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,

Square and add up

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,

Answer:

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,

Answer:

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

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