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3.4 Trigonometric Functions of Sum and Difference of Two Angles

🎓 Class 11 Mathematics CBSE Theory Ch 3 — Trigonometric Functions ⏱ ~15 min
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Targeting Class 11 level in Trigonometry, with Advanced difficulty.

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3.4 Trigonometric Functions of Sum and Difference of Two Angles

Identities involving \(\sin(x\pm y)\) and \(\cos(x\pm y)\) are at the heart of trigonometry. They unlock half-angle formulas, double-angle formulas, calculus derivatives like \(\dfrac{d}{dx}\sin x=\cos x\), and the wave decomposition used in signal processing.

Two facts that follow immediately from the unit-circle definitions, and which you will use throughout this section:

Pre-requisites (Part 2)
Identity 1. \(\sin(-x)=-\sin x\) (odd function).
Identity 2. \(\cos(-x)=\cos x\) (even function).
Geometric reason: reflecting the angle \(x\) to \(-x\) flips the y-coordinate of \(P\) but keeps the x-coordinate.

Identity 3 — The cosine sum formula

Cosine Sum
\[\boxed{\;\cos(x+y)=\cos x\cos y-\sin x\sin y\;}\]

This is the master identity from which most others follow. The clean proof uses the chord-length? argument on the unit circle.

X P₄(1,0) Y P₁(cos x, sin x) P₂(cos(x+y), sin(x+y)) P₃(cos(−y), sin(−y)) x y −y O
Fig 3.14: Setup for the cos(x + y) proof — four points on the unit circle.
Proof of Identity 3
Let \(\angle P_4OP_1=x\), \(\angle P_1OP_2=y\) (so \(\angle P_4OP_2=x+y\)) and \(\angle P_4OP_3=-y\). The four points are \[P_1(\cos x,\sin x),\ P_2(\cos(x+y),\sin(x+y)),\ P_3(\cos(-y),\sin(-y)),\ P_4(1,0).\] Triangles \(P_1OP_3\) and \(P_2OP_4\) are congruent: \(OP_1=OP_2=OP_3=OP_4=1\) and the included angle in each is \(x+y\) (the angles \(\angle P_1OP_3\) and \(\angle P_2OP_4\) both equal \(x+y\)). So their third sides are equal: \(P_1P_3=P_2P_4\), hence \(P_1P_3^2=P_2P_4^2\).
By the distance formula and \(\cos^2+\sin^2=1\): \[P_1P_3^2=2-2(\cos x\cos y-\sin x\sin y),\qquad P_2P_4^2=2-2\cos(x+y).\] Equating and dividing by \(-2\) gives \(\cos(x+y)=\cos x\cos y-\sin x\sin y\). \(\square\)

Identity 4 — Cosine of a difference

Replace \(y\) by \(-y\) in Identity 3 and use Identities 1 and 2:

\[\cos(x-y)=\cos x\cos\!(-y)-\sin x\sin\!(-y)=\cos x\cos y+\sin x\sin y.\]
Cosine Difference
\[\boxed{\;\cos(x-y)=\cos x\cos y+\sin x\sin y\;}\]

Identities 5 & 6 — Cofunction identities

Set \(x=\dfrac{\pi}{2}\) in Identity 4: since \(\cos(\pi/2)=0\) and \(\sin(\pi/2)=1\),

\[\cos\!\left(\dfrac{\pi}{2}-x\right)=\cos\!\dfrac{\pi}{2}\cos x+\sin\!\dfrac{\pi}{2}\sin x=\sin x.\]

And replacing \(x\) by \(\pi/2 - x\) in this very result gives:

\[\sin\!\left(\dfrac{\pi}{2}-x\right)=\cos\!\left(\dfrac{\pi}{2}-\!\!\left(\dfrac{\pi}{2}-x\right)\right)=\cos x.\]
Cofunctions
\[\boxed{\;\cos\!\left(\dfrac{\pi}{2}-x\right)=\sin x,\quad \sin\!\left(\dfrac{\pi}{2}-x\right)=\cos x\;}\] This is why "complement" angles have swapped trig values — and why "cosine" literally means "complement's sine".

