# Trigonometric questions

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#1
I must be having a birdbrain day today.

Question: Find the values of x in the interval 0<x<360 for which Sin2x + Cos2x =0

I have attempted to square and then sub in using the identity: sin^2x + cos^2x = 1, but this just ends with 1 as both sin and cos just cancel each other out, and I don’t see a clear way of using tan x=sin x/ cos x .

Is there something that I am missing?
1
1 month ago
#2
Try and rearrange to the form “tan 2x= ?”
1
#3
(Original post by JGLM)
Try and rearrange to the form “tan 2x= ?”
I have attempted but if I move cos x over, to become -cos x there’s no way to move sin x underneath, or to what I can see at least see anyway
0
1 month ago
#4
Use some identities for cos2x and sin2x?
1
1 month ago
#5
Find the values of x in the interval 0<x<360 for which Sin2x + Cos2x =0
cos2x = -sin2x
-tan2x =1, 1+tan2x =0
0<x<360
0<2x<720
tan0<tan2x<tan720
then it may get the answer
1
#6
(Original post by laurawatt)
Use some identities for cos2x and sin2x?
Can you elaborate?
0
#7
(Original post by deskochan)
Find the values of x in the interval 0<x<360 for which Sin2x + Cos2x =0
cos2x = -sin2x
-tan2x =1, 1+tan2x =0
0<x<360
0<2x<720
tan0<tan2x<tan720
then it may get the answer
Can you show me how you took cos2x=-sin2x to be -tan2x=1? ? I am only aware of the identity tan x=sin x/cos x. I must be missing something
0
1 month ago
#8
(Original post by KingRich)
Can you show me how you took cos2x=-sin2x to be -tan2x=1? ? I am only aware of the identity tan x=sin x/cos x. I must be missing something
cos2x = -sin2x (divided both sides by cos2x)
Last edited by deskochan; 1 month ago
0
1 month ago
#9
Identities mentioned earlier

Sin2x = 2sinxcosx

Cos2x = cosx^2 - sinx^2
Cos2x = 2cosx^2 - 1
Cos2x= 1-2sinx^2
0
1 month ago
#10
(Original post by deskochan)
Find the values of x in the interval 0<x<360 for which Sin2x + Cos2x =0
cos2x = -sin2x
We only give hints - edit your post
0
1 month ago
#11
(Original post by deskochan)
cos2x = -sin2x
divide both sides by cos2x
0
#12
(Original post by deskochan)
cos2x = -sin2x (divided both sides by cos2x)
Okay, I see what you did. According to the book I am working from the first x value is 67.5 degrees, therefore the solution above doesn't work out as it gives an x value of -22.5
0
1 month ago
#13
(Original post by KingRich)
Okay, I see what you did. According to the book I am working from the first x value is 67.5 degrees, therefore the solution above doesn't work out as it gives an x value of -22.5
In the case of the negative value of the tangent region, we can consider two quadrants (II and IV) such that for quadrant II (sine region) is 90+(-22.5) = 67.5 and quadrant IV (cosine region) is 360+(-22.5) = 337.5.
Last edited by deskochan; 1 month ago
1
#14
(Original post by deskochan)
In the case of the negative value of the tangent region, we can consider two quadrants (II and IV) such that quadrant II (sine region) is 90+(-22.5) = 67.5 and quadrant IV (cosine region) is 360+(-22.5) = 337.5.
Superstar! I need to stop for the day because this seems so simple now that you have explained it. Thank you for this
0
1 month ago
#15
(Original post by KingRich)
Superstar! I need to stop for the day because this seems so simple now that you have explained it. Thank you for this
You are welcome.
0
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