Acceso
Prima
Hogar Banco de preguntas Matemáticas Trigonometría
Pregunta
upstudy study bank question image url

Evaluate, without using a calculator: \( X \) \[ \sum_{y=43^{\circ}}^{47^{\circ}} \sin ^{2} y \]

Ask by Cruz Bond. in South Africa
Mar 12,2025

Solución de inteligencia artificial de Upstudy

Respuesta verificada por el tutor

Responder

The sum \( X = \frac{5}{2} \).

Solución

We start with \[ X=\sum_{y=43^{\circ}}^{47^{\circ}} \sin ^{2} y = \sin^2 43^\circ + \sin^2 44^\circ + \sin^2 45^\circ + \sin^2 46^\circ + \sin^2 47^\circ. \] **Step 1. Use the complementary angle identity** Recall that \[ \sin^2 \theta = \cos^2 (90^\circ - \theta). \] Thus, notice that: \[ \sin^2 43^\circ = \cos^2 47^\circ \quad \text{and} \quad \sin^2 47^\circ = \cos^2 43^\circ. \] Similarly, \[ \sin^2 44^\circ = \cos^2 46^\circ \quad \text{and} \quad \sin^2 46^\circ = \cos^2 44^\circ. \] **Step 2. Group the complementary terms** Grouping the terms, we have: - For \(43^\circ\) and \(47^\circ\): \[ \sin^2 43^\circ + \sin^2 47^\circ = \cos^2 47^\circ + \cos^2 43^\circ. \] But observe that \(\cos^2 47^\circ + \sin^2 47^\circ = 1\) if the argument were the same. Instead, we apply the identity in a different way: since we already have the pair \(\sin^2 43^\circ\) and \(\cos^2 43^\circ\) if we use \(\sin^2 47^\circ = \cos^2 43^\circ\), so \[ \sin^2 43^\circ + \sin^2 47^\circ = \sin^2 43^\circ + \cos^2 43^\circ = 1. \] - For \(44^\circ\) and \(46^\circ\): \[ \sin^2 44^\circ + \sin^2 46^\circ = \sin^2 44^\circ + \cos^2 44^\circ = 1. \] - The remaining term is: \[ \sin^2 45^\circ = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}. \] **Step 3. Sum up the values** Adding the groups together: \[ X = 1 + 1 + \frac{1}{2} = 2 + \frac{1}{2} = \frac{5}{2}. \] Thus, the evaluated sum is: \[ \frac{5}{2}. \]

Respondido por UpStudy AI y revisado por un tutor profesional

error msg
Explicar
Simplifique esta solución

Bonus Knowledge

To evaluate the sum \( X = \sum_{y=43^{\circ}}^{47^{\circ}} \sin^2 y \), we will compute \( \sin^2 43^{\circ} \), \( \sin^2 44^{\circ} \), \( \sin^2 45^{\circ} \), \( \sin^2 46^{\circ} \), and \( \sin^2 47^{\circ} \). First, we recall that \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \) from the double angle identity. Therefore, we can rewrite each term as follows: \[ \sin^2 43^{\circ} = \frac{1 - \cos(86^{\circ})}{2} \] \[ \sin^2 44^{\circ} = \frac{1 - \cos(88^{\circ})}{2} \] \[ \sin^2 45^{\circ} = \frac{1 - \cos(90^{\circ})}{2} = \frac{1}{2} \] \[ \sin^2 46^{\circ} = \frac{1 - \cos(92^{\circ})}{2} \] \[ \sin^2 47^{\circ} = \frac{1 - \cos(94^{\circ})}{2} \] Now we can compute the total sum: \[ X = \frac{1 - \cos(86^{\circ})}{2} + \frac{1 - \cos(88^{\circ})}{2} + \frac{1}{2} + \frac{1 - \cos(92^{\circ})}{2} + \frac{1 - \cos(94^{\circ})}{2} \] Combining these, we have: \[ X = \frac{5}{2} - \frac{\cos(86^{\circ}) + \cos(88^{\circ}) + \cos(92^{\circ}) + \cos(94^{\circ})}{2} \] Next, we use the fact that \( \cos(90^{\circ} + x) = -\sin(x) \) to simplify: \[ \cos(86^{\circ}) = \sin(4^{\circ},\quad \cos(88^{\circ}) = \sin(2^{\circ}),\quad \cos(92^{\circ}) = \sin(2^{\circ}),\quad \cos(94^{\circ}) = \sin(4^{\circ}) \] These provide us with the equalities: \[ \cos(86^{\circ}) + \cos(94^{\circ}) = \sin(4^{\circ}) + \sin(4^{\circ}) = 2\sin(4^{\circ}) \] \[ \cos(88^{\circ}) + \cos(92^{\circ}) = \sin(2^{\circ}) + \sin(2^{\circ}) = 2\sin(2^{\circ}) \] Thus, we have: \[ X = \frac{5}{2} - \frac{2\sin(4^{\circ}) + 2\sin(2^{\circ})}{2} = \frac{5}{2} - (\sin(4^{\circ}) + \sin(2^{\circ})) \] Finally, without precise values for \( \sin(2^{\circ}) \) and \( \sin(4^{\circ}) \), we conclude that the sum \( \sum_{y=43^{\circ}}^{47^{\circ}} \sin^2 y \) can be expressed as: \[ X = \frac{5}{2} - (\sin(4^{\circ}) + \sin(2^{\circ})). \] While we do not have exact numerical answers, we summarize: \[ \boxed{X = \frac{5}{2} - (\sin(4^{\circ}) + \sin(2^{\circ}))} \]

