step working rule:

  1. Function: with π\pi or 2π2\pi, the function survives; with π2\frac{\pi}{2} or 3π2\frac{3\pi}{2}, it swaps (sin ↔ cos, tan ↔ cot, sec ↔ cosec).
  2. Sign: taken from the quadrant of the ORIGINAL angle via ASTC (treating x as acute).

Example: sin⁡5π6\sin\frac{5\pi}{6} — write 5π6=π−π6\frac{5\pi}{6} = \pi - \frac{\pi}{6} (Q II, sine positive, function survives): =+sin⁡π6=12= +\sin\frac{\pi}{6} = \frac{1}{2}.

[JEE Tip] Prefer reductions through π\pi and 2π2\pi when you have the choice — no function swap means one less place to slip.

Sum and Difference for tan and cot

Dividing sin(x + y) by cos(x + y) and then dividing every term by cos x cos y:

tan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡xtan⁡ytan⁡(x−y)=tan⁡x−tan⁡y1+tan⁡xtan⁡y\tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \qquad \tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}

(valid when none of x, y, x ± y is an odd multiple of π2\frac{\pi}{2}). Similarly, dividing by sin x sin y:

cot⁡(x+y)=cot⁡xcot⁡y−1cot⁡y+cot⁡xcot⁡(x−y)=cot⁡xcot⁡y+1cot⁡y−cot⁡x\cot(x + y) = \frac{\cot x \cot y - 1}{\cot y + \cot x} \qquad \cot(x - y) = \frac{\cot x \cot y + 1}{\cot y - \cot x}

(valid when none of x, y, x ± y is a multiple of π\pi).

The 15°-75° dividend

The formulas immediately deliver new exact values:

tan⁡15°=tan⁡(45°−30°)=1−131+13=3−13+1=2−3\tan 15° = \tan(45° - 30°) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3}

and likewise tan⁡75°=2+3\tan 75° = 2 + \sqrt{3}, cos⁡75°=3−122\cos 75° = \frac{\sqrt{3} - 1}{2\sqrt{2}}, sin⁡75°=3+122\sin 75° = \frac{\sqrt{3} + 1}{2\sqrt{2}}, sin⁡15°=3−122\sin 15° = \frac{\sqrt{3} - 1}{2\sqrt{2}}.

Key Point: tan⁡15°⋅tan⁡75°=(2−3)(2+3)=1\tan 15° \cdot \tan 75° = (2 - \sqrt{3})(2 + \sqrt{3}) = 1 — as it must be, since 15° and 75° are complementary.

[JEE Tip] The tan sum formula rearranges into tan⁡x+tan⁡y=tan⁡(x+y)(1−tan⁡xtan⁡y)\tan x + \tan y = \tan(x + y)(1 - \tan x \tan y) — the engine behind classics like proving tan⁡3A−tan⁡2A−tan⁡A=tan⁡3Atan⁡2Atan⁡A\tan 3A - \tan 2A - \tan A = \tan 3A \tan 2A \tan A (write 3A=2A+A3A = 2A + A and cross-multiply).

Solved Examples

Example 1: Exact values of cos 75° and sin 15°

Evaluate cos 75° and sin 15° using sum-difference formulas.

Solution:

Step 1 — Split 75° over known angles. cos⁡75°=cos⁡(45°+30°)=cos⁡45°cos⁡30°−sin⁡45°sin⁡30°\cos 75° = \cos(45° + 30°) = \cos 45°\cos 30° - \sin 45°\sin 30°.

Step 2 — Substitute the standard values. =12⋅32−12⋅12=3−122= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}.

Step 3 — Same split for sin 15°. sin⁡15°=sin⁡(45°−30°)=sin⁡45°cos⁡30°−cos⁡45°sin⁡30°=3−122\sin 15° = \sin(45° - 30°) = \sin 45°\cos 30° - \cos 45°\sin 30° = \frac{\sqrt{3} - 1}{2\sqrt{2}}.

Step 4 — Cross-check with co-functions. cos 75° = sin 15° — as it must be, since 75° and 15° are complementary. ✓

Takeaway: Split 75° and 15° over the 45°-30° pair; the co-function check certifies the answer.