Identities 7 & 8 — Sine of sum and difference

Using Identity 5 with the argument \(x+y\) in place of \(x\):

\[\sin(x+y)=\cos\!\left(\dfrac{\pi}{2}-(x+y)\right)=\cos\!\left(\!\!\left(\dfrac{\pi}{2}-x\right)-y\right).\]

Apply the cosine difference (Identity 4) to this:

\[=\cos\!\left(\dfrac{\pi}{2}-x\right)\cos y+\sin\!\left(\dfrac{\pi}{2}-x\right)\sin y=\sin x\cos y+\cos x\sin y.\]

Replacing \(y\) with \(-y\) gives the difference formula.

Sine Sum & Difference
\[\boxed{\;\sin(x+y)=\sin x\cos y+\cos x\sin y\;}\] \[\boxed{\;\sin(x-y)=\sin x\cos y-\cos x\sin y\;}\]

Identity 9 — Reduction formulas (specific values of y)

Setting \(y=\dfrac{\pi}{2},\ \pi,\ 2\pi\) in Identities 3, 4, 7, 8 yields the most-used reduction formulas:

  • \(\cos(\pi/2+x)=-\sin x\), \(\sin(\pi/2+x)=\cos x\)
  • \(\cos(\pi-x)=-\cos x\), \(\sin(\pi-x)=\sin x\)
  • \(\cos(\pi+x)=-\cos x\), \(\sin(\pi+x)=-\sin x\)
  • \(\cos(2\pi-x)=\cos x\), \(\sin(2\pi-x)=-\sin x\)

These let us reduce any trig value to one of an angle in \([0,\pi/2]\). Together with the ASTC sign rule from Part 2, this is the standard "reference-angle" technique.

Identities 10 & 11 — Tangent of sum and difference

Provided no denominator is zero (i.e. \(x,y,x+y\) are not odd multiples of \(\pi/2\)),

\[\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}.\]

Divide every term in numerator and denominator by \(\cos x\cos y\):

Tangent Sum & Difference
\[\boxed{\;\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\;}\] \[\boxed{\;\tan(x-y)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}\;}\] The difference form follows by replacing \(y\) with \(-y\) and using \(\tan(-y)=-\tan y\).

Identities 12 & 13 — Cotangent of sum and difference

Similarly, dividing by \(\sin x\sin y\) gives

Cotangent Sum & Difference
\[\boxed{\;\cot(x+y)=\dfrac{\cot x\cot y-1}{\cot y+\cot x}\;}\] \[\boxed{\;\cot(x-y)=\dfrac{\cot x\cot y+1}{\cot y-\cot x}\;}\] Valid when none of \(x,\ y,\ x\pm y\) is a multiple of \(\pi\).
Interactive: Sum-of-Angles Visualiser

Adjust \(x\) and \(y\). The simulation places \(P_1\) at angle \(x\) and \(P_2\) at angle \(x+y\) on the unit circle, then prints both sides of \(\cos(x+y)=\cos x\cos y-\sin x\sin y\) — they always agree (within rounding).

40° 29°
P₁(x) P₂(x+y)
LHS: cos(x+y) = cos(69°) = 0.3584
RHS: cos x cos y − sin x sin y = (0.766)(0.875) − (0.643)(0.485) = 0.3584
Δ = 0.0000 ✓

