Clase 11 Soluciones RD Sharma – Capítulo 8 Fórmulas de transformación – Ejercicio 8.1

Posted on by Rudeus Greyrat

Pregunta 1. Exprese cada uno de los siguientes como la suma o diferencia de senos y cosenos:

(i) 2 sen 3θ cos θ

Solución:

Usando la identidad trigonométrica,

2 sen A cos B = sen (A+B) + sen (AB)

Tomando A = 3θ y B = θ

2 sen 3θ cos θ = sen (3θ+θ) + sen (3θ-θ)

= sen 4θ + sen 2θ

(ii) 2 cos 3θ sen 2θ

Solución:

Usando la identidad trigonométrica,

2 sen A cos B = sen (A+B) + sen (AB)

Tomando A = 2θ y B = 3θ

2 cos 3θ sen 2θ = sen (3θ+2θ) + sen (2θ-3θ)

= sen 5θ + sen (-θ)

= sen 5θ – sen θ

(iii) 2 sen 4θ sen 3θ

Solución:

Usando la identidad trigonométrica,

2 sen A sen B = cos (AB) – cos (A+B)

Tomando A = 4θ y B = 3θ

2 sen 4θ sen 3θ = coseno (4θ-3θ) – coseno (4θ+3θ)

= cos θ – cos 7θ

(iv) 2 cos 7θ cos 3θ

Solución:

Usando la identidad trigonométrica,

2 cos A cos B = cos (A+B) + cos (AB)

Tomando A = 7θ y B = 3θ

2 cos 7θ cos 3θ = cos (7θ+3θ) + cos (7θ-3θ)

= cos 10θ – cos 4θ

Pregunta 2. Demuestra que:

(i) 2 \hspace{0.1cm}sin \frac{5\pi}{12}\hspace{0.1cm} sin \frac{\pi}{12} = \frac{1}{2}

Solución:

Usando la identidad trigonométrica,

2 sen A sen B = cos (AB) – cos (A+B)

Tomando A =  \frac{5\pi}{12} y B = \frac{\pi}{12}

2 \hspace{0.1cm}sin \frac{5\pi}{12}\hspace{0.1cm} sin \frac{\pi}{12} = coseno (\frac{5\pi}{12}-\frac {\pi}{12}) - coseno (\frac{5\pi}{12}+\frac{\pi}{12})\\ = coseno (\frac{4\pi}{12}) - coseno (\frac{6\pi}{12})\\ = coseno (\frac{\pi}{3}) - coseno (\frac{\pi}{2})\\ = \frac{1}{2 } - 0\\ = \frac{1}{2}

Por lo tanto, LHS = RHS

(ii) 2\hespacio{0,1cm} cos \frac{5\pi}{12} \hespacio{0,1cm}cos \frac{\pi}{12} = \frac{1}{2}

Solución:

Usando la identidad trigonométrica,

2 cos A cos B = cos (A+B) + cos (AB)

Tomando A =  \frac{5\pi}{12} y B = \frac{\pi}{12}

2\hspace{0.1cm} cos \frac{5\pi}{12} \hspace{0.1cm}cos \frac{\pi}{12} = cos (\frac{5\pi}{12}+\frac{\pi}{12}) + cos (\frac{5\pi}{12}-\frac{\pi}{12})

= cos (\frac{6\pi}{12}) + cos (\frac{4\pi}{12})\\ = cos (\frac{\pi}{2}) + cos (\frac{\pi}{3})\\ = 0 + \frac{1}{2}\\ = \frac{1}{2}

Por lo tanto, LHS = RHS

(iii) 2 \hspace{0.1cm}sin \frac{5\pi}{12}\hspace{0.1cm} cos \frac{\pi}{12} = \frac{(\sqrt{3} + 2)}{2}

Solución:

Usando la identidad trigonométrica,

2 sen A cos B = sen (A+B) + sen (AB)

Tomando A =  \frac{5\pi}{12}  y B = \frac{\pi}{12}

2 \hspace{0.1cm}sin \frac{5\pi}{12}\hspace{0.1cm} cos \frac{\pi}{12} = sin (\frac{5\pi}{12}+\frac{\pi}{12}) + sin (\frac{5\pi}{12}-\frac{\pi}{12})

= sin (\frac{6\pi}{12}) + sin (\frac{4\pi}{12})

= pecado  (\frac{\pi}{2})  + pecado (\frac{\pi}{3})

