Difference between revisions of "Specimen Cyprus Seniors Provincial/2nd grade/Problem 2"

m (correction)
(added solution)
 
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
If <math>\alpha=sinx_{1}</math>,<math>\beta=cosx_{1}</math><math>sinx_{2}</math>, <math>\gamma=cosx_{1}cosx_{2} sinx_{3}</math> and <math>\delta=cosx_{1}cosx_{2}cosx_{3}</math> prove that <math>\alpha^2+\beta^2+\gamma^2+\delta^2=1</math>
+
If <math>\alpha=\sin x_{1}</math>,<math>\beta=\cos x_{1}\sin x_{2}</math>, <math>\gamma=\cos x_{1}\cos x_{2} \sin x_{3}</math> and <math>\delta=\cos x_{1}\cos x_{2}\cos x_{3}</math> prove that <math>\alpha^2+\beta^2+\gamma^2+\delta^2=1</math>
  
  
 
== Solution ==
 
== Solution ==
 +
Plug in the expressions for <math>\alpha</math>, <math>\beta</math>, <math>\gamma</math>, and <math>\delta</math>.
  
 +
<math>\alpha^2+\beta^2+\gamma^2+\delta^2=</math> <math>\sin^2x_1 + \cos^2x_{1}\sin^2x_{2} + \cos^2x_{1}\cos^2x_{2} \sin^2x_{3} + \cos^2x_{1}\cos^2x_{2} \cos^2x_{3}</math>
 +
 +
 +
Factor the last two terms:
 +
 +
<math>\alpha^2+\beta^2+\gamma^2+\delta^2=</math> <math>\sin^2x_1 + \cos^2x_{1}\sin^2x_{2} + (\cos^2x_{1}\cos^2x_{2})(\sin^2x_{3}+\cos^2x_{3})</math>
 +
 +
 +
Use the identity <math>\cos^2x + \sin^2x = 1</math>:
 +
 +
<math>\alpha^2+\beta^2+\gamma^2+\delta^2=</math> <math>\sin^2x_1 + \cos^2x_{1}\sin^2x_{2} + \cos^2x_{1}\cos^2x_{2}</math>
 +
 +
 +
Factor the last two terms:
 +
 +
<math>\alpha^2+\beta^2+\gamma^2+\delta^2=</math> <math>\sin^2x_1 + (\cos^2x_{1})(\sin^2x_{2}+\cos^2x_{2})</math>
 +
 +
 +
Use the identity <math>\cos^2x + \sin^2x = 1</math>:
 +
 +
<math>\alpha^2+\beta^2+\gamma^2+\delta^2=</math> <math>\sin^2x_1 + \cos^2x_{1}</math>
 +
 +
 +
Use the identity <math>\cos^2x + \sin^2x = 1</math>:
 +
 +
<math>\alpha^2+\beta^2+\gamma^2+\delta^2=</math> <math>1</math>
 +
 +
<math>\boxed{\mathbb{Q.E.D}}</math>
  
  

Latest revision as of 23:11, 22 May 2009

Problem

If $\alpha=\sin x_{1}$,$\beta=\cos x_{1}\sin x_{2}$, $\gamma=\cos x_{1}\cos x_{2} \sin x_{3}$ and $\delta=\cos x_{1}\cos x_{2}\cos x_{3}$ prove that $\alpha^2+\beta^2+\gamma^2+\delta^2=1$


Solution

Plug in the expressions for $\alpha$, $\beta$, $\gamma$, and $\delta$.

$\alpha^2+\beta^2+\gamma^2+\delta^2=$ $\sin^2x_1 + \cos^2x_{1}\sin^2x_{2} + \cos^2x_{1}\cos^2x_{2} \sin^2x_{3} + \cos^2x_{1}\cos^2x_{2} \cos^2x_{3}$


Factor the last two terms:

$\alpha^2+\beta^2+\gamma^2+\delta^2=$ $\sin^2x_1 + \cos^2x_{1}\sin^2x_{2} + (\cos^2x_{1}\cos^2x_{2})(\sin^2x_{3}+\cos^2x_{3})$


Use the identity $\cos^2x + \sin^2x = 1$:

$\alpha^2+\beta^2+\gamma^2+\delta^2=$ $\sin^2x_1 + \cos^2x_{1}\sin^2x_{2} + \cos^2x_{1}\cos^2x_{2}$


Factor the last two terms:

$\alpha^2+\beta^2+\gamma^2+\delta^2=$ $\sin^2x_1 + (\cos^2x_{1})(\sin^2x_{2}+\cos^2x_{2})$


Use the identity $\cos^2x + \sin^2x = 1$:

$\alpha^2+\beta^2+\gamma^2+\delta^2=$ $\sin^2x_1 + \cos^2x_{1}$


Use the identity $\cos^2x + \sin^2x = 1$:

$\alpha^2+\beta^2+\gamma^2+\delta^2=$ $1$

$\boxed{\mathbb{Q.E.D}}$



See also

Invalid username
Login to AoPS