Example 2: tan 15°

Show that tan⁡15°=2−3\tan 15° = 2 - \sqrt{3}.

Solution:

Step 1 — Split over 45° and 30°. tan⁡15°=tan⁡(45°−30°)=tan⁡45°−tan⁡30°1+tan⁡45°tan⁡30°=1−131+13\tan 15° = \tan(45° - 30°) = \frac{\tan 45° - \tan 30°}{1 + \tan 45°\tan 30°} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}.

Step 2 — Clear the inner fractions. Multiply top and bottom by 3\sqrt{3}: 3−13+1\frac{\sqrt{3} - 1}{\sqrt{3} + 1}.

Step 3 — Rationalise. Multiply by 3−13−1\frac{\sqrt{3} - 1}{\sqrt{3} - 1}: (3−1)23−1=4−232=2−3\frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}. ∎

Takeaway: Rationalising the surd quotient is the second half of the marks — do not stop at 3−13+1\frac{\sqrt{3}-1}{\sqrt{3}+1}.

Example 3: A disguised sin(x + y)

Prove that cos⁡(π4−x)cos⁡(π4−y)−sin⁡(π4−x)sin⁡(π4−y)=sin⁡(x+y)\cos\left(\frac{\pi}{4} - x\right)\cos\left(\frac{\pi}{4} - y\right) - \sin\left(\frac{\pi}{4} - x\right)\sin\left(\frac{\pi}{4} - y\right) = \sin(x + y).

Solution:

Step 1 — Recognise the pattern. The left side has the exact shape cos A cos B −- sin A sin B = cos(A + B), with A=π4−xA = \frac{\pi}{4} - x and B=π4−yB = \frac{\pi}{4} - y.

Step 2 — Add the angles. A+B=π2−(x+y)A + B = \frac{\pi}{2} - (x + y).

Step 3 — Apply the co-function rule. cos⁡(π2−(x+y))=sin⁡(x+y)\cos\left(\frac{\pi}{2} - (x + y)\right) = \sin(x + y). ∎

Takeaway: Spot the addition-formula pattern BEFORE expanding anything — pattern recognition turns four terms into one line.

Example 4: Allied angle workout

Prove that cos⁡(3π2+x)cos⁡(2π+x)[cot⁡(3π2−x)+cot⁡(2π+x)]=1\cos\left(\frac{3\pi}{2} + x\right)\cos(2\pi + x)\left[\cot\left(\frac{3\pi}{2} - x\right) + \cot(2\pi + x)\right] = 1.

Solution:

Step 1 — Reduce each allied angle. cos⁡(3π2+x)=sin⁡x\cos\left(\frac{3\pi}{2} + x\right) = \sin x (function swaps for a 3π2\frac{3\pi}{2} shift; Q IV makes cosine positive, and the swap lands on +sin⁡x+\sin x); cos⁡(2π+x)=cos⁡x\cos(2\pi + x) = \cos x; cot⁡(3π2−x)=tan⁡x\cot\left(\frac{3\pi}{2} - x\right) = \tan x; cot⁡(2π+x)=cot⁡x\cot(2\pi + x) = \cot x.

Step 2 — Substitute. The left side becomes sin⁡xcos⁡x (tan⁡x+cot⁡x)\sin x \cos x\,(\tan x + \cot x).

Step 3 — Combine the bracket over one denominator. tan⁡x+cot⁡x=sin⁡xcos⁡x+cos⁡xsin⁡x=sin⁡2x+cos⁡2xsin⁡xcos⁡x=1sin⁡xcos⁡x\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}.

Step 4 — Multiply. sin⁡xcos⁡x⋅1sin⁡xcos⁡x=1\sin x \cos x \cdot \frac{1}{\sin x \cos x} = 1. ∎

Takeaway: Reduce every allied angle FIRST; the algebra that remains is usually one identity long.

Example 5: tan of a shifted angle

Find tan⁡(π4+x)\tan\left(\frac{\pi}{4} + x\right) in terms of tan x, and evaluate tan⁡105°\tan 105°.

