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3.4 (continued) — Multiple-Angle and Product-to-Sum Identities

🎓 Class 11 Mathematics CBSE Theory Ch 3 — Trigonometric Functions ⏱ ~15 min
🌐 Language: [gtranslate]

This MCQ module is based on: 3.4 (continued) — Multiple-Angle and Product-to-Sum Identities

This mathematics assessment will be based on: 3.4 (continued) — Multiple-Angle and Product-to-Sum Identities
Targeting Class 11 level in Trigonometry, with Advanced difficulty.

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3.4 (continued) — Multiple-Angle and Product-to-Sum Identities

Setting \(y=x\) in the sum formulas of Part 3 produces the powerful family of double-angle? identities. Iterating once more gives triple-angle formulas. Adding and subtracting sum/difference identities produces the sum-to-product family used routinely in integration and equation-solving.

Identity 14 — Three forms of cos 2x

From Identity 3, \(\cos(x+y)=\cos x\cos y-\sin x\sin y\), with \(y=x\):

\[\cos 2x=\cos^2 x-\sin^2 x.\]

Using \(\sin^2x=1-\cos^2 x\) and \(\cos^2 x=1-\sin^2 x\) gives the other two forms:

Double angle for cosine — three equivalent forms
\[\boxed{\;\cos 2x=\cos^2 x-\sin^2 x=2\cos^2 x-1=1-2\sin^2 x\;}\] Dividing the first by \(\cos^2 x\) (and adjusting denominator): \[\cos 2x=\dfrac{1-\tan^2 x}{1+\tan^2 x},\qquad x\neq (2n+1)\dfrac{\pi}{2}.\]

The forms are interchangeable; pick whichever matches the data you have. Half-angle formulas come from rearranging the second and third forms:

\[\sin^2 x=\dfrac{1-\cos 2x}{2},\qquad \cos^2 x=\dfrac{1+\cos 2x}{2}.\]

These are the workhorses for integrating \(\sin^2 x\) and \(\cos^2 x\) in Class 12.

Identity 15 — sin 2x

From Identity 7 with \(y=x\):

Double angle for sine
\[\boxed{\;\sin 2x=2\sin x\cos x\;}=\dfrac{2\tan x}{1+\tan^2 x}.\]

Identity 16 — tan 2x

From Identity 10 with \(y=x\):

Double angle for tangent
\[\boxed{\;\tan 2x=\dfrac{2\tan x}{1-\tan^2 x}\;},\qquad x\neq (2n+1)\dfrac{\pi}{4}\text{ etc.}\]

Identity 17 — sin 3x

Write \(3x=2x+x\) and apply Identity 7:

\[\sin 3x=\sin(2x+x)=\sin 2x\cos x+\cos 2x\sin x.\]

Substitute \(\sin 2x=2\sin x\cos x\) and \(\cos 2x=1-2\sin^2 x\):

\[=2\sin x\cos^2 x+(1-2\sin^2 x)\sin x=2\sin x(1-\sin^2 x)+\sin x-2\sin^3 x=3\sin x-4\sin^3 x.\]
Triple angle for sine
\[\boxed{\;\sin 3x=3\sin x-4\sin^3 x\;}\]

Identity 18 — cos 3x

Similarly, \(\cos 3x=\cos(2x+x)=\cos 2x\cos x-\sin 2x\sin x\). Substituting \(\cos 2x=2\cos^2 x-1\) and \(\sin 2x=2\sin x\cos x\):

\[=(2\cos^2 x-1)\cos x-2\sin^2 x\cos x=2\cos^3 x-\cos x-2(1-\cos^2 x)\cos x=4\cos^3 x-3\cos x.\]
Triple angle for cosine
\[\boxed{\;\cos 3x=4\cos^3 x-3\cos x\;}\]

Identity 19 — tan 3x

\[\tan 3x=\tan(2x+x)=\dfrac{\tan 2x+\tan x}{1-\tan 2x\tan x}.\]

Substituting Identity 16 and simplifying:

Triple angle for tangent
\[\boxed{\;\tan 3x=\dfrac{3\tan x-\tan^3 x}{1-3\tan^2 x}\;},\qquad 3x\neq (2n+1)\dfrac{\pi}{2}\text{ etc.}\]

