Photo AI

6. (a) (i) By writing $3\theta = (2\beta + \theta)$, show that $$\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta.$$ (ii) Hence, or otherwise, for $0 < \theta < \frac{\pi}{3}$, solve $$8 \sin^3 \theta - 6 \sin \theta + 1 = 0.$$ Give your answers in terms of $\pi$ - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 2

Question icon

Question 7

6.-(a)-(i)-By-writing-$3\theta-=-(2\beta-+-\theta)$,-show-that-$$\sin-3\theta-=-3-\sin-\theta---4-\sin^3-\theta.$$---(ii)-Hence,-or-otherwise,-for-$0-<-\theta-<-\frac{\pi}{3}$,-solve-$$8-\sin^3-\theta---6-\sin-\theta-+-1-=-0.$$---Give-your-answers-in-terms-of-$\pi$-Edexcel-A-Level Maths Pure-Question 7-2009-Paper 2.png

6. (a) (i) By writing $3\theta = (2\beta + \theta)$, show that $$\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta.$$ (ii) Hence, or otherwise, for $0 < \theta < \fra... show full transcript

Worked Solution & Example Answer:6. (a) (i) By writing $3\theta = (2\beta + \theta)$, show that $$\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta.$$ (ii) Hence, or otherwise, for $0 < \theta < \frac{\pi}{3}$, solve $$8 \sin^3 \theta - 6 \sin \theta + 1 = 0.$$ Give your answers in terms of $\pi$ - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 2

Step 1

By writing $3\theta = (2\beta + \theta)$, show that $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$.

96%

114 rated

Answer

To show that sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta, we start with the identity for the sine of a sum:

sin⁡(3θ)=sin⁡(2θ+θ)=sin⁡(2θ)cos⁡(θ)+cos⁡(2θ)sin⁡(θ).\sin(3\theta) = \sin(2\theta + \theta) = \sin(2\theta)\cos(\theta) + \cos(2\theta)\sin(\theta).

Using the double angle formulas, we have:

sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)\sin(2\theta) = 2 \sin(\theta) \cos(\theta)
cos⁡(2θ)=1−2sin⁡2(θ).\cos(2\theta) = 1 - 2 \sin^2(\theta).

Substituting these into our equation gives:

sin⁡(3θ)=(2sin⁡(θ)cos⁡(θ))cos⁡(θ)+(1−2sin⁡2(θ))sin⁡(θ).\sin(3\theta) = (2 \sin(\theta) \cos(\theta))\cos(\theta) + (1 - 2 \sin^2(\theta))\sin(\theta).

Simplifying this results in:

sin⁡(3θ)=2sin⁡(θ)(cos⁡2(θ))+sin⁡(θ)−2sin⁡3(θ).\sin(3\theta) = 2\sin(\theta)(\cos^2(\theta)) + \sin(\theta) - 2\sin^3(\theta).

Now, using the Pythagorean identity cos⁡2(θ)=1−sin⁡2(θ)\cos^2(\theta) = 1 - \sin^2(\theta), we can replace cos⁡2(θ)\cos^2(\theta):

sin⁡(3θ)=2sin⁡(θ)(1−sin⁡2(θ))+sin⁡(θ)−2sin⁡3(θ).\sin(3\theta) = 2\sin(\theta)(1 - \sin^2(\theta)) + \sin(\theta) - 2\sin^3(\theta).

Distributing the terms leads to:

sin⁡(3θ)=2sin⁡(θ)−2sin⁡3(θ)+sin⁡(θ)−2sin⁡3(θ)=3sin⁡(θ)−4sin⁡3(θ).\sin(3\theta) = 2\sin(\theta) - 2\sin^3(\theta) + \sin(\theta) - 2\sin^3(\theta) = 3\sin(\theta) - 4\sin^3(\theta).

Step 2

Hence, or otherwise, for $0 < \theta < \frac{\pi}{3}$, solve $8 \sin^3 \theta - 6 \sin \theta + 1 = 0$.

99%

104 rated

Answer

To solve the equation 8sin⁡3θ−6sin⁡θ+1=08 \sin^3 \theta - 6 \sin \theta + 1 = 0, we can let x=sin⁡θx = \sin \theta. The equation becomes:

8x3−6x+1=0.8x^3 - 6x + 1 = 0.

We can use the Rational Root Theorem to test possible roots. Testing x=−12x = -\frac{1}{2}:

Plugging in: 8(−12)3−6(−12)+1=−1+3+1=3≠0.8(-\frac{1}{2})^3 - 6(-\frac{1}{2}) + 1 = -1 + 3 + 1 = 3 \neq 0.

Testing x=12x = \frac{1}{2}: 8(12)3−6(12)+1=8(18)−3+1=1−3+1=−1≠0.8(\frac{1}{2})^3 - 6(\frac{1}{2}) + 1 = 8(\frac{1}{8}) - 3 + 1 = 1 - 3 + 1 = -1 \neq 0.

After testing several roots, we find that x=12x = \frac{1}{2} works. So sin⁡θ=12\sin \theta = \frac{1}{2} gives:

θ=π6.\theta = \frac{\pi}{6}.

Thus, since 0<θ<π30 < \theta < \frac{\pi}{3}, this is a valid solution.

Step 3

Using $\sin(\theta - \alpha) = \sin \theta \cos \alpha - \cos \theta \sin \alpha$, show that $\sin 15^\circ = \frac{1}{4}(\sqrt{6} - \sqrt{2})$.

96%

101 rated

Answer

To show that sin⁡15∘=14(6−2)\sin 15^\circ = \frac{1}{4}(\sqrt{6} - \sqrt{2}), we can use the angle difference formula:

sin⁡(θ−α)=sin⁡θcos⁡α−cos⁡θsin⁡α.\sin(\theta - \alpha) = \sin \theta \cos \alpha - \cos \theta \sin \alpha.

Let θ=45∘\theta = 45^\circ and α=30∘\alpha = 30^\circ:

sin⁡15∘=sin⁡(45∘−30∘)=sin⁡45∘cos⁡30∘−cos⁡45∘sin⁡30∘.\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ.

We know:

  • sin⁡45∘=22\sin 45^\circ = \frac{\sqrt{2}}{2},
  • cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2},
  • cos⁡45∘=22\cos 45^\circ = \frac{\sqrt{2