= 1 + \frac{\sqrt{3}}{2}

= \frac{\sqrt{3}+2}{2}

Por lo tanto, LHS = RHS

Pregunta 3. Demuestre que:

(i) sen 50° cos 85° = \frac{(1 - \sqrt{2}sin 35\degree)}{2\sqrt{2}}

Solución:

Usando la identidad trigonométrica,

2 sen A cos B = sen (A+B) + sen (AB)

sen A cos B =  \frac{1}{2} (sen (A+B) + sen (AB))

Tomando A = 50° y B = 85°

sen 50° cos 85° =  \frac{1}{2} (sen (50°+85°) + sen (50°-85°))

=  \frac{1}{2} (pecado (135°) + seno (-35°))

=  \frac{1}{2} (sin (180°-45°) – sin (35°))      (sin(-θ)=-sin θ)

=  \frac{1}{2} (sen (45°) – sen (35°))      (sen(π-θ)=sen θ)

= \frac{1}{2}(\frac{1}{\sqrt{2}} - sin (35\degree))\\ = \frac{1}{2}(\frac{(1 - \sqrt{2}sin 35\degree)}{\sqrt{2}})\\ = \frac{(1 - \sqrt{2}sin 35\degree)}{2\sqrt{2}}

Por lo tanto, LHS = RHS

(ii) sen 25° cos 115° = \frac{1}{2}(sin 40 \degree - 1)

Solución:

Usando la identidad trigonométrica,

2 sen A cos B = sen (A+B) + sen (AB)

sen A cos B =  \frac{1}{2} (sen (A+B) + sen (AB))

Tomando A = 25° y B = 115°

sen 25° cos 115° =  \frac{1}{2} (sen (25°+85°) + sen (25°-115°))

=  \frac{1}{2} (pecado (140°) + seno (-90°))

=  \frac{1}{2} (sin (180-40°) – sin (90°))      (sin(-θ)=-sin θ)

=  \frac{1}{2} (sen (40°) – sen (90°))      (sen(π-θ)=sen θ)

=  \frac{1}{2} (pecado (40°) – 1)

Por lo tanto, LHS = RHS

Pregunta 4. Demostrar que: 4 \hspace{0.1cm}cos \hspace{0.1cm}\theta \hspace{0.1cm}cos\hspace{0.1cm} (\frac{\pi}{3} + \theta) \hspace{0.1cm}cos\hspace{0.1cm} (\frac{\pi}{3} - \theta) = cos 3\theta

Solución:

4 \hspace{0.1cm}cos \hspace{0.1cm}\theta \hspace{0.1cm}cos\hspace{0.1cm} (\frac{\pi}{3} + \theta) \hspace{0.1cm}cos\hspace{0.1cm} (\frac{\pi}{3} - \theta) = 2 \hspace{0.1cm}cos \hspace{0.1cm}\theta(2 \hspace{0.1cm}cos\hspace{0.1cm} (\frac{\pi}{3} + \theta) \hspace{0.1cm}cos\hspace{0.1cm} (\frac{\pi}{3} - \theta))

Usando la identidad trigonométrica,

2 cos A cos B = cos (A+B) + cos (AB)

Tomando A =  \frac{\pi}{3} +θ y B =  \frac{\pi}{3} -θ

= 2 \hspace{0.1cm}cos \hspace{0.1cm}\theta(cos\hspace{0.1cm} (\frac{\pi}{3} + \theta+\frac{\pi}{3} - \theta) +cos\hspace{0.1cm} (\frac{\pi}{3} + \theta)-(\frac{\pi}{3} - \theta))\\ = 2 \hspace{0.1cm}cos \hspace{0.1cm}\theta(cos\hspace{0.1cm} (\frac{2\pi}{3}) +cos\hspace{0.1cm} (2\theta))\\ = 2 \hspace{0.1cm}cos \hspace{0.1cm}\theta(cos\hspace{0.1cm} (\pi+\frac{\pi}{3}) +cos\hspace{0.1cm} (2\theta))\\ = 2 \hspace{0.1cm}cos \hspace{0.1cm}\theta(- cos\hspace{0.1cm} (\frac{\pi}{3}) +cos\hspace{0.1cm} (2\theta))\\ = 2 \hspace{0.1cm}cos \hspace{0.1cm}\theta(- \frac{1}{2} +cos\hspace{0.1cm} (2\theta))\\ = - cos \hspace{0.1cm}\theta + [2 \hspace{0.1cm}cos \hspace{0.1cm}\theta\hspace{0.1cm} cos\hspace{0.1cm} (2\theta))]