Worked Examples

Example 10. Find the value of \(\sin 15°\).
\(\sin 15°=\sin(45°-30°)=\sin 45°\cos 30°-\cos 45°\sin 30°\) \[=\dfrac{1}{\sqrt 2}\cdot\dfrac{\sqrt 3}{2}-\dfrac{1}{\sqrt 2}\cdot\dfrac{1}{2}=\dfrac{\sqrt 3-1}{2\sqrt 2}.\]
Example 11. Find the value of \(\tan\dfrac{13\pi}{12}\).
\(\dfrac{13\pi}{12}=\pi+\dfrac{\pi}{12}\), so \(\tan\dfrac{13\pi}{12}=\tan\dfrac{\pi}{12}\) (period \(\pi\)). Now \(\dfrac{\pi}{12}=\dfrac{\pi}{3}-\dfrac{\pi}{4}\): \[\tan\!\dfrac{\pi}{12}=\dfrac{\tan(\pi/3)-\tan(\pi/4)}{1+\tan(\pi/3)\tan(\pi/4)}=\dfrac{\sqrt 3-1}{1+\sqrt 3}=\dfrac{(\sqrt 3-1)^2}{(\sqrt 3+1)(\sqrt 3-1)}=\dfrac{4-2\sqrt 3}{2}=2-\sqrt 3.\] Hence \(\tan\dfrac{13\pi}{12}=2-\sqrt 3\).
Example 12. Prove that \(\dfrac{\sin(x+y)}{\sin(x-y)}=\dfrac{\tan x+\tan y}{\tan x-\tan y}\).
RHS \(=\dfrac{\sin x/\cos x+\sin y/\cos y}{\sin x/\cos x-\sin y/\cos y}=\dfrac{\sin x\cos y+\cos x\sin y}{\sin x\cos y-\cos x\sin y}=\dfrac{\sin(x+y)}{\sin(x-y)}\) = LHS. \(\square\)
Example 13. Show that \(\cos\!\left(\dfrac{\pi}{4}+x\right)+\cos\!\left(\dfrac{\pi}{4}-x\right)=\sqrt 2\,\cos x\).
LHS \(= [\cos(\pi/4)\cos x-\sin(\pi/4)\sin x]+[\cos(\pi/4)\cos x+\sin(\pi/4)\sin x]\) \[=2\cos(\pi/4)\cos x=2\cdot\dfrac{1}{\sqrt 2}\cdot\cos x=\sqrt 2\,\cos x.\quad\square\]
Activity: Build the Sum Formula From a Picture
L4 Analyse
Materials: Compass, ruler, protractor, graph paper.
Predict: If you draw a unit circle and place \(P_1\) at angle 30° and \(P_2\) at angle \(30°+45°=75°\), can you visually convince yourself that the chord \(P_1P_3\) (where \(P_3\) is at \(-45°\)) has the same length as the chord \(P_2P_4\) (where \(P_4=(1,0)\))?
  1. Draw a unit circle (radius 5 cm) and mark \(P_4=(1,0)\), the rightmost point.
  2. Use the protractor to mark \(P_1\) at 30° (anti-clockwise from positive x-axis), \(P_2\) at 75°, and \(P_3\) at \(-45°\) (i.e. 315°).
  3. Measure chord \(P_1P_3\) and chord \(P_2P_4\) with a ruler. Record both.
  4. The two should be equal (≈ 4.5 cm). Compute the values predicted by the proof: \(P_1P_3^2=P_2P_4^2=2-2\cos 75°\).
  5. Verify your measurement squared matches \(2-2\cos 75°\approx 1.482\).
Both chords have length \(\sqrt{2-2\cos 75°}\approx 1.218\). The reason is the SAS congruence used in the formal proof: triangles \(P_1OP_3\) and \(P_2OP_4\) share two sides of length 1 (radii) and an included angle of 75°. So their third sides are equal — and from this single equality the entire \(\cos(x+y)\) formula falls out via algebra.

Competency-Based Questions

Scenario: A surveyor stands at a fixed point and measures the elevation angle to the top of a building as \(\alpha\). She walks 50 m closer and the new angle is \(\beta\). She wants to express the building's height \(h\) in a closed form involving \(\tan(\beta-\alpha)\) so a calculator can finish the work.
Q1. \(\sin 75°\) equals:
L3 Apply
  • (a) \(\dfrac{\sqrt 3-1}{2\sqrt 2}\)
  • (b) \(\dfrac{\sqrt 3+1}{2\sqrt 2}\)
  • (c) \(\dfrac{1}{\sqrt 2}\)
  • (d) \(\dfrac{\sqrt 3}{2}\)
Answer: (b). \(\sin 75°=\sin(45°+30°)=\sin 45°\cos 30°+\cos 45°\sin 30°=\dfrac{1}{\sqrt 2}\cdot\dfrac{\sqrt 3}{2}+\dfrac{1}{\sqrt 2}\cdot\dfrac{1}{2}=\dfrac{\sqrt 3+1}{2\sqrt 2}\).
Q2. (Fill in the blank) The expression \(\cos 80°\cos 20°+\sin 80°\sin 20°\) equals \(\cos\) ___ which has the value ____.
L2 Understand
Answer: By Identity 4, this equals \(\cos(80°-20°)=\cos 60°=1/2\).
Q3. (True/False) "\(\sin(A+B)=\sin A+\sin B\) for all angles A, B." Justify with a counter-example or proof.
L5 Evaluate
False. Counter-example: \(A=B=30°\). LHS \(=\sin 60°=\sqrt 3/2\approx 0.866\). RHS \(=\sin 30°+\sin 30°=1\). Different. The correct identity is \(\sin(A+B)=\sin A\cos B+\cos A\sin B\) — there is no purely additive expansion.
Q4. Apply Identity 11 to derive a closed-form expression for the surveyor's building height \(h\) in the scenario. (Hint: \(h=d\tan\alpha\) at first position, where \(d\) is distance to base; use the second observation to eliminate \(d\).)
L4 Analyse
Solution: Let \(d_1\) and \(d_2=d_1-50\) be the two distances. \(h=d_1\tan\alpha=d_2\tan\beta\). Solving: \(d_1=\dfrac{50\tan\beta}{\tan\beta-\tan\alpha}\), so \[h=\dfrac{50\tan\alpha\tan\beta}{\tan\beta-\tan\alpha}.\] This is essentially \(50\sin\alpha\sin\beta/\sin(\beta-\alpha)\) after multiplying top and bottom by \(\cos\alpha\cos\beta\) and using \(\sin(\beta-\alpha)=\sin\beta\cos\alpha-\cos\beta\sin\alpha\) (Identity 8).
Q5. Create your own pair of identities by setting \(y=x\) in the sum formulas (sin and cos). What two new identities do you obtain? (You will see these called "double-angle formulas" in Part 4.)
L6 Create
Solution: Setting \(y=x\) in Identity 7: \(\sin 2x=\sin x\cos x+\cos x\sin x=2\sin x\cos x\). Setting \(y=x\) in Identity 3: \(\cos 2x=\cos^2 x-\sin^2 x\). These are exactly the double-angle formulas — Part 4 will explore them and their three equivalent forms.

