A-Level Edexcel Maths Pure Trigonometric Functions: 4. (a) Show that the equation

4. (a) Show that the equation 3 sin² θ - 2 cos² θ = 1 can be written as 5 sin² θ = 3 - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 2

Question 6

4. (a) Show that the equation 3 sin² θ - 2 cos² θ = 1 can be written as 5 sin² θ = 3. (b) Hence solve, for 0° < θ < 360°, the equation 3 sin² θ - 2 cos²... show full transcript

Worked Solution & Example Answer:4. (a) Show that the equation 3 sin² θ - 2 cos² θ = 1 can be written as 5 sin² θ = 3 - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 2

Step 1

Show that the equation 3 sin² θ - 2 cos² θ = 1 can be written as 5 sin² θ = 3

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Answer

To show that we can rewrite the equation, start with the original equation:

3 ext{ sin}^2 θ - 2 ext{ cos}^2 θ = 1

Using the identity $ext{cos}^2 θ = 1 - ext{sin}^2 θ$ , substitute for $ext{cos}^2 θ$ :

3 ext{ sin}^2 θ - 2(1 - ext{sin}^2 θ) = 1

Expanding the equation gives:

3 ext{ sin}^2 θ - 2 + 2 ext{ sin}^2 θ = 1

Combining like terms results in:

5 ext{ sin}^2 θ - 2 = 1

Adding 2 to both sides leads to:

5 ext{ sin}^2 θ = 3

This completes the showing process.

Step 2

Hence solve, for 0° < θ < 360°, the equation 3 sin² θ - 2 cos² θ = 1

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Answer

Using the rewritten equation from part (a):

5 ext{ sin}^2 θ = 3

Divide both sides by 5:

ext{ sin}^2 θ = rac{3}{5}

Taking the square root gives:

ext{ sin} θ = rac{ ext{\sqrt}3}{ ext{\sqrt}5} = rac{ ext{\sqrt}15}{5}

Now, the possible angles are:

First Quadrant: $heta = ext{sin}^{-1}igg( rac{ ext{√}15}{5}igg)$
Second Quadrant: $heta = 180° - ext{sin}^{-1}igg( rac{ ext{√}15}{5}igg)$

Calculating the angles:

First Quadrant: Approximate value:

$heta ≈ 36.9° ext{ (to 1 decimal place)}$
Second Quadrant:

$heta ≈ 180° - 36.9° ≈ 143.1° ext{ (to 1 decimal place)}$

Thus, the solutions are approximately:

$heta ≈ 36.9°, 143.1°$

4. (a) Show that the equation 3 sin² θ - 2 cos² θ = 1 can be written as 5 sin² θ = 3 - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 2

Worked Solution & Example Answer:4. (a) Show that the equation 3 sin² θ - 2 cos² θ = 1 can be written as 5 sin² θ = 3 - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 2

Show that the equation 3 sin² θ - 2 cos² θ = 1 can be written as 5 sin² θ = 3

Hence solve, for 0° < θ < 360°, the equation 3 sin² θ - 2 cos² θ = 1

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