Usando de nuevo la identidad, tenemos

Tomando A = 2θ y B = θ

= - cos \hspace{0.1cm}\theta + [cos\hspace{0.1cm} (2\theta + \theta) +cos\hspace{0.1cm} (2\theta - \theta)]\\ = - cos \hspace{0.1cm}\theta + cos\hspace{0.1cm} (3\theta) +cos\hspace{0.1cm} (\theta)\\ = cos\hspace{0.1cm} (3\theta)

Por lo tanto, LHS = RHS

Pregunta 5. Demostrar que:

(i) cos 10° cos 30° cos 50° cos 70° = \frac{3}{16}

Solución:

cos 10° cos 30° cos 50° cos 70° = cos 30° cos 10° cos 50° cos 70°

=  \frac{\sqrt{3}}{2}  (cos 10° cos 50°) cos 70°

Usando la identidad trigonométrica,

2 cos A cos B = cos (A+B) + cos (AB)

cos A cos B =  \frac{1}{2} [cos (A+B) + cos (AB)]

Tomando A = 10° y B = 50°

=  \frac{\sqrt{3}}{2}  ( \frac{1}{2} [cos (10°+50°) + cos (10°-50°)]) cos 70°

=  \frac{\sqrt{3}}{4}  (cos (60°) + cos (-40°)) cos 70°

=  \frac{\sqrt{3}}{4}  ( \frac{1}{2}  + coseno (40°)) coseno 70°

=  \frac{\sqrt{3}}{8}  cos 70° +  \frac{\sqrt{3}}{4}  (cos 70° cos (40°))

Nuevamente usando la identidad, obtenemos

=  \frac{\sqrt{3}}{8}  cos 70° +  \frac{\sqrt{3}}{4}  ( \frac{1}{2} [cos (70°+40°) + cos (70°-40°)])

=  \frac{\sqrt{3}}{8}  cos 70° +  \frac{\sqrt{3}}{8}  [cos (110°) + cos (30°)]

=  \frac{\sqrt{3}}{8}  cos 70° +  \frac{\sqrt{3}}{8}  [cos (110°) +  \frac{\sqrt{3}}{2} ]

=  \frac{\sqrt{3}}{8}  coseno 70° +  \frac{\sqrt{3}}{8}  coseno (110°) + \frac{3}{16}

=  \frac{\sqrt{3}}{8}  (cos 70° + cos (110°)) + \frac{3}{16}

=  \frac{\sqrt{3}}{8}  (cos 70° + cos (180°-70°)) + \frac{3}{16}

=  \frac{\sqrt{3}}{8}  (cos 70° – cos (70°)) + \frac{3}{16}

= \frac{3}{16}

Por lo tanto, LHS = RHS

(ii) cos 40° cos 80° cos 160° = -\frac{1}{8}

Solución:

cos 40° cos 80° cos 160° = cos 80° (cos 40° cos 160°)

Usando la identidad trigonométrica,

2 cos A cos B = cos (A+B) + cos (AB)

cos A cos B =  \frac{1}{2} [cos (A+B) + cos (AB)]

Tomando A = 160° y B = 40°

= cos 80° ( \frac{1}{2} [cos (160°+40°) + cos (160°-40°)])

= cos 80° ( \frac{1}{2} [cos (200°) + cos (120°)])

= cos 80° ( \frac{1}{2} [cos (180°+20°) + cos (180°-60°)])

= cos 80° ( \frac{1}{2} [- cos (20°) + (- cos (60°))])

= cos 80° ( \frac{1}{2} [- cos (20°) – cos (60°)])

= cos 80° ( \frac{1}{2} [- cos (20°) –  \frac{1}{2} ])

=  -\frac{1}{2}  (cos 80° cos (20°) +  \frac{1}{2}  cos 80°])

Nuevamente usando la identidad, obtenemos

= -\frac{1}{2} ((\frac{1}{2}[cos (80\degree+20\degree) + cos (80\degree-20\degree)]) + \frac{1}{2} cos 80\degree)\\ = -\frac{1}{4} ((cos (100\degree) + cos (60\degree)) + cos 80\degree)\\ = -\frac{1}{4} (cos (180\degree-80\degree) + cos (60\degree) + cos 80\degree)\\ = -\frac{1}{4} (- cos (80\degree) + cos (60\degree) + cos 80\degree)\\ = -\frac{1}{4} (cos (60\degree))\\ = -\frac{1}{4} (\frac{1}{2})\\ = -\frac{1}{8}

