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i need help as i dont get what method or topic is this  but i am meant to be able to  do this question please explain the answers or tell me the topics names

 Apr 17, 2019
 #1
avatar+8341 
+2

This is trigonometry.

4(a)

\(\text{As }\sin x = \cos\left(\dfrac{\pi}{2}-x\right),\\ \sin \left(\dfrac{\pi}{3}\right) = \cos\left(\dfrac{\pi}{2}-\dfrac{\pi}3\right)=\cos\left(\pi\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\right)=\cos\left(\pi\left(\dfrac{1}{6}\right)\right) =\cos\left(\dfrac{\pi}{6}\right)\\ \sin \left(\dfrac{\pi}{3}\right) = \cos\left(\dfrac{\pi}{k}\right)\\ \cos\left(\dfrac{\pi}{6}\right) = \cos\left(\dfrac{\pi}{k}\right)\\ \text{Comparing the left hand side and the right hand side of the equation,}\\ k =6\)

 Apr 18, 2019
 #4
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thank you, but can you please explain the first line and where you got that from, apart from that part it all makes sense

YEEEEEET  Apr 18, 2019
 #6
avatar+8341 
0

You can derive this from a right triangle.

MaxWong  Apr 18, 2019
 #2
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+1

4(b)

\(\cos\left(2x-\dfrac{5\pi}{6}\right) = \sin\left(\dfrac{\pi}{3}\right)\\ \cos\left(2x-\dfrac{5\pi}{6}\right) = \cos\left(\dfrac{\pi}{6}\right)\\ \text{As the cosine function is periodic with period }2\pi\text{, }\\ \cos\left(2x-\dfrac{5\pi}{6}\right) = \cos\left(\dfrac{\pi}{6}+2n\pi\right)\\ \text{As }\cos x = \cos\left(-x\right)\text{, }\\ \cos\left(2x-\dfrac{5\pi}{6}\right) = \cos\left(-\dfrac{\pi}{6}+2n\pi\right)\\ \text{Comparing the left hand side and the right hand side of the equation, }\\ 2x-\dfrac{5\pi}{6} = \dfrac{\pi}{6} + 2n\pi \text{ or } 2x-\dfrac{5\pi}{6} = -\dfrac{\pi}{6} + 2n\pi\\ 2x = \pi+2n\pi\text{ or }2x=\dfrac{2\pi}{3}+2n\pi\\ x = \dfrac{\pi}{2}+n\pi\text{ or }x=\dfrac{\pi}{3}+n\pi\)

 Apr 18, 2019
 #5
avatar+845 
+1

could you explain how the 2n appeared please

YEEEEEET  Apr 18, 2019
 #7
avatar+8341 
+1

Actually, (n)(2pi), not (2n)(pi). After n periods, which is n(2pi), the value of the function is still the same.

MaxWong  Apr 18, 2019
 #3
avatar+8341 
+2

4(c)

\(\text{As shown above in 4(b), } \cos\left(2x-\dfrac{5\pi}{6}\right) = \sin\left(\dfrac{\pi}{3}\right)\implies x = n\pi+\dfrac{\pi}{2}\text{ or }x=n\pi+\dfrac{\pi}{3}\\ \text{As tan} x \text{ has a period of }\pi,\\ \tan x = \tan\left(\dfrac{\pi}{2}\right)\text{ or }\tan x = \tan\left(\dfrac{\pi}{3}\right)\\ \text{But considering the fact that}\tan x \text{ has an asymptote at } x = \dfrac{\pi}{2},\text{ the only finite value is }\tan\left(\dfrac{\pi}{3}\right).\\ \text{The required value is }\tan\left(\dfrac{\pi}{3}\right) = \sqrt3\)

 Apr 18, 2019

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