preguntas relacionadas

\( 12.3 \quad \) 12.3.1 \( \quad \) Prove that, for angles \( A \) and \( B \) \[ \frac{\sin A}{\sin B}-\frac{\cos A}{\cos B}=\frac{2 \sin (A-B)}{\sin 2 B} \] 12.3.2 Hence, or otherwise, without using a calculator, show that: (a) \( \frac{\sin 5 B}{\sin B}-\frac{\cos 5 B}{\cos B}=4 \cos 2 B \) (b) \( \quad \frac{1}{\sin 18^{\circ}}=4 \cos 36^{\circ} \)
Trigonometría South Africa Mar 12, 2025
prove the following identities: \( \begin{array}{ll}\text { (1) } \cos x \sin x+\sin ^{2} x \tan x=\tan x & \text { (2) } \frac{\tan A \cdot \sin 2 A}{1-\cos A}=2+2 \cos A \\ \text { (3) } \frac{\cos \alpha+\sin \alpha}{\cos 2 \alpha}=\frac{\cos ^{2} \alpha+\sin ^{2} \alpha}{\cos \alpha-\sin \alpha} & \text { (4) } \frac{\sin 2 x-2 \sin x}{\cos 2 x-1}=\frac{1}{\sin x}-\frac{1}{\tan x} \\ \text { (5) } \frac{\sin \theta-\sin 2 \theta}{\cos \theta-\cos 2 \theta-1}=\tan \theta & \text { (6) } \frac{\cos 2 \alpha+\cos \alpha}{\sin 2 \alpha-\sin \alpha}=\frac{\cos \alpha+1}{\sin \alpha}\end{array} \)
Trigonometría South Africa Mar 12, 2025
6. Rezolvaṭi următoarele ecuatii trigonometrice elementare: \( \begin{array}{ll}\text { a) } \sin 5 x=\sin 3 x ; & \text { b) } \cos x=\cos 4 x \\ \text { c) } \cos 5 x=-\sin 3 x ; & \text { d) } \operatorname{tg} 4 x=\operatorname{ctg}\left(3 x+\frac{\pi}{4}\right) ; \\ \begin{array}{ll}\text { e) } \sin \left(\frac{3 \pi}{2}+x\right)+\cos 3 x=0 ; & \text { f) } \cos \left(\frac{3 x}{2}+\frac{\pi}{4}\right)+\sin \left(\frac{2 x}{3}-\frac{\pi}{4}\right)=0\end{array} \\ \text { 7. Rezolvați următoarele ecuații trigonometrice elementare: } \\ \begin{array}{lll}\text { a) } \sin x=\sin 7 x ; & \text { b) }-\sin 2 x=\sin x ; & \text { e) } \operatorname{tg} 4 x=\operatorname{tg} 2 x ;\end{array} & \text { c) } \cos 4 x=\cos 5 x \\ \text { d) } \cos 2 x=-\cos 6 x ; & \text { f) } \operatorname{ctg} 2 x=\operatorname{ctg} x\end{array} \)
Trigonometría Romania Mar 12, 2025
мданыз: \( \frac{\sin (2 \pi-\alpha)}{\operatorname{ctg}\left(\frac{\pi}{2}-\alpha\right)} \cdot \frac{\operatorname{tg}\left(\frac{\pi}{2}+\alpha\right)}{\cos \left(\frac{\pi}{2}+\alpha\right)}+\frac{\cos (2 \pi+\alpha)}{\sin \left(\frac{3 \pi}{2}+\alpha\right)} \)
Trigonometría Kazakhstan Mar 12, 2025
шамданыз: \( \left(\operatorname{ctg} \alpha+\frac{\sin \alpha}{1+\cos \alpha}\right)^{2}-\operatorname{ctg}^{2} \alpha \)
Trigonometría Kazakhstan Mar 12, 2025

