Question

Express 6cosx-sinx in the form Rcos(x+a), where R>0 and 0 14 years ago

Answer 1

1st find the R value \[R = \sqrt{6^2 + (-1)^2} = \sqrt{37}\]


then \[\sqrt{37} (6/\sqrt{37} \cos(x) - 1/\sqrt{37} \sin(x))\]


then using the sum of 2 angle2 expansion for cos

\[\cos(x + \alpha) = \cos(x)\cos(\alpha) - \sin(x) \sin (\alpha)\]

then \[\cos(\alpha) = 6/\sqrt{37}\]
\[\sin(\alpha) = 1/\sqrt{37}\]

tan = sin/cos

then \[\tan \alpha = 1/6\]

\[\sqrt{6}\cos(x) - \sin(x) = \sqrt{37} \cos(x + 0.165)\]

than angle is expressed in radians

Answer 2

I understand everything except the first part. Why is it (-1)?

Answer 3

because the double angle expansion of cos(a + b) is cos(a)cos(b) - sin(a)sin(b)

I hope that makes sense

Answer 4

Ok, I know that's the expansion, but how did you arrive at (-1)^2?

Answer 5

its pythagoras's theorem R^2 = a^2 + b^2 a = 6 and b = -1

Answer 6

Ah OK. :) Thanks

Answer 7

Could you help with the latest one I put up?

Answer 8

I did the derivative...

Answer 9

Of which?