Solution:

Step 1 — Specialise the sum formula. With tan⁡π4=1\tan\frac{\pi}{4} = 1: tan⁡(π4+x)=1+tan⁡x1−tan⁡x\tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan x}{1 - \tan x}.

Step 2 — Split 105°. tan⁡105°=tan⁡(60°+45°)=tan⁡60°+tan⁡45°1−tan⁡60°tan⁡45°=3+11−3\tan 105° = \tan(60° + 45°) = \frac{\tan 60° + \tan 45°}{1 - \tan 60°\tan 45°} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}.

Step 3 — Rationalise. Multiply by 1+31+3\frac{1 + \sqrt{3}}{1 + \sqrt{3}}: (3+1)21−3=4+23−2=−(2+3)\frac{(\sqrt{3} + 1)^2}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2} = -(2 + \sqrt{3}).

Step 4 — Sign check. 105° lies in Q II, where tan is negative — and the answer is negative. ✓

Takeaway: 105° sits in Q II where tan is negative — the sign of the final answer is a built-in check.

Example 6: cos(x + y) from data

If sin⁡x=35\sin x = \frac{3}{5} (x in Q I) and cos⁡y=−1213\cos y = -\frac{12}{13} (y in Q II), find cos⁡(x+y)\cos(x + y) and sin⁡(x+y)\sin(x + y).

Solution:

Step 1 — Complete the missing values with quadrant signs. From the 3-4-5 triangle: cos⁡x=+45\cos x = +\frac{4}{5} (Q I). From the 5-12-13 triangle: sin⁡y=+513\sin y = +\frac{5}{13} (Q II keeps sine positive).

Step 2 — Expand cos(x + y). cos⁡xcos⁡y−sin⁡xsin⁡y=45(−1213)−35⋅513=−48−1565=−6365\cos x \cos y - \sin x \sin y = \frac{4}{5}\left(-\frac{12}{13}\right) - \frac{3}{5} \cdot \frac{5}{13} = \frac{-48 - 15}{65} = -\frac{63}{65}.

Step 3 — Expand sin(x + y). sin⁡xcos⁡y+cos⁡xsin⁡y=35(−1213)+45⋅513=−36+2065=−1665\sin x \cos y + \cos x \sin y = \frac{3}{5}\left(-\frac{12}{13}\right) + \frac{4}{5} \cdot \frac{5}{13} = \frac{-36 + 20}{65} = -\frac{16}{65}.

Step 4 — Consistency check. (6365)2+(1665)2=3969+2564225=1\left(\frac{63}{65}\right)^2 + \left(\frac{16}{65}\right)^2 = \frac{3969 + 256}{4225} = 1. ✓

Takeaway: Fill in the missing sin/cos values (with quadrant signs!) before touching the addition formulas — the 3-4-5 and 5-12-13 triangles do the arithmetic.

Example 7: Co-function juggling

Prove that sin⁡(π+x)cos⁡(π2+x)tan⁡(3π2−x)cot⁡(2π−x)sin⁡(2π−x)cos⁡(2π+x)csc⁡(−x)sin⁡(3π2+x)=1\frac{\sin(\pi + x)\cos\left(\frac{\pi}{2} + x\right)\tan\left(\frac{3\pi}{2} - x\right)\cot(2\pi - x)}{\sin(2\pi - x)\cos(2\pi + x)\csc(-x)\sin\left(\frac{3\pi}{2} + x\right)} = 1.

Solution:

Step 1 — Reduce the numerator factor by factor. sin⁡(π+x)=−sin⁡x\sin(\pi + x) = -\sin x; cos⁡(π2+x)=−sin⁡x\cos\left(\frac{\pi}{2} + x\right) = -\sin x; tan⁡(3π2−x)=cot⁡x\tan\left(\frac{3\pi}{2} - x\right) = \cot x; cot⁡(2π−x)=−cot⁡x\cot(2\pi - x) = -\cot x. Product: (−sin⁡x)(−sin⁡x)(cot⁡x)(−cot⁡x)=−sin⁡2xcot⁡2x(-\sin x)(-\sin x)(\cot x)(-\cot x) = -\sin^2 x \cot^2 x.

