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# Solve Trigonometry equations with a cos x^0 + b sin x^0

Change the form a cos x0 + b sin x0 to be k cos (x – a)
With the summation formula that has been learned in school like

cos (a – b) = cos a cos b + sian a sin b
we can declare form 4 cos (x – 30)0  to form
4 cos x0 cos 300 + 4 sin x0 sin 300
Or
2√3 cos x0 + 2 sin x0.

The question is can you change the form of the equation into the previous equation
be a form of equation like 4 cos (x – 30)0 ?
For that, consider the following discussion.

For example, a cos x0 + b sin x0 = k cos (x – α)
= k (cos x0 cos α0 + sin x0 sin α0)
= k cos x0 cos α0 + k sin x0 sin α0

The right segment is identical to the left segment, so the sentence is always true for each x value. So we can equate the cos x0 and sin x0 coefficients as follows,

The values ​​k and α can be determined by the following steps,

Do the division as follows,

So that the form a cos x0 + b sin x0 can be expressed as k cos (x - α)0, with

The angle of α is determined by the sign sin α and cos α

Example 1.
State the form cos x0 + √3 sin x0 in the form k cos (x - α)0,

Sin α0 and cos α0 are both positive, then α is located in quadrant 1 so that α = 60. So, cos x0 + √3 sin x0 = 2 cos (x – 60)0.

Example 2.
Change form 4 cos x0 – 3 sin x0 to form k cos (x - α)0,

Sin α0 and cos α0 each are marked negative and positive, so α is located in quadrant IV, so
α = 360 – 36.9 = 323.1

so, sin x0 - √3 sin x0 = 5 cos(x – 323.1)0