Miscellaneous Exercise — Trigonometric Functions

1. 3.M.14 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   Prove that 2cos(π/13)cos(9π/13) + cos(3π/13) + cos(5π/13) = 0.

   Hint. Convert the product 2cos(pi/13)cos(9pi/13) to a sum first, then look for angles that are supplementary to the two remaining cosine terms.

   2cosAcosB = cos(A-B)+cos(A+B), with A=pi/13, B=9pi/13: 2cos(pi/13)cos(9pi/13) = cos(-8pi/13)+cos(10pi/13) = cos(8pi/13)+cos(10pi/13). Now 8pi/13 = pi-5pi/13, so cos(8pi/13) = -cos(5pi/13); and 10pi/13 = pi-3pi/13, so cos(10pi/13) = -cos(3pi/13). Substituting: LHS = -cos(5pi/13)-cos(3pi/13)+cos(3pi/13)+cos(5pi/13) = 0.

   ✦ Working through each part gives: lHS = 0, matching the RHS.

2. 3.M.23 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   Prove that (sin 3x + sin x) sin x + (cos 3x - cos x) cos x = 0.

   Hint. Convert both brackets to products using the sum-to-product formulas before multiplying through.

   sin3x+sinx = 2sin2x cosx. cos3x-cosx = -2sin2x sinx. LHS = (2sin2x cosx)(sinx) + (-2sin2x sinx)(cosx) = 2sin2x sinx cosx - 2sin2x sinx cosx = 0.

   ✦ Working through each part gives: lHS = 0, matching the RHS.

3. 3.M.34 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   Prove that (cos x + cos y)^2 + (sin x - sin y)^2 = 4cos^2((x+y)/2).

   Hint. Expand both squares, group using the Pythagorean identity, and collapse the remaining cross terms with the cos(x+y) expansion.

   Expanding: cos^2x+2cosxcosy+cos^2y+sin^2x-2sinxsiny+sin^2y = (cos^2x+sin^2x)+(cos^2y+sin^2y)+2(cosxcosy-sinxsiny) = 1+1+2cos(x+y) = 2+2cos(x+y). Using 1+cos(x+y)=2cos^2((x+y)/2): 2+2cos(x+y) = 2[1+cos(x+y)] = 4cos^2((x+y)/2).

   ✦ Working through each part gives: lHS = 4cos^2((x+y)/2), matching the RHS.

4. 3.M.44 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   Prove that (cos x - cos y)^2 + (sin x - sin y)^2 = 4sin^2((x-y)/2).

   Hint. Expand both squares, group using the Pythagorean identity, and collapse the remaining cross terms with the cos(x-y) expansion.

   Expanding: cos^2x-2cosxcosy+cos^2y+sin^2x-2sinxsiny+sin^2y = 1+1-2(cosxcosy+sinxsiny) = 2-2cos(x-y). Using 1-cos(x-y)=2sin^2((x-y)/2): 2-2cos(x-y) = 2[1-cos(x-y)] = 4sin^2((x-y)/2).

   ✦ Working through each part gives: lHS = 4sin^2((x-y)/2), matching the RHS.

5. 3.M.54 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   Prove that sin x + sin 3x + sin 5x + sin 7x = 4cos x cos 2x sin 4x.

   Hint. Pair the outermost terms (x and 7x) and the inner terms (3x and 5x) separately, then factor the common sin4x.

   (sinx+sin7x)+(sin3x+sin5x) = 2sin4x cos3x + 2sin4x cosx = 2sin4x(cos3x+cosx). Using cos3x+cosx=2cos2x cosx: 2sin4x x 2cos2x cosx = 4sin4x cos2x cosx.

   ✦ Working through each part gives: lHS = 4cos x cos2x sin4x, matching the RHS.

6. 3.M.65 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   Prove that [(sin 7x + sin 5x) + (sin 9x + sin 3x)] / [(cos 7x + cos 5x) + (cos 9x + cos 3x)] = tan 6x.

   Hint. Convert each bracketed pair in the numerator and denominator to a product first; a common factor of (cosx+cos3x) appears in both and cancels.

   sin7x+sin5x = 2sin6x cosx, and sin9x+sin3x = 2sin6x cos3x, so the numerator is 2sin6x(cosx+cos3x). cos7x+cos5x = 2cos6x cosx, and cos9x+cos3x = 2cos6x cos3x, so the denominator is 2cos6x(cosx+cos3x). Dividing cancels the shared 2(cosx+cos3x) factor, leaving sin6x/cos6x = tan6x.

   ✦ Working through each part gives: lHS = tan6x, matching the RHS.

7. 3.M.74 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   Prove that sin 3x + sin 2x - sin x = 4sin x cos(x/2) cos(3x/2).