Latest Trigonometry Questions

6. Rezolvaṭi următoarele ecuatii trigonometrice elementare: \( \begin{array}{ll}\text { a) } \sin 5 x=\sin 3 x ; & \text { b) } \cos x=\cos 4 x \\ \text { c) } \cos 5 x=-\sin 3 x ; & \text { d) } \operatorname{tg} 4 x=\operatorname{ctg}\left(3 x+\frac{\pi}{4}\right) ; \\ \begin{array}{ll}\text { e) } \sin \left(\frac{3 \pi}{2}+x\right)+\cos 3 x=0 ; & \text { f) } \cos \left(\frac{3 x}{2}+\frac{\pi}{4}\right)+\sin \left(\frac{2 x}{3}-\frac{\pi}{4}\right)=0\end{array} \\ \text { 7. Rezolvați următoarele ecuații trigonometrice elementare: } \\ \begin{array}{lll}\text { a) } \sin x=\sin 7 x ; & \text { b) }-\sin 2 x=\sin x ; & \text { e) } \operatorname{tg} 4 x=\operatorname{tg} 2 x ;\end{array} & \text { c) } \cos 4 x=\cos 5 x \\ \text { d) } \cos 2 x=-\cos 6 x ; & \text { f) } \operatorname{ctg} 2 x=\operatorname{ctg} x\end{array} \)
Trigonometría Romania Mar 12, 2025
STION 6 Without using a calculator, simplify the following expression to a single trigonometric term: \( \frac{\sin 10^{\circ}}{\cos 440^{\circ}}+\tan \left(360^{\circ}-\theta\right) \cdot \sin 2 \theta \) Given: \( \sin \left(60^{\circ}+2 x\right)+\sin \left(60^{\circ}-2 x\right) \) 6.2 .1 \( 6.2 .2 \quad \) Calculate the value of \( k \) if \( \sin \left(60^{\circ}+2 x\right)+\sin \left(60^{\circ}-2 x\right)=k \cos 2 x \). If \( \cos x=\sqrt{t} \), without using a calculator, determine the value of \( \tan 60^{\circ}\left[\sin \left(60^{\circ}+2 x\right)+\sin \left(60^{\circ}-2 x\right)\right] \) in terms of \( t \).
Trigonometría South Africa Mar 12, 2025
мданыз: \( \frac{\sin (2 \pi-\alpha)}{\operatorname{ctg}\left(\frac{\pi}{2}-\alpha\right)} \cdot \frac{\operatorname{tg}\left(\frac{\pi}{2}+\alpha\right)}{\cos \left(\frac{\pi}{2}+\alpha\right)}+\frac{\cos (2 \pi+\alpha)}{\sin \left(\frac{3 \pi}{2}+\alpha\right)} \)
Trigonometría Kazakhstan Mar 12, 2025
prove the following identities: \( \begin{array}{ll}\text { (1) } \cos x \sin x+\sin ^{2} x \tan x=\tan x & \text { (2) } \frac{\tan A \cdot \sin 2 A}{1-\cos A}=2+2 \cos A \\ \text { (3) } \frac{\cos \alpha+\sin \alpha}{\cos 2 \alpha}=\frac{\cos ^{2} \alpha+\sin ^{2} \alpha}{\cos \alpha-\sin \alpha} & \text { (4) } \frac{\sin 2 x-2 \sin x}{\cos 2 x-1}=\frac{1}{\sin x}-\frac{1}{\tan x} \\ \text { (5) } \frac{\sin \theta-\sin 2 \theta}{\cos \theta-\cos 2 \theta-1}=\tan \theta & \text { (6) } \frac{\cos 2 \alpha+\cos \alpha}{\sin 2 \alpha-\sin \alpha}=\frac{\cos \alpha+1}{\sin \alpha}\end{array} \)
Trigonometría South Africa Mar 12, 2025
\( { }^{\circ} \cos 50^{\circ}+\sin 80^{\circ} \sin 50^{\circ} \) даныз: \( \left(\frac{\sin \alpha}{1+\cos \alpha}+\frac{\sin \alpha}{1-\cos \alpha}\right) \cdot \sin 2 \alpha \) \( \frac{\cos \left(\frac{2 \pi+\alpha)}{3 \pi+\alpha)}\right.}{\left.\frac{3 \pi}{\pi}+\alpha\right)} \)
Trigonometría Kazakhstan Mar 12, 2025
Anterior Próximo
¡Prueba Premium ahora!
¡Prueba Premium y hazle a Thoth AI preguntas de matemáticas ilimitadas ahora!
Quizas mas tarde Hazte Premium
Estudiar puede ser una verdadera lucha
¿Por qué no estudiarlo en UpStudy?
Seleccione su plan a continuación
Prima

Puedes disfrutar

Empieza ahora
  • Explicaciones paso a paso
  • Tutores expertos en vivo 24/7
  • Número ilimitado de preguntas
  • Sin interrupciones
  • Acceso completo a Respuesta y Solución
  • Acceso completo al chat de PDF, al chat de UpStudy y al chat de navegación
Básico

Totalmente gratis pero limitado

  • Solución limitada
Bienvenido a ¡Estudia ahora!
Inicie sesión para continuar con el recorrido de Thoth AI Chat
Continuar con correo electrónico
O continuar con
Al hacer clic en "Iniciar sesión", acepta nuestros términos y condiciones. Términos de Uso & Política de privacidad