Step 2 — Reduce the denominator. sin⁡(2π−x)=−sin⁡x\sin(2\pi - x) = -\sin x; cos⁡(2π+x)=cos⁡x\cos(2\pi + x) = \cos x; csc⁡(−x)=−csc⁡x\csc(-x) = -\csc x; sin⁡(3π2+x)=−cos⁡x\sin\left(\frac{3\pi}{2} + x\right) = -\cos x. Product: (−sin⁡x)(cos⁡x)(−csc⁡x)(−cos⁡x)=−cos⁡2x(-\sin x)(\cos x)(-\csc x)(-\cos x) = -\cos^2 x (using sin⁡x⋅csc⁡x=1\sin x \cdot \csc x = 1).

Step 3 — Divide. −sin⁡2xcot⁡2x−cos⁡2x=sin⁡2x⋅cos⁡2xsin⁡2xcos⁡2x=1\frac{-\sin^2 x \cot^2 x}{-\cos^2 x} = \frac{\sin^2 x \cdot \frac{\cos^2 x}{\sin^2 x}}{\cos^2 x} = 1. ∎

Takeaway: Long allied-angle chains: reduce factor by factor, carry every minus sign separately, and only simplify at the end.

Example 8: The tan-sum rearrangement

Prove that tan⁡3A−tan⁡2A−tan⁡A=tan⁡3Atan⁡2Atan⁡A\tan 3A - \tan 2A - \tan A = \tan 3A \tan 2A \tan A.

Solution:

Step 1 — Write 3A as a sum. tan⁡3A=tan⁡(2A+A)=tan⁡2A+tan⁡A1−tan⁡2Atan⁡A\tan 3A = \tan(2A + A) = \frac{\tan 2A + \tan A}{1 - \tan 2A \tan A}.

Step 2 — Cross-multiply. tan⁡3A (1−tan⁡2Atan⁡A)=tan⁡2A+tan⁡A\tan 3A\,(1 - \tan 2A \tan A) = \tan 2A + \tan A, i.e. tan⁡3A−tan⁡3Atan⁡2Atan⁡A=tan⁡2A+tan⁡A\tan 3A - \tan 3A \tan 2A \tan A = \tan 2A + \tan A.

Step 3 — Rearrange. Move tan⁡2A+tan⁡A\tan 2A + \tan A left and the triple product right: tan⁡3A−tan⁡2A−tan⁡A=tan⁡3Atan⁡2Atan⁡A\tan 3A - \tan 2A - \tan A = \tan 3A \tan 2A \tan A. ∎

Takeaway: Whenever an identity mixes tan of angles summing conveniently (3A = 2A + A, or A + B + C = π\pi), start from the tan-sum formula and cross-multiply.

Example 9: Reducing to the 15° family

Show that tan⁡13π12=2−3\tan\frac{13\pi}{12} = 2 - \sqrt{3}.

Solution:

Step 1 — Strip a half-revolution. 13π12=π+π12\frac{13\pi}{12} = \pi + \frac{\pi}{12}, and tan survives a π\pi shift: tan⁡13π12=tan⁡π12\tan\frac{13\pi}{12} = \tan\frac{\pi}{12}.

Step 2 — Recognise the degree value. π12=15°\frac{\pi}{12} = 15°, and tan⁡15°=2−3\tan 15° = 2 - \sqrt{3} (Example 2).

Step 3 — Sign check. 13π12\frac{13\pi}{12} lies in Q III, where tan is positive — and 2−3>02 - \sqrt{3} > 0. ✓

Takeaway: Big fractional multiples of π\pi almost always reduce to a 15°-family angle — reduce first, then recognise.

Example 10: cos(x − y) with sign care

If cos⁡x=−13\cos x = -\frac{1}{3} with x in Q III, find cos⁡(x−π3)\cos\left(x - \frac{\pi}{3}\right) in exact form.

Solution:

Step 1 — Complete the data. sin⁡2x=1−19=89\sin^2 x = 1 - \frac{1}{9} = \frac{8}{9}, and Q III makes sine negative: sin⁡x=−223\sin x = -\frac{2\sqrt{2}}{3}.

