StudentAidUnit 10: Trigonometric Identities — Exercise 10 4
11th Class Mathematics · Unit 10: Trigonometric Identities
1.(vii) Express sin 12° sin 46° as a sum or difference
Given
(vii) Product
sin 12° sin 46°
Formula
Product-to-sum
sin A sin B = 12[cos(A-B) - cos(A+B)]
Working
Substituting
= -12[-2sin 46° sin 12°]
= -12[cos(46°+12°) - cos(46°-12°)]
= -12[cos 58° - cos 34°]
Result
Final form
= 12[cos 34° - cos 58°]
2.(v) Express cos 12° + cos 48° as a product
Given
(v) Sum
cos 12° + cos 48°
Formula
Sum-to-product
cos A + cos B = 2cosA+B2cosA-B2
Working
Substituting
= 2cos12°+48°2cos12°-48°2
= 2cos 30° cos(-18°)
Result
Final form
= 2cos 30° cos 18°
4.(i) Prove that cos 15° + cos 105° + cos 195° + cos 285° = 0
Given
(i) Expression
cos 15° + cos 105° + cos 195° + cos 285°
Formula
Rewrite using angle relations
cos 105° = cos(90°+15°) = -sin 15°; cos 195° = cos(180°+15°) = -cos 15°; cos 285° = cos(270°+15°) = sin 15°
Working
Substituting
= cos 15° - sin 15° - cos 15° + sin 15°
Simplifying
= (cos 15° - cos 15°) + (-sin 15° + sin 15°)
Result
Result
= 0
5.(i) Prove that cos 20° cos 40° cos 60° cos 80° = 1/16 (ii) Prove that sin(π/9) sin(2π/9) sin(π/3) sin(4π/9) = 3/16 (iii) Prove that sin 10° sin 30° sin 50° sin 70° = 1/16
Given
(i) Product
cos 20° cos 40° cos 60° cos 80°
Formula
Product rule
2cos A cos B = cos(A+B) + cos(A-B)
= 12cos 80° cos 40° cos 20° = 14[cos(80°+40°) + cos(80°-40°)]cos 20°
= 14[cos 120° + cos 40°]cos 20° = 14[-12 + cos 40°]cos 20°
= -18cos 20° + 14cos 40° cos 20°
= -18cos 20° + 18[cos 60° + cos 20°] = -18cos 20° + 116 + 18cos 20°
Result
Result
= 116
Given
(ii) Product
sin 20° sin 40° sin 60° sin 80°
Using product formula
= frac{sqrt{3}{2}[sin 80° sin 40°]sin 20° = frac{sqrt{3}{4}[-2sin 80° sin 40°]sin 20°
= -frac{sqrt{3}{4}[cos(80°+40°) - cos(80°-40°)]sin 20° = -frac{sqrt{3}{4}[cos 120° - cos 40°]sin 20°
= -frac{sqrt{3}{4}[-12 - cos 40°]sin 20° = frac{sqrt{3}{8}sin 20° + frac{sqrt{3}{4}cos 40° sin 20°
= frac{sqrt{3}{8}sin 20° + frac{sqrt{3}{8}[sin(40°+20°) - sin(40°-20°)]
= frac{sqrt{3}{8}sin 20° + frac{sqrt{3}{8}[sin 60° - sin 20°] = frac{sqrt{3}{8}sin 20° + frac{sqrt{3}{8}[frac{sqrt{3}{2} - sin 20°]
= frac{sqrt{3}{8}sin 20° + 316 - frac{sqrt{3}{8}sin 20° = 316
Result
Result
= 316
Given
(iii) Product
sin 10° sin 30° sin 50° sin 70°
Using product formula
= 12[sin 70° sin 50°]sin 10° = 14[-2sin 70° sin 50°]sin 10°
= -14[cos(70°+50°) - cos(70°-50°)]sin 10° = -14[cos 120° - cos 20°]sin 10°
= -14[-12 - cos 20°]sin 10° = 18sin 10° + 14cos 20° sin 10°
= 18sin 10° + 18[sin(20°+10°) - sin(20°-10°)]
= 18sin 10° + 18[sin 30° - sin 10°] = 18sin 10° + 116 - 18sin 10°
Result
Result
= 116
6.Prove that sin 3θ/(1+2cos 2θ) = sin θ; deduce the value of sin 15°
Given
Expression
sin 3θ1+2cos 2θ
Formula
sin 3θ formula
sin 3θ = 3sin θ - 4sin3 θ
Numerator
= sin θ(3 - 4sin2 θ)1 + 2(1 - 2sin2 θ)
Denominator
= sin θ(3 - 4sin2 θ)3 - 4sin2 θ
Result
Simplify
= sin θ = R.H.S.
For sin 15°: Put θ = 15°
sin 15° = sin 45°1+2cos 30° = frac{frac{sqrt{2}{2}{1 + 2(frac{sqrt{3}{2})} = frac{frac{sqrt{2}{2}{1+sqrt{3}
Rationalize
= frac{sqrt{2}{2(1+sqrt{3})} · frac{sqrt{3}-1}{sqrt{3}-1} = frac{sqrt{2}(sqrt{3}-1)}{2(3-1)} = frac{sqrt{2}(sqrt{3}-1)}{4} = frac{sqrt{6}-sqrt{2}{4}
sin 15°
sin 15° = frac{sqrt{3}-1}{2sqrt{2} or frac{sqrt{6}-sqrt{2}{4}
7.Prove that tan 75° - tan 15° = 2√3
Given
Expression
tan 75° - tan 15°
Formula
tan(90°-x) = cot(x)
tan 75° = cot 15° = cos 15°sin 15°
Working
Substitute
= cot 15° - tan 15° = cos 15°sin 15° - sin 15°cos 15°
Combine
= cos2 15° - sin2 15°sin 15° cos 15°
Double angle formula
= frac{cos 30°}{12sin 30°}
Substitute values
= 2cos 30°sin 30° = frac{2 · frac{sqrt{3}{2}{12 = frac{sqrt{3}{12
Result
Result
= 2sqrt{3}
8.Prove that cos 15° - sin 15° = 1/√2
Given
Expression
cos 15° - sin 15°
Formula
Sum-to-product
sin A - cos B = 2cosA+B2sinA-B2
Rewrite
= cos(90° - 75°) - sin 15° = sin 75° - sin 15°
Working
Apply formula
= 2cos75°+15°2sin75°-15°2
Simplify
= 2cos 45° sin 30°
Substitute
= 2 · frac{1}{sqrt{2} · 12
Result
Result
= frac{1}{sqrt{2} = R.H.S.