Por lo tanto, LHS = RHS

(iii) sen 20° sen 40° sen 80° = \frac{\sqrt{3}}{8}

Solución:

sen 20° sen 40° sen 80° = (sen 20° sen 40°) sen 80°

Usando la identidad trigonométrica,

2 sen A sen B = cos (AB) – cos (A+B)

sen A sen B =  \frac{1}{2} [cos (AB) – cos (A+B)]

Tomando A = 40° y B = 20°

= ( \frac{1}{2} [cos (40°-20°) – cos (40°+20°)]) sen 80°

=  \frac{1}{2}  sen 80° [cos (20°) – cos (60°)]

=  \frac{1}{2}  sen 80° [cos (20°) –  \frac{1}{2} ]

=  \frac{1}{2}  [sen 80° cos (20°) –  \frac{1}{2}  sen 80°]

Usando la identidad trigonométrica,

2 sen A cos B = sen (A+B) + sen (AB)

sen A cos B =  \frac{1}{2} [sen (A+B) + sen (AB)]

Tomando A = 80° y B = 20°

= \frac{1}{2} [(\frac{1}{2}[sin (80\degree+20\degree) + sin (80\degree-20\degree)]) - \frac{1}{2} sin 80\degree]\\ = \frac{1}{4} [(sin (80\degree+20\degree) + sin (80\degree-20\degree)) - sin 80\degree]\\ = \frac{1}{4} [sin (100\degree) + sin (60\degree) - sin 80\degree]\\ = \frac{1}{4} [sin (180\degree-80\degree) + \frac{\sqrt{3}}{2} - sin 80\degree]\\ = \frac{1}{4} [sin (80\degree) + \frac{\sqrt{3}}{2} - sin 80\degree]\\ = \frac{1}{4} [\frac{\sqrt{3}}{2}]\\ = \frac{\sqrt{3}}{8}

Por lo tanto, LHS = RHS

(iv) cos 20° cos 40° cos 80° = \frac{1}{8}

Solución:

cos 20° cos 40° cos 80° = cos 40° (cos 20° cos 80°)

Usando la identidad trigonométrica,

2 cos A cos B = cos (A+B) + cos (AB)

cos A cos B =  \frac{1}{2} [cos (A+B) + cos (AB)]

Tomando A = 80° y B = 20°

= cos 40\degree (\frac{1}{2}[cos (80\degree+20\degree) + cos (80\degree-20\degree)])\\ = \frac{1}{2} cos 40\degree [cos (100\degree) + cos (60\degree)]\\ = \frac{1}{2} cos 40\degree[cos (180\degree-80\degree) + cos (60\degree)]\\ = \frac{1}{2} cos 40\degree [- cos (80\degree) + \frac{1}{2}]\\ = \frac{1}{2} cos 40\degree [\frac{1}{2}- cos (80\degree)]\\ = \frac{1}{2} [\frac{1}{2} cos 40\degree - cos (80\degree) cos 40\degree]\\

Nuevamente usando la identidad, obtenemos

= \frac{1}{2} [\frac{1}{2} cos 40\degree - \frac{1}{2}(cos (80\degree+40\degree) + cos (80\degree-40\degree))]\\ = \frac{1}{2} [\frac{1}{2} cos 40\degree - \frac{1}{2}(cos (120\degree) + cos (40\degree))]\\ = \frac{1}{2} [\frac{1}{2} cos 40\degree - \frac{1}{2}(cos (180\degree-60\degree) + cos (40\degree))]\\ = \frac{1}{2} [\frac{1}{2} cos 40\degree - \frac{1}{2}(- cos (60\degree) + cos (40\degree))]\\ = \frac{1}{2} [\frac{1}{2} cos 40\degree + \frac{1}{2} cos (60\degree) - \frac{1}{2} cos (40\degree)]\\ = \frac{1}{2} [\frac{1}{2} \frac{1}{2}]\\ = \frac{1}{8}

Por lo tanto, LHS = RHS

(v) tan 20° tan 40° tan 60° tan 80° = 3

Solución:

tan 20° tan 40° tan 60° tan 80° = tan 60° \frac{sin 20\degree sin 40\degree sin 80\degree}{cos 20\degree cos 40\degree cos 80\degree}

= \sqrt{3} \frac{sin 20\degree sin 40\degree sin 80\degree}{cos 20\degree cos 40\degree cos 80\degree}