Assertion–Reason Questions

Assertion (A): \(\cos(x+y)=\cos x\cos y-\sin x\sin y\) is a direct consequence of the chord-distance equality \(P_1P_3=P_2P_4\) on the unit circle.
Reason (R): Triangles \(P_1OP_3\) and \(P_2OP_4\) are congruent by SAS with included angle \(x+y\) and equal sides of length 1.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). R is exactly why the chord lengths are equal, which is exactly the algebraic step that yields A.
Assertion (A): \(\cos(\pi/2-x)=\sin x\).
Reason (R): The complementary angles in a right triangle have swapped sine and cosine values.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). The right-triangle picture (where the two acute angles sum to π/2 and their sin/cos swap) is exactly the geometric content of the cofunction identity.
Assertion (A): \(\tan(x+y)=\tan x+\tan y\) for all valid \(x,y\).
Reason (R): Tangent is an additive function on the real line.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (d). Wait — A is false. The correct formula is \(\tan(x+y)=(\tan x+\tan y)/(1-\tan x\tan y)\); the denominator matters. R is also false: tangent is not additive (no trig function is). So actually neither A nor R is true. The closest single-letter option is (d) "A false, R true" — but R is also false. Best graded answer: both A and R are false; if forced to choose from a/b/c/d, the question itself contains an error and should be flagged.

Frequently Asked Questions

What is the formula for cos(x + y)?
cos(x + y) = cos x cos y − sin x sin y. The minus sign is critical — it is a common error to forget it. The companion identity is cos(x − y) = cos x cos y + sin x sin y.
What is the formula for sin(x + y)?
sin(x + y) = sin x cos y + cos x sin y. Both terms add — there is no sign change. By contrast, sin(x − y) = sin x cos y − cos x sin y has a minus sign in the middle.
How is the cos(x + y) identity proved?
Place x and y as central angles on the unit circle, mark the points P1, P2, P3, P4. Show triangles P1OP3 and P2OP4 are congruent (SAS, with included angle x + y). Equate the squared chord lengths P1P3² = P2P4² and simplify using cos²θ + sin²θ = 1.
What is cos(π/2 − x)?
cos(π/2 − x) = sin x. This is one of the cofunction identities, derived by setting y = π/2 in cos(x − y) and using cos(π/2) = 0, sin(π/2) = 1.
What is the tangent addition formula?
tan(x + y) = (tan x + tan y) / (1 − tan x · tan y), valid whenever none of x, y, x + y is an odd multiple of π/2.
How do we derive the difference formulas from the sum formulas?
Replace y with −y. Using sin(−y) = −sin y and cos(−y) = cos y, the sum formulas immediately yield the difference formulas. This is faster than re-running the chord-distance proof.
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