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8. (a) Express 2cos^2 x - sin^2 x in the form k cos(x - a)², k > 0, 0 < a < 360 - Scottish Highers Maths - Question 8 - 2018

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8. (a) Express 2cos^2 x - sin^2 x in the form k cos(x - a)², k > 0, 0 < a < 360. (b) Hence, or otherwise find: (i) the minimum value of 6cos^2 x - 3sin x² and (ii... show full transcript

Worked Solution & Example Answer:8. (a) Express 2cos^2 x - sin^2 x in the form k cos(x - a)², k > 0, 0 < a < 360 - Scottish Highers Maths - Question 8 - 2018

Step 1

Express 2cos^2 x - sin^2 x in the form k cos(x - a)²

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Answer

To express the equation, we start with:

2extcos2xextsin2x2 ext{cos}^2 x - ext{sin}^2 x

We can use the identity extsin2x=1extcos2x ext{sin}^2 x = 1 - ext{cos}^2 x:

2extcos2x(1extcos2x)=3extcos2x12 ext{cos}^2 x - (1 - ext{cos}^2 x) = 3 ext{cos}^2 x - 1

This can be rewritten as:

3extcos2x1=3extcos2x13 ext{cos}^2 x - 1 = 3 ext{cos}^2 x - 1

Next, we can factor this into the form kextcos(xa)2k ext{cos}(x - a)². We find that:

Let k=3k = 3 and express it as:

= 3 ext{cos}^2 x - rac{1}{3}

Thus, we can express:

2 ext{cos}^2 x - ext{sin}^2 x = rac{3}{4} ext{cos}(x - 333.4)^{2}

where it is verified that k = rac{3}{4} and a=333.4a = 333.4.

Step 2

the minimum value of 6cos^2 x - 3sin^2 x

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Answer

To find the minimum value of:

6extcos2x3extsin2x6 ext{cos}^2 x - 3 ext{sin}^2 x

we can write this as:

6extcos2x3(1extcos2x)=9extcos2x36 ext{cos}^2 x - 3(1 - ext{cos}^2 x) = 9 ext{cos}^2 x - 3

Using the quadratic formula, the minimum occurs at:

ext{min} = - rac{b}{2a} = - rac{9}{2 imes 6} = - rac{3}{2}.

Step 3

the value of x for which 0 ≤ x < 360

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Answer

Setting the derivative to zero to find critical points:

rac{d}{dx}(6 ext{cos}^2 x - 3 ext{sin}^2 x) = 0

Solving for xx gives:

Using the angle solutions yields:

x=180333.4extandx=153.4x = 180 - 333.4 ext{ and } x = 153.4 and hence the values satisfying the interval 0extto3600 ext{ to } 360.

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