= \sqrt{3} \frac{(sin 20\degree sin 40\degree) sin 80\degree}{(cos 20\degree cos 40\degree) cos 80\degree}

Usando la identidad trigonométrica,

2 cos A cos B = cos (A+B) + cos (AB)

cos A cos B =  \frac{1}{2} [cos (A+B) + cos (AB)]

y, 2 sen A sen B = cos (AB) – cos (A+B)

sen A sen B =  \frac{1}{2} [cos (AB) – cos (A+B)]

Tomando A = 40° y B = 20°

= \sqrt{3} (\frac{(\frac{1}{2}[cos ( 40\degree- 20\degree) - cos ( 40\degree+ 20\degree)]) sin 80\degree}{(\frac{1}{2}[cos ( 40\degree+ 20\degree) + cos ( 40\degree- 20\degree)]) cos 80\degree})\\ = \sqrt{3} (\frac{[cos ( 20\degree) - cos (60\degree)] sin 80\degree}{([cos ( 60\degree) + cos (20\degree)) cos 80\degree})\\ = \sqrt{3} (\frac{[cos ( 20\degree) - \frac{1}{2}] sin 80\degree}{(\frac{1}{2} + cos (20\degree)) cos 80\degree})\\ = \sqrt{3} (\frac{(cos ( 20\degree) sin 80\degree - \frac{1}{2} sin 80\degree)}{\frac{1}{2} cos 80\degree + cos (20\degree) cos 80\degree})\\

Nuevamente usando la identidad, obtenemos

2 sen A cos B = sen (A+B) + sen (AB)

sen A cos B =  \frac{1}{2} [sen (A+B) + sen (AB)]

Tomando A = 80° y B = 20°

= \sqrt{3} (\frac{(\frac{1}{2}[sin (80\degree+20\degree) + sin (80\degree-20\degree)] - \frac{1}{2} sin 80\degree)}{\frac{1}{2} cos 80\degree + \frac{1}{2}[cos (80\degree+20\degree) + cos (80\degree-20\degree)]})\\ = \sqrt{3} (\frac{(\frac{1}{2}[sin (100\degree) + sin (60\degree)] - \frac{1}{2} sin 80\degree)}{\frac{1}{2} cos 80\degree + \frac{1}{2}[cos (100\degree) + cos (60\degree)]})\\ = \sqrt{3} (\frac{(\frac{1}{2}[sin (180\degree-80\degree) + sin (60\degree)] - \frac{1}{2} sin 80\degree)}{\frac{1}{2} cos 80\degree + \frac{1}{2}[cos (180\degree-80\degree) + cos (60\degree)]})\\ = \sqrt{3} (\frac{(\frac{1}{2} sin (80\degree) + \frac{1}{2} sin (60\degree) - \frac{1}{2} sin 80\degree)}{\frac{1}{2} (- cos 80\degree) + \frac{1}{2} cos (80\degree) + \frac{1}{2} cos (60\degree)})\\ = \sqrt{3} (\frac{(\frac{1}{2} sin (60\degree)}{\frac{1}{2} cos (60\degree)})\\ = \sqrt{3} (tan (60\degree))\\ = \sqrt{3} (\sqrt{3})\\ = 3

Por lo tanto, LHS = RHS

(vi) tan 20° tan 30° tan 40° tan 80° = 1

Solución:

bronceado 20° bronceado 30° bronceado 40° bronceado 80° = bronceado 30° \frac{sin 20\degree sin 40\degree sin 80\degree}{cos 20\degree cos 40\degree cos 80\degree}

= \frac{1}{\sqrt{3}} \frac{sin 20\degree sin 40\degree sin 80\degree}{cos 20\degree cos 40\degree cos 80\degree}

= \frac{1}{\sqrt{3}}\frac{(sin 20\degree sin 40\degree) sin 80\degree}{(cos 20\degree cos 40\degree) cos 80\degree}

Usando la identidad trigonométrica,

2 cos A cos B = cos (A+B) + cos (AB)

cos A cos B =  \frac{1}{2} [cos (A+B) + cos (AB)]

y, 2 sen A sen B = cos (AB) – cos (A+B)

sen A sen B =  \frac{1}{2} [cos (AB) – cos (A+B)]