Step 2 — Expand with the difference formula. cos⁡(x−π3)=cos⁡xcos⁡π3+sin⁡xsin⁡π3=−13⋅12+(−223)32\cos\left(x - \frac{\pi}{3}\right) = \cos x \cos\frac{\pi}{3} + \sin x \sin\frac{\pi}{3} = -\frac{1}{3} \cdot \frac{1}{2} + \left(-\frac{2\sqrt{2}}{3}\right)\frac{\sqrt{3}}{2}.

Step 3 — Combine over one denominator. =−16−266=−1+266= -\frac{1}{6} - \frac{2\sqrt{6}}{6} = -\frac{1 + 2\sqrt{6}}{6}.

Takeaway: Surd answers are normal here — resist the urge to decimalise; exact form is what the examiner wants.

Example 11: Proving an identity with π/4 shifts

Prove that cos⁡(π4+x)+cos⁡(π4−x)=2cos⁡x\cos\left(\frac{\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} - x\right) = \sqrt{2}\cos x.

Solution:

Step 1 — Expand both terms. (cos⁡π4cos⁡x−sin⁡π4sin⁡x)+(cos⁡π4cos⁡x+sin⁡π4sin⁡x)\left(\cos\frac{\pi}{4}\cos x - \sin\frac{\pi}{4}\sin x\right) + \left(\cos\frac{\pi}{4}\cos x + \sin\frac{\pi}{4}\sin x\right).

Step 2 — Watch the sine terms cancel. The ±sin⁡π4sin⁡x\pm\sin\frac{\pi}{4}\sin x pair vanishes, leaving 2cos⁡π4cos⁡x2\cos\frac{\pi}{4}\cos x.

Step 3 — Substitute the value. 2⋅12cos⁡x=2cos⁡x2 \cdot \frac{1}{\sqrt{2}}\cos x = \sqrt{2}\cos x. ∎

Takeaway: Symmetric ±x\pm x pairs make half the terms cancel — expand both and watch the collapse. (Also provable by sum-to-product next section.)

Example 12: The cot-sum formula in action

If cot⁡x=34\cot x = \frac{3}{4} and cot⁡y=17\cot y = \frac{1}{7}, both x and y acute, show that x+y=3π4x + y = \frac{3\pi}{4}.

Solution:

Step 1 — Compute cot(x + y). cot⁡(x+y)=cot⁡xcot⁡y−1cot⁡y+cot⁡x=34⋅17−117+34\cot(x + y) = \frac{\cot x \cot y - 1}{\cot y + \cot x} = \frac{\frac{3}{4} \cdot \frac{1}{7} - 1}{\frac{1}{7} + \frac{3}{4}}.

Step 2 — Clear denominators. Numerator: 3−2828=−2528\frac{3 - 28}{28} = -\frac{25}{28}; denominator: 4+2128=2528\frac{4 + 21}{28} = \frac{25}{28}. Quotient: −1-1.

Step 3 — Locate the sum. x and y are acute, so 0<x+y<π0 < x + y < \pi; the only angle there with cot = −1-1 is 3π4\frac{3\pi}{4}. ∎

Takeaway: Computing a trig function OF the sum is the standard route to finding the sum itself — pick tan or cot to match the given data.

Example 13: A subtraction that reveals equality

Prove that sin⁡(n+1)xsin⁡(n+2)x+cos⁡(n+1)xcos⁡(n+2)x=cos⁡x\sin(n + 1)x \sin(n + 2)x + \cos(n + 1)x \cos(n + 2)x = \cos x.

Solution:

Step 1 — Match the pattern. cos A cos B + sin A sin B = cos(A −- B), with A = (n + 2)x and B = (n + 1)x.

Step 2 — Subtract the angles. A−B=(n+2)x−(n+1)x=xA - B = (n + 2)x - (n + 1)x = x.

Step 3 — Conclude. The whole expression collapses to cos⁡x\cos x, independent of n. ∎

Takeaway: The difference formula runs equally well in reverse, collapsing four products into a single cosine of the angle GAP.