Circular functions

[Louise Johnson]

Can someone please look over these and point me in the right direction if my answers are incorrect?
Thank you
Louise

#1 If tanØ=7/13 and cosØ<0,find the exact value of sinØ
Answer sinØ is neg in quadrant 3 so sinØ = -7/sqrt218

#2 State the amplitude of the curve with equation y=4 sin(2x-Pi)+5
Answer 9 I graphed this one on my calcualtor and it came out to 8.9996
Shouldn't the amplitude be 4?
Given amplitude = a from the function a sin (bx+c)+d ?

#3 Write the expression cos^2xsin^2x/tan^2x(1-sin^2x) in terms of only one circular function.
How in the heck do you get only one circular function? (1-sin^2x) = cos^2x but that leaves you with Sin^2x/tan^2x. Any help would really ......help.

Thank you
Louise

 

[skeeter]

#1 is correct

#2 ... amplitude is 4, the vertical shift does not change amplitude.

another way to look at amplitude is (largest y value - smallest y-value)/2

for your graph ... (9 - 1)/2 = 4

#3 ...
cos²xsin²x/[tan²x(1 - sin²x)]

note that 1 - sin²x = cos²x ...

cos²xsin²x/(tan²xcos²x) =

note that tan²x = sin²x/cos²x ...

cos²xsin²x/[(sin²x/cos²x)cos²x] =

cos²xsin²x/sin²x =

cos²x

 

[Louise Johnson]

Thank you Skeeter,
I have made a note from number two how to remember what the amplitude really is. Nice having someone to explain it for a change!!


In question three I guess my problem really is in this line

cos2xsin2x/[(sin2x/cos2x)cos2x] =

and cancelling the cos^2x to get the final line of cos2xsin2x/sin2x .

Thanks again for all your help!