Tomando A = 40° y B = 20°

= \frac{1}{\sqrt{3}}\frac{(\frac{1}{2}[cos ( 40\degree- 20\degree) - cos ( 40\degree+ 20\degree)]) sin 80\degree}{(\frac{1}{2}[cos ( 40\degree+ 20\degree) + cos ( 40\degree- 20\degree)]) cos 80\degree}\\ = \frac{1}{\sqrt{3}}\frac{[cos ( 20\degree) - cos (60\degree)] sin 80\degree}{([cos ( 60\degree) + cos (20\degree)) cos 80\degree}\\ = \frac{1}{\sqrt{3}} \frac{[cos ( 20\degree) - \frac{1}{2}] sin 80\degree}{(\frac{1}{2}+ cos (20\degree)) cos 80\degree}\\ = \frac{1}{\sqrt{3}} \frac{(cos ( 20\degree) sin 80\degree - \frac{1}{2}sin 80\degree)}{\frac{1}{2} cos 80\degree + cos (20\degree) cos 80\degree}

Nuevamente usando la identidad, obtenemos

2 sen A cos B = sen (A+B) + sen (AB)

sen A cos B = \frac{1}{2}[sen (A+B) + sen (AB)]

Tomando A = 80° y B = 20°

= \frac{1}{\sqrt{3}} (\frac{1}{2}[sin (80\degree+20\degree) + sin (80\degree-20\degree)] - \frac{1}{2} sin 80\degree){\frac{1}{2} cos 80\degree + \frac{1}{2}[cos (80\degree+20\degree) + cos (80\degree-20\degree)]}\\ = \frac{1}{\sqrt{3}} (\frac{1}{2}[sin (100\degree) + sin (60\degree)] - \frac{1}{2} sin 80\degree){\frac{1}{2} cos 80\degree + \frac{1}{2}[cos (100\degree) + cos (60\degree)]}\\ = \frac{1}{\sqrt{3}} (\frac{1}{2}[sin (180\degree-80\degree) + sin (60\degree)] - \frac{1}{2} sin 80\degree){\frac{1}{2} cos 80\degree + \frac{1}{2}[cos (180\degree-80\degree) + cos (60\degree)]}\\ = \frac{1}{\sqrt{3}} (\frac{1}{2} sin (80\degree) + \frac{1}{2} sin (60\degree) - \frac{1}{2} sin 80\degree){\frac{1}{2} (- cos 80\degree) + \frac{1}{2} cos (80\degree) + \frac{1}{2} cos (60\degree)}\\ = \frac{1}{\sqrt{3}} (\frac{1}{2} sin (60\degree){\frac{1}{2} cos (60\degree)}\\ = \frac{1}{\sqrt{3}} (tan (60\degree))\\ = \frac{1}{\sqrt{3}} (\sqrt{3})\\ = 1

Por lo tanto, LHS = RHS

(vii) sen 10° sen 50° sen 60° sen 70° = \frac{\sqrt{3}}{16}

Solución:

sin 10° sin 50° sin 60° sin 70° = sin 60° (sin 10° sin 50° sin 70°)

= \frac{\sqrt{3}}{2} (sin (90-80°) sin (90-40°) sin (90-20°))

=  \frac{\sqrt{3}}{2}  (cos (80°) cos (40°) cos (20°))

=  \frac{\sqrt{3}}{2}  cos 40° (cos 80° cos 20°)

Usando la identidad trigonométrica,

2 cos A cos B = cos (A+B) + cos (AB)

cos A cos B =  \frac{1}{2} [cos (A+B) + cos (AB)]

Tomando A = 80° y B = 20°

= \frac{\sqrt{3}}{2} cos 40\degree (\frac{1}{2}[cos (80\degree+20\degree) + cos (80\degree-20\degree)])\\ = \frac{\sqrt{3}}{2} \frac{1}{2} cos 40\degree [cos (100\degree) + cos (60°)]\\ = \frac{\sqrt{3}}{4} cos 40\degree[cos (180\degree-80\degree) + cos (60\degree)]\\ = \frac{\sqrt{3}}{4} cos 40\degree [- cos (80\degree) + \frac{1}{2}]\\ = \frac{\sqrt{3}}{4} cos 40\degree [\frac{1}{2}- cos (80\degree)]\\ = \frac{\sqrt{3}}{4} [\frac{1}{2} cos 40\degree - cos (80\degree) cos 40\degree]

