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Need help with trig graphs

Hi, could someone explain why sin(90+x)=cosx and cosx= sin(90-x)? Also by sin(90+x), dp you move th graph along to the rhs and sin(90-x) is moving the graph to the left? Please explain as simple as posible thank you.
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Avatar for Zacken
Zacken
22
Original post by coconut64
Hi, could someone explain why sin(90+x)=cosx and cosx= sin(90-x)? Also by sin(90+x), dp you move th graph along to the rhs and sin(90-x) is moving the graph to the left? Please explain as simple as posible thank you.


sin⁡(90+x)\sin (90 +x) is moving the sine graph to the left by 90 degrees. So the maximum at sin (90) = 1 now occurs when x=0 (and you know cos x starts from 0)

sin⁡(90−x)\sin (90-x) is a reflection of the sine graph in the y/x-axis (it's odd) then translates it to the left by 90 degrees.

You may want to use the addition angle formulas to expand them out and see that they equal what they equal.
Reply 2
Avatar for coconut64
coconut64
OP
16
Original post by Zacken
sin⁡(90+x)\sin (90 +x) is moving the sine graph to the left by 90 degrees. So the maximum at sin (90) = 1 now occurs when x=0 (and you know cos x starts from 0)

sin⁡(90−x)\sin (90-x) is a reflection of the sine graph in the y/x-axis (it's odd) then translates it to the left by 90 degrees.

You may want to use the addition angle formulas to expand them out and see that they equal what they equal.


I think I get it now, thank you.

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