Identity 20 — Sum-to-product (and product-to-sum)

Adding \(\cos(x+y)+\cos(x-y)\) gives \(2\cos x\cos y\). Subtracting gives \(-2\sin x\sin y\). Doing the same with \(\sin(x+y)\pm\sin(x-y)\) yields:

Product-to-sum (LHS) and sum-to-product (RHS)
\[2\cos x\cos y=\cos(x-y)+\cos(x+y),\] \[-2\sin x\sin y=\cos(x+y)-\cos(x-y),\] \[2\sin x\cos y=\sin(x+y)+\sin(x-y),\] \[2\cos x\sin y=\sin(x+y)-\sin(x-y).\] With the substitution \(x+y=\theta,\ x-y=\phi\) (so \(x=\dfrac{\theta+\phi}{2},\ y=\dfrac{\theta-\phi}{2}\)) the last four become: \[\cos\theta+\cos\phi=2\cos\!\dfrac{\theta+\phi}{2}\cos\!\dfrac{\theta-\phi}{2},\] \[\cos\theta-\cos\phi=-2\sin\!\dfrac{\theta+\phi}{2}\sin\!\dfrac{\theta-\phi}{2},\] \[\sin\theta+\sin\phi=2\sin\!\dfrac{\theta+\phi}{2}\cos\!\dfrac{\theta-\phi}{2},\] \[\sin\theta-\sin\phi=2\cos\!\dfrac{\theta+\phi}{2}\sin\!\dfrac{\theta-\phi}{2}.\]

Memory aid: "sin + sin" and "sin − sin" each give a 2 × (sin or cos)(cos or sin) product; "cos + cos" gives a positive cos·cos product; only "cos − cos" carries a leading minus sign.

