Question

Need help with Trigonometry -- Sum & Differences Formulas. Please see the attachment for the problem. Thanks!

Answer 1

I know that I have to use the second formula of the two given. The part that begins to confuse me a bit is arctan (a/b)

Answer 2

ok arc tan is given to u so u can find the value of a and b
7cos (x-pi/4)
c=pi/4 agreed?

Answer 3

@MoonlitFate

Answer 4

Yes,

Answer 5

A = 7 and B = 1, at least I think. Am I right, @nubeer ?

Answer 6

ok good.. now
hmm maybe .. i dont know i havent solved it.. this can be solution as u want to make 7 but is it really 7.. we can check that
arctan (a/b) = pi/4
a/b = tan(pi/4)
a/b = 1
a=b
as u can see A and B have the same value.. so it is not 7 and 1..

Answer 7

Hmm. That' just confusing. :/

Answer 8

u just have to find A and B.. hit and trial can be your last option .. well tell me what part u dont get..???
we agreed that c=pi/4
c=arctan(a/b)
pi/4 = arctan (a/b)
tan(pi/4) = a/b..
do u get this much?

Answer 9

Yes, I get that. :)

Answer 10

ok tan(pi/4) = 1
so
1 = a/b
a=b
now see the formula.. do u agree that
(a^2+b^2)^0.5 = 7?

Answer 11

Yes, that makes sense.

Answer 12

\[(a ^{2}+b ^{2})^{\frac{ 1 }{ 2 }\times2} = 7^{2}\]

Answer 13

remember a=b
so
b^2 +b^2 =49
2b^2 = 49
b^2 =49/2
b=(49/2)^0.5

Answer 14

a = b
so u found everything i guess rest is easy :)

Answer 15

from the expression B=1,C=pi/4 and a^2+b^2=49
arc tana/b=pi/4 implies a/b=1
so a=b gives a=7/rt2,b=7/rt2
we get the answer 7cos(bt-pi/4)=7/rt2sint+7/rt2cost