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Trigonometric questions

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?
Reply 1
Try and rearrange to the form “tan 2x= ?” :smile:
Reply 2
Original post by JGLM
Try and rearrange to the form “tan 2x= ?” :smile:

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
Use some identities for cos2x and sin2x? :smile:
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
Reply 5
Original post by laurawatt
Use some identities for cos2x and sin2x? :smile:


Can you elaborate?
Reply 6
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
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)
(edited 2 years ago)
Identities mentioned earlier

Sin2x = 2sinxcosx

Cos2x = cosx^2 - sinx^2
Cos2x = 2cosx^2 - 1
Cos2x= 1-2sinx^2
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
Original post by deskochan
cos2x = -sin2x
divide both sides by cos2x

HINTS only - edit please
Reply 11
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
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.
(edited 2 years ago)
Reply 13
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
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.

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