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9. (a) Express 7 cos² x - 3 sin² x in the form k sin(x + a)² where k > 0, 0 < a < 360 - Scottish Highers Maths - Question 9 - 2023

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9.-(a)-Express-7-cos²-x---3-sin²-x-in-the-form-k-sin(x-+-a)²-where-k->-0,-0-<-a-<-360-Scottish Highers Maths-Question 9-2023.png

9. (a) Express 7 cos² x - 3 sin² x in the form k sin(x + a)² where k > 0, 0 < a < 360. (b) Hence, or otherwise, find: (i) the maximum value of 14 cos² x - 6 sin ... show full transcript

Worked Solution & Example Answer:9. (a) Express 7 cos² x - 3 sin² x in the form k sin(x + a)² where k > 0, 0 < a < 360 - Scottish Highers Maths - Question 9 - 2023

Step 1

Express 7 cos² x - 3 sin² x in the form k sin(x + a)²

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Answer

To express the above equation in the desired form, we can start by using the identity

extsin(x+a)2=extsin2xextcos2a+extcos2xextsin2a+2extsinxextcosxextcosaextsina ext{sin}(x + a)^2 = ext{sin}^2 x ext{cos}^2 a + ext{cos}^2 x ext{sin}^2 a + 2 ext{sin} x ext{cos} x ext{cos} a ext{sin} a

Next, we rewrite 7 cos² x - 3 sin² x:

  1. Apply the identity:

    • Note: Using the compound angle formula, we have:

    7extcos2x3extsin2x=7extcexto2x+(3)extsexti2x7 ext{cos}^2 x - 3 ext{sin}^2 x = 7 ext{c} ext{o}^2 x + (-3) ext{s} ext{i}^2 x

  2. We compare coefficients with the terms kextsin(x+a)2k ext{sin}(x + a)^2. Therefore, we get:

    k = rac{ ext{sqrt}(58)}{7} ext{ and } a = 113.19^ heta

Thus, kextsin(x+a)2=extsqrt(58)extsin(x+113.19)hetak ext{sin}(x + a)^2 = ext{sqrt}(58) ext{sin}(x + 113.19)^{ heta}.

Step 2

the maximum value of 14 cos² x - 6 sin x

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Answer

To find the maximum value of 14extcos2x6extsinx14 ext{cos}^2 x - 6 ext{sin} x, we substitute xx into the equation derived. Calculating the derivative will yield the maximum point:

Max value is:

rac{2}{ ext{sqrt}(58)}

Step 3

the value of x for which 0 ≤ x < 360

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Answer

We then solve for xx:

  1. Setting the maximum at x=113.19+90ightarrowx=23hetax = 113.19 + 90 ightarrow x = 23^ heta.
  2. The value occurs at x=336.80x = 336.80 due to the periodic nature of the trigonometric functions or through specifying conditions on the angle.

In conclusion, the value of xx is:

  • x113x ≈ 113 degrees for the primary solution.

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