Nuevamente usando la identidad, obtenemos

= \frac{\sqrt{3}}{4} [\frac{1}{2} cos 40\degree - \frac{1}{2}(cos (80\degree+40\degree) + cos (80\degree-40\degree))]\\ = \frac{\sqrt{3}}{4} [\frac{1}{2} cos 40\degree - \frac{1}{2}(cos (120\degree) + cos (40\degree))]\\ = \frac{\sqrt{3}}{4} [\frac{1}{2} cos 40\degree - \frac{1}{2}(cos (180\degree-60\degree) + cos (40\degree))]\\ = \frac{\sqrt{3}}{4} [\frac{1}{2} cos 40\degree - \frac{1}{2}(- cos (60\degree) + cos (40\degree))]\\ = \frac{\sqrt{3}}{4} [\frac{1}{2} cos 40\degree + \frac{1}{2} cos (60\degree) - \frac{1}{2} cos (40\degree)]\\ = \frac{\sqrt{3}}{4} [\frac{1}{2} \frac{1}{2}]\\ = \frac{\sqrt{3}}{16}

Por lo tanto, LHS = RHS

(viii) sen 20° sen 40° sen 60° sen 80° = \frac{3}{16}

Solución:

sin 20° sin 40° sin 60° sin 80° = sin 60° (sin 20° sin 40° sin 80°)

=  \frac{\sqrt{3}}{2} (sen 20° sen 40°) sen 80°

Usando la identidad trigonométrica,

2 sen A sen B = cos (AB) – cos (A+B)

sen A sen B =  \frac{1}{2} [cos (AB) – cos (A+B)]

Tomando A = 40° y B = 20°

= \frac{\sqrt{3}}{2} (\frac{1}{2}[cos (40\degree-20\degree) - cos (40\degree+20\degree)]) sin 80\degree\\ = \frac{\sqrt{3}}{2} \frac{1}{2} sin 80\degree [cos (20\degree) - cos (60\degree)]\\ = \frac{\sqrt{3}}{4} sin 80\degree [cos (20\degree) - \frac{1}{2}]\\ = \frac{\sqrt{3}}{4} [sin 80\degree cos (20\degree) - \frac{1}{2} sin 80\degree]\\

Usando la identidad trigonométrica,

2 sen A cos B = sen (A+B) + sen (AB)

sen A cos B =  \frac{1}{2} [sen (A+B) + sen (AB)]

Tomando A = 80° y B = 20°

= \frac{\sqrt{3}}{4} [(\frac{1}{2}[sin (80\degree+20\degree) + sin (80\degree-20\degree)]) - \frac{1}{2} sin 80\degree]\\ = \frac{\sqrt{3}}{8} [(sin (80\degree+20\degree) + sin (80\degree-20\degree)) - sin 80\degree]\\ = \frac{\sqrt{3}}{8} [sin (100\degree) + sin (60\degree) - sin 80\degree]\\ = \frac{\sqrt{3}}{8} [sin (180\degree-80\degree) + \frac{\sqrt{3}}{2} - sin 80\degree]\\ = \frac{\sqrt{3}}{8} [sin (80\degree) + \frac{\sqrt{3}}{2} - sin 80\degree]\\ = \frac{\sqrt{3}}{8} [\frac{\sqrt{3}}{2}]\\ = \frac{3}{16}

Por lo tanto, LHS = RHS

Pregunta 6. Demuestra que

(i) sen A sen (BC) + sen B sen (CA) + sen C sen (AB) = 0

Solución:

Usando la identidad trigonométrica,

2 sen θ sen Φ = coseno (θ-Φ) – coseno (θ+Φ)

sen θ sen Φ =  \frac{1}{2} [cos (θ-Φ) – cos (θ+Φ)]

sen A sen (BC) + sen B sen (CA) + sen C sen (AB) = ( \frac{1}{2} [cos (A-(BC)) – cos (A+(BC))]) + ( \frac{1}{2} [cos (B-(CA) )) – cos (B+(CA))]) + ( \frac{1}{2} [cos (C-(AB)) – cos (C+(AB))])

=  \frac{1}{2} (cos (A-B+C)) – cos (A+BC) + cos (B-C+A) – cos (B+CA) + cos (C-A+B) – cos (C+AB) )

=  \frac{1}{2} (cos (A-B+C)) – cos (C+AB) – cos (A+BC) + cos (B-C+A) – cos (B+CA) + cos (C-A+B) )

= \frac{1}{2}(0)

 = 0

Por lo tanto, LHS = RHS

(ii) sen (BC) cos (AD) + sen (CA) cos (BD) + sen (AB) cos (CD) = 0

Solución:

