help with proving trig stuff (1 Viewer)

[totallybord]

hi,
this is from the jones and couchman 3 unit book 1 (ill use @ as theter)
prove:
1) (sin@ + cos@)^2 +(sin@ -cos@)^2 = 2
is it just me or does it not equal to 2?
2) sin@/(1+cos@) = 1-cos@/sin@
i just dunno how to go abt it
thank you to whoever replies

 

[watatank]

hi,
this is from the jones and couchman 3 unit book 1 (ill use @ as theter)
prove:
1) (sin@ + cos@)^2 +(sin@ -cos@)^2 = 2
is it just me or does it not equal to 2?

LHS = (sin@ + cos@)^2 + (sin@ -cos@)^2

= (sin@)^2 + (cos@)^2 + 2sin@cos@ + (sin@)^2 + (cos@)^2 - 2sin@cos@

= (sin@)^2 + (cos@)^2 + (sin@)^2 + (cos@)^2

ok now i dont know if you guys have learned trig identities yet, but (sin@)^2 + (cos@)^2 = 1

so its just 2*(sin@)^2 + (cos@)^2) = 2*1 =2

= 2

 

[watatank]

2) sin@/(1+cos@) = 1-cos@/sin@

i'd use proof by contradition method. assume the two are not equal. if you can prove they are in fact equal then your assuption is wrong so the opposite must be true (they are equal)

so assume sin@/(1+cos@) =/= 1-cos@/sin@

cross multiply

(1+cos@)(1-cos@) = sin@^2

1-(cos@)^2 = sin@^2

(cos@)^2 + sin@^2 = 1

this is a true statement

so your assumption was false, therefore

sin@/(1+cos@) = 1-cos@/sin@

 

[totallybord]

ok...i admit the first question was stupid for me not to be able to do but yeah is there another way to do the 2nd one?
thank you

 

[watatank]

i can assure you its the easiest way to do that question. get someone to explain the proof by contradiction method, i think that's what you are having trouble understanding...

 

[SoulSearcher]

For the second question, multiply the LHS by (1-cos@)/(1-cos@), the result should come out from there.