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

Change the form a cos x

^{0}+ b sin x^{0}^{ }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 x

^{0}cos 30^{0}+ 4 sin x^{0}sin 30^{0}
Or

2√3 cos x

^{0}+ 2 sin x^{0}.
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 x

^{0}+ b sin x^{0}= k cos (x – α)
= k (cos x

^{0 }cos α^{0}+ sin x^{0}sin α^{0})
= k cos x

^{0 }cos α^{0}+ k sin x^{0}sin α^{0}^{}

^{0}and sin x

^{0}coefficients as follows,

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

Square and add up

Do the division as follows,

So that the form a cos x

^{0}+ b sin x^{0}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 x

^{0}+ √3 sin x^{0}in the form k cos (x - α)^{0},
Answer:

Sin α

^{0}and cos α^{0}are both positive, then α is located in quadrant 1 so that α = 60. So, cos x^{0}+ √3 sin x^{0}= 2 cos (x – 60)^{0}.**Example 2.**

Change
form 4 cos x

^{0}– 3 sin x^{0}to form k cos (x - α)^{0},
Answer:

Sin α

^{0}and cos α^{0}each are marked negative and positive, so α is located in quadrant IV, so
α = 360 – 36.9 = 323.1

so, sin x

^{0}- √3 sin x^{0}= 5 cos(x – 323.1)^{0}SUBSCRIBE TO OUR NEWSLETTER

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