# tricky trig equations

• Sep 12th 2009, 05:58 PM
requal
tricky trig equations
Express 3cosx-4sinx in the form Rcos(x+a) for some R>0 and 0<a<90 giving the value of a correct to the nearest minute

Hence solve the equation 3cosx-4sinx=5 for in the domain of [0,360]
• Sep 12th 2009, 08:25 PM
songoku
Hi requal

I'm not sure what you mean by "nearest minute". Maybe it means nearest degree?

Let :
p cos x - q sin x = R cos (x+a)

$R=\sqrt{p^2+q^2}$

Then, expand RHS and compare it with LHS to find cos(a) and sin(a). Hence, find a :)
• Sep 12th 2009, 10:50 PM
mr fantastic
Quote:

Originally Posted by songoku
Hi requal

I'm not sure what you mean by "nearest minute". Maybe it means nearest degree?

[snip]

60 minutes = 1 degree: $60 \, ' = 1^0$.

60 seconds = 1 minute: $60 \, '' = 1 \, '$
• Sep 13th 2009, 04:00 AM
songoku
Hi mr fantastic

Oh I see. Find (a) in degree first then convert it to minute.

Thanks
• Sep 14th 2009, 02:57 AM
requal
sorry, to bump but I'm still confused. Why does R equal the square root of p^2+q^2
• Sep 14th 2009, 04:13 AM
songoku
Hi requal

p cos x - q sin x = R cos (x+a)
p cos x - q sin x = R cos(x) cos(a) - R sin(x) sin(a)

Comparing coefficient :
p cos x = R cos(x) cos(a)
p = R cos(a) .........................(1)

q sin x = R sin(x) sin(a)
q = R sin(a)...........................(2)

Square (1) and (2) and add them :
p^2 = R^2 [cos(a)]^2
q^2 = R^2 [sin(a)]^2
--------------------------- +
p^2 + q^2 = R^2