How do you write y=5cos(7x)+2sin(7x) using only one cosine function?

Let 5 cos (7x) + 2 sin (7x) = a cos (bx + c)

= a [cos(bx) cos c - sin (bx) sin c]

= (a cos c) cos (bx) - (a sin c) sin (bx)

We require

b = 7

a cos c = 5

- a sin c = 2

cos c = 5/a

sin c = -2/a

so (5/a)^2 + (-2/a)^2 = 1

5^2 + (-2)^2 = a^2

a = ± sqrt 29

a cos (bx + c) = - a cos (bx + c + pi)

so we need only consider 1 value of a.

If a = sqrt 29 then cos c is positive and sin c is negative so c is in quadrant 4.

c = - cos^(-1) (5/sqrt 29)

So

5 cos (7x) + 2 sin (7x) =

(sqrt 29) cos [(7x) - cos^(-1) (5/sqrt 29)]

= approx. 5.3852 cos [(7x) - 0.3805]

.

before reading please ANSWER my question ... anybody?

/question/index?qid=20161...

y = 5 cos(7x) + 2 sin(7x)

trig identities:

sin(x + π/2) = cos(x)

sin(x) = cos(x - π/2)

a sin(x) + b sin(x + Θ) = c sin(x + Φ)

c = √(a^2 + b^2 + 2ab cos Θ), Φ = arctan(b sin Θ / (a + b cos Θ))

y = 2 sin(7x) + 5 sin(7x + π/2)

a = 2, b = 5, Θ = π/2

c = √29, Φ = 1.19029 radians

y = √29 sin(7x + 1.19029)

y = √29 cos(7x - 0.380506)

hope this is CLEAR

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and please answer mine

sin ( a + b ) = sin a cos b + cos a sin b...use...y = √29 [ 2/√29 sin (7x) + 5 / √29 cos (7x) ]

......tan b = 5 / 2....a = 7x....sorry , you wanted cosine...cos ( a - b ) = cos a cos b + sin a sin b...use it