   Hint. Pair sin3x with -sinx first (a difference), then bring in sin2x and factor out the common 2sinx.

   sin3x-sinx = 2cos2x sinx. Adding sin2x=2sinxcosx: LHS = 2cos2x sinx + 2sinx cosx = 2sinx(cos2x+cosx). Using cos2x+cosx = 2cos(3x/2)cos(x/2): 2sinx x 2cos(3x/2)cos(x/2) = 4sinx cos(x/2)cos(3x/2).

   ✦ Working through each part gives: lHS = 4sin x cos(x/2) cos(3x/2), matching the RHS.

8. 3.M.85 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   tan x = -4/3, x in quadrant II. Find sin(x/2), cos(x/2), and tan(x/2).

   Hint. Get sinx and cosx first from the quadrant, then check which quadrant x/2 itself falls into before applying the half-angle formulas.

   Using a 3-4-5 reference triangle with QII: sinx=4/5, cosx=-3/5. Since pi/2<x<pi, halving gives pi/4<x/2<pi/2 -- x/2 is in QI, so both sin(x/2) and cos(x/2) are positive. sin^2(x/2)=(1-cosx)/2=(1+3/5)/2=4/5, so sin(x/2)=2/root5. cos^2(x/2)=(1+cosx)/2=(1-3/5)/2=1/5, so cos(x/2)=1/root5. tan(x/2)=2/root5 divided by 1/root5=2, confirmed also by tan(x/2)=sinx/(1+cosx)=(4/5)/(2/5)=2.

   ✦ Working through each part gives: sin(x/2) = 2/root5, cos(x/2) = 1/root5, tan(x/2) = 2.

9. 3.M.95 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

   cos x = -1/3, x in quadrant III. Find sin(x/2), cos(x/2), and tan(x/2).

   Hint. Since pi<x<3pi/2, halving gives pi/2<x/2<3pi/4 -- x/2 lands in QII, not QIII, so its sine and cosine have opposite signs.

   Since pi<x<3pi/2, x/2 lies between pi/2 and 3pi/4, i.e. in QII: sin(x/2) is positive, cos(x/2) is negative. sin^2(x/2)=(1-cosx)/2=(1+1/3)/2=2/3, so sin(x/2)=root6/3. cos^2(x/2)=(1+cosx)/2=(1-1/3)/2=1/3, so cos(x/2)=-root3/3 (negative, as x/2 is in QII). tan(x/2)=sinx/(1+cosx); with sinx=-2root2/3 (negative, since x is in QIII) and 1+cosx=2/3: tan(x/2)=(-2root2/3)/(2/3)=-root2.

   ✦ Working through each part gives: sin(x/2) = root6/3, cos(x/2) = -root3/3, tan(x/2) = -root2.

   Where students slip. Assuming x/2 stays in the third quadrant because x is in the third quadrant -- halving the interval pi<x<3pi/2 actually lands x/2 in the second quadrant, which is why cos(x/2) comes out negative while sin(x/2) stays positive.

10. 3.M.105 marksNCERT Class 11 Mathematics, Trigonometric Functions, Reprint 2026-27

    sin x = 1/4, x in quadrant II. Find sin(x/2), cos(x/2), and tan(x/2).

    Hint. Get cosx first from the quadrant, then apply the half-angle formulas directly -- the numbers will not simplify to a tidy fraction here, and that is expected.

    cos^2x=1-1/16=15/16, and since x is in QII, cosx=-root15/4. Since pi/2<x<pi, x/2 is in QI, so both sin(x/2) and cos(x/2) are positive. sin^2(x/2)=(1-cosx)/2=(1+root15/4)/2=(4+root15)/8, so sin(x/2)=root[(4+root15)/8]. cos^2(x/2)=(1+cosx)/2=(4-root15)/8, so cos(x/2)=root[(4-root15)/8]. tan(x/2)=sinx/(1+cosx)=(1/4)/(1-root15/4)=1/(4-root15), which rationalises to 4+root15.

    ✦ Working through each part gives: sin(x/2) = root[(4+root15)/8], cos(x/2) = root[(4-root15)/8], tan(x/2) = 4+root15.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh103.pdf) — three numbered exercises (3.1-3.3, 42 questions) plus the chapter's Miscellaneous Exercise (10 questions); the chapter has no Exercise 3.4 and no general-solution-of-trigonometric-equations content, contrary to an earlier stub. Miscellaneous Exercise Q1's stacked-fraction identity was re-rendered directly from the PDF at 300dpi after raw text extraction garbled the numerators, confirming it as 2cos(pi/13)cos(9pi/13)+cos(3pi/13)+cos(5pi/13)=0. Questions are referenced from the NCERT textbook for identification.