Usando la identidad trigonométrica,

2 sen θ cos Φ = sen (θ+Φ) + sen (θ-Φ)

sen θ cos Φ =  \frac{1}{2} [sen (θ+Φ) + sen (θ-Φ)]

sen (BC) cos (AD) + sen (CA) cos (BD) + sen (AB) cos (CD) = ( \frac{1}{2} [sen (B-C+(AD)) + sen (BC-(AD))]) + ( \frac{1}{2} [sin (C-A+(BD)) + sin (CA-(BD))]) +( \frac{1}{2} [sin (A-B+(CD)) + sin (AB-(CD))])

= ( \frac{1}{2} [sin (A+BCD) + sin (-A+B-C+D)]) + ( \frac{1}{2} [sin (-A+B+CD) + sin (-A-B+C+D)]) + ( \frac{1}{2} [pecado (A-B+CD) + pecado (AB-C+-D)])

=  \frac{1}{2} (sin (A+BCD) + sin (-(A-B+CD)) + sin (-(AB-C+D)) + sin (-(A+BCD)) +sin (A-B+CD ) + pecado (AB-C+-D))

= \frac{1}{2} (sin (A+BCD) – sin(A-B+CD) – sin (AB-C+D) – sin (A+BCD) +sin (A-B+CD) + sin (AB-C+-D ))

= \frac{1}{2}(0)

= 0

Por lo tanto, LHS = RHS

Pregunta 7. Demostrar que : tan θ tan (60°-θ) tan (60°+θ) = tan 3θ

Solución:

tan θ tan (60°-θ) tan (60°+θ) = tan θ (tan (60°-θ)) (tan (60°+θ))

Usando la identidad trigonométrica,

bronceado (a+b) = \frac{tan\hspace{0.1cm} a + tan\hspace{0.1cm} b}{1 - tan\hspace{0.1cm} a \hspace{0.1cm}tan\hspace{0.1cm} b}

bronceado (a+b) = \frac{tan\hspace{0.1cm} a - tan\hspace{0.1cm} b}{1 + tan\hspace{0.1cm} a \hspace{0.1cm}tan\hspace{0.1cm} b}

= tan\hspace{0.1cm} θ (\frac{tan 60\degree - tan θ}{1 + tan 60\degree tan θ}) ( \frac{tan 60\degree + tan θ}{1 - tan 60\degree tan θ})\\ = tan\hspace{0.1cm} θ (\frac{(tan 60\degree)^2 - (tan θ)^2}{1^2 - (tan 60\degree° tan θ)^2})\hspace{0.1cm}\hspace{0.1cm}\hspace{0.1cm}((a+b)(a-b)=a^2-b^2)\\ = tan\hspace{0.1cm} θ (\frac{(tan 60\degree)^2 - tan^2 θ}{1 - (tan 60\degree)^2 tan^2 θ})\\ = tan \hspace{0.1cm}θ (\frac{(\sqrt{3})^2 - tan^2 θ}{1 - (\sqrt{3})^2 tan^2 θ})\\ = tan\hspace{0.1cm} θ (\frac{3 - tan^2 θ}{1 - 3 tan^2 θ})\\ = \frac{3 \hspace{0.1cm}tan\hspace{0.1cm} \theta - tan^3 \theta}{1 - 3 \hspace{0.1cm}tan^2 \theta}

= tan 3θ

Por lo tanto, LHS = RHS

Pregunta 8. Si α + β = 90°, demuestre que el valor máximo de cos(α) cos(β) es \frac{1}{2}.

Solución:

cos(α) cos(β) = y

Usando la identidad trigonométrica,

2 cos A cos B = cos (A+B) + cos (AB)

cos A cos B =  \frac{1}{2} [cos (A+B) + cos (AB)]

Tomando A = α y B = β

cos(α) cos(β) =  \frac{1}{2} [cos (α+β) + cos (α-β)]

Como, α + β = 90°

y =  \frac{1}{2} [cos (90°) + cos (α-β)]

y =  \frac{1}{2} [0 + coseno (α-β)]

y =  \frac{1}{2} (cos (α-β))

AS, sabemos que el rango de la función cos es [-1,1]

-1\leq (cos (\alpha-\beta)) \leq 1

\frac{-1}{2} \leq \frac{1}{2}(cos (\alpha-\beta)) \leq \frac{1}{2}

\frac{-1}{2} \leq y \leq \frac{1}{2}

Por lo tanto, el valor máximo de cos(α) cos(β) es \frac{1}{2}.

Publicación traducida automáticamente

Artículo escrito por _shinchancode y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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