Worked Examples

Example 14. Prove that \(\dfrac{\cos 7x+\cos 5x}{\sin 7x-\sin 5x}=\cot x\).
Apply sum-to-product to numerator and denominator: \[\cos 7x+\cos 5x=2\cos 6x\cos x,\qquad \sin 7x-\sin 5x=2\cos 6x\sin x.\] Hence the ratio is \(\dfrac{2\cos 6x\cos x}{2\cos 6x\sin x}=\dfrac{\cos x}{\sin x}=\cot x\). \(\square\)
Example 15. Prove that \(\dfrac{\sin 5x-2\sin 3x+\sin x}{\cos 5x-\cos x}=\tan x\).
Numerator: \((\sin 5x+\sin x)-2\sin 3x=2\sin 3x\cos 2x-2\sin 3x=2\sin 3x(\cos 2x-1)=-4\sin 3x\sin^2 x\). Denominator: \(\cos 5x-\cos x=-2\sin 3x\sin 2x=-4\sin 3x\sin x\cos x\). The ratio is \[\dfrac{-4\sin 3x\sin^2 x}{-4\sin 3x\sin x\cos x}=\dfrac{\sin x}{\cos x}=\tan x.\quad\square\]
Example 16. Find the value of \(\tan\dfrac{\pi}{8}\).
Let \(y=\tan(\pi/8)\). Then \(2y/(1-y^2)=\tan(2\cdot\pi/8)=\tan(\pi/4)=1\), giving \[y^2+2y-1=0\Rightarrow y=\dfrac{-2\pm\sqrt 8}{2}=-1\pm\sqrt 2.\] Since \(\pi/8\) is in the first quadrant, \(y>0\), so \(\tan(\pi/8)=\sqrt 2-1\).
Example 17. If \(\tan x=\dfrac{3}{4}\) and \(\pi
\(x\in(\pi,3\pi/2)\Rightarrow\) Q3, so cos x is negative. \(\sec^2 x=1+\tan^2 x=1+9/16=25/16\Rightarrow \cos x=-4/5\). Range of \(x/2\) is \((\pi/2,3\pi/4)\Rightarrow\) Q2, so \(\sin(x/2)>0,\ \cos(x/2)<0\). \(2\sin^2(x/2)=1-\cos x=1+4/5=9/5\Rightarrow\sin(x/2)=3/\sqrt{10}\). \(2\cos^2(x/2)=1+\cos x=1-4/5=1/5\Rightarrow\cos(x/2)=-1/\sqrt{10}\). \(\tan(x/2)=\sin(x/2)/\cos(x/2)=-3\).
Example 18. Prove that \(\cos^2 x+\cos^2\!\left(x+\dfrac{\pi}{3}\right)+\cos^2\!\left(x-\dfrac{\pi}{3}\right)=\dfrac{3}{2}\).
Use \(\cos^2\theta=\dfrac{1+\cos 2\theta}{2}\) on each term: \[\text{LHS}=\dfrac{1+\cos 2x}{2}+\dfrac{1+\cos(2x+2\pi/3)}{2}+\dfrac{1+\cos(2x-2\pi/3)}{2}.\] The constant terms add to \(3/2\). The \(\cos\) terms: \(\cos 2x+\cos(2x+2\pi/3)+\cos(2x-2\pi/3)\). Apply sum-to-product on the last two: \(2\cos 2x\cos(2\pi/3)=2\cos 2x\cdot(-1/2)=-\cos 2x\). So the three \(\cos\) terms sum to \(\cos 2x-\cos 2x=0\). Hence LHS = \(3/2\). \(\square\)
Activity: Three Forms of cos 2x — Pick the Right One
L4 Analyse
Materials: Calculator, paper, pen.
Predict: If you only know \(\cos x\), which of the three forms of \(\cos 2x\) gives the answer in one step? What if you only know \(\sin x\)? Or only \(\tan x\)?
  1. Take the value \(\cos x=0.6\) (with \(x\) acute) and compute \(\cos 2x\) using each of the three forms:
    (a) \(2\cos^2 x-1\);   (b) \(1-2\sin^2 x\) (you'll need \(\sin x\) first);   (c) \((1-\tan^2 x)/(1+\tan^2 x)\) (you'll need \(\tan x\) first).
  2. Verify all three give the same answer (≈ \(-0.28\)).
  3. Now repeat with only \(\sin x=0.6\) given. Which form is fastest now?
  4. And with only \(\tan x=0.75\) given. Which form is fastest?
  5. Conclusion: form (a) is fastest if you know cos; form (b) if you know sin; form (c) if you know tan.
All three forms produce the identical numerical value (≈ \(-0.28\) for \(\cos x=0.6\), or \(0.28\) for \(\sin x=0.6\)). Match the form to the data you have — that's the entire utility of having three equivalent expressions.

Competency-Based Questions

Scenario: A pendulum of length 1 m is released from a horizontal position (angle 90° from vertical). Its angular displacement at time \(t\) is approximately \(\theta(t)=\dfrac{\pi}{2}\cos(\omega t)\) for small \(\omega t\) — though we keep the first-order behaviour for this exercise. The vertical height of the bob above its lowest point is \(h(t)=1-\cos\theta(t)=2\sin^2\!\left(\dfrac{\theta(t)}{2}\right)\).
Q1. The expression \(\sin 2x\) when \(\sin x=\dfrac{4}{5}\) and \(x\) is in the second quadrant equals:
L3 Apply
  • (a) \(24/25\)
  • (b) \(-24/25\)
  • (c) \(7/25\)
  • (d) \(-7/25\)
Answer: (b). In Q2, \(\cos x=-3/5\). \(\sin 2x=2\sin x\cos x=2(4/5)(-3/5)=-24/25\). (Note: \(2x\in(\pi,2\pi)\), so sin 2x can be either sign — its actual sign depends on the exact value.)
Q2. (Fill in the blank) \(\cos 2x=2\cos^2 x-1\) is most useful for converting \(\cos 2x\) into an expression in \(\cos x\) only — but the dual form ____ is preferred when the data is in terms of \(\sin x\).
L2 Understand
Answer: \(\cos 2x=1-2\sin^2 x\).
Q3. (True/False) "\(\sin 3x=\sin(2x+x)\) and a single use of the sum formula gives the closed-form \(3\sin x-4\sin^3 x\) without using any double-angle identity." Explain.
L5 Evaluate
False. One application of the sum formula gives \(\sin 2x\cos x+\cos 2x\sin x\), which still contains \(\sin 2x\) and \(\cos 2x\). To reach a polynomial in \(\sin x\) alone, you must substitute the double-angle formulas. So the proof essentially uses both the sum formula and the double-angle identities — they cooperate, neither alone is enough.
Q4. Use the half-angle relation \(2\sin^2(\theta/2)=1-\cos\theta\) to compute \(h(t)\) when \(\theta=60°\), the bob's height in metres above its lowest point.
L3 Apply
Solution: \(h=2\sin^2(30°)=2\cdot(1/2)^2=2\cdot 1/4=0.5\) m. Cross-check: \(1-\cos 60°=1-1/2=0.5\) m. ✓
Q5. Design: An audio engineer wants to convert the product \(\sin 440\pi t\cdot\sin 220\pi t\) (two pure tones) into a sum of sines/cosines, to identify the audible "beat" frequencies. Use a product-to-sum identity to do so.
L6 Create
Solution: \(2\sin A\sin B=\cos(A-B)-\cos(A+B)\). With \(A=440\pi t,\ B=220\pi t\): \[\sin 440\pi t\sin 220\pi t=\dfrac{1}{2}\bigl[\cos(220\pi t)-\cos(660\pi t)\bigr].\] The output contains a low-frequency 110 Hz term (the audible beat / difference tone) and a high-frequency 330 Hz term (sum tone). This is the mathematical foundation of AM-radio modulation.

Assertion–Reason Questions

Assertion (A): \(\cos 2x=2\cos^2 x-1\) implies \(\cos^2 x=\dfrac{1+\cos 2x}{2}\).
Reason (R): The half-angle formula is the algebraic rearrangement of one of the three forms of \(\cos 2x\).
(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). Solve \(2\cos^2 x-1=\cos 2x\) for \(\cos^2 x\). R is the precise mechanism behind A.
Assertion (A): \(\sin 3x=3\sin x-4\sin^3 x\) is a cubic in \(\sin x\).
Reason (R): Triple-angle identities can always be expressed as polynomials in \(\sin x\) (or \(\cos x\)) of degree 3.
(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). Both true. The general fact (R) is exactly the Chebyshev polynomial result; the specific case of degree 3 is A.
Assertion (A): Sum-to-product identities can convert \(\sin 7x+\sin 3x\) into \(2\sin 5x\cos 2x\).
Reason (R): \(\sin\theta+\sin\phi=2\sin\dfrac{\theta+\phi}{2}\cos\dfrac{\theta-\phi}{2}\).
(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). Apply R with \(\theta=7x,\phi=3x\): \((7x+3x)/2=5x\) and \((7x-3x)/2=2x\). Direct.

Frequently Asked Questions

What are the three forms of cos 2x?
cos 2x = cos²x − sin²x = 2cos²x − 1 = 1 − 2sin²x. The first follows from cos(x+x). The second uses sin²x = 1 − cos²x; the third uses cos²x = 1 − sin²x.
What is the formula for sin 3x?
sin 3x = 3 sin x − 4 sin³x. It is derived from sin(2x + x) using the sin sum formula and the double-angle identities.
What is the formula for cos 3x?
cos 3x = 4 cos³x − 3 cos x. Derived from cos(2x + x) and the double-angle identities.
What is a sum-to-product identity?
sin x + sin y = 2 sin((x+y)/2) cos((x−y)/2). Similar formulas exist for sin x − sin y, cos x + cos y, and cos x − cos y. They convert sums into products and are useful in integration, in solving trig equations, and in proving identities.
How do you express cos 2x using tan x?
cos 2x = (1 − tan²x) / (1 + tan²x), provided x is not an odd multiple of π/2. This is obtained by dividing the numerator and denominator of cos²x − sin²x by cos²x.
What is tan(π/8)?
tan(π/8) = √2 − 1. Set y = tan(π/8); then tan(π/4) = 2y/(1 − y²) = 1, giving y² + 2y − 1 = 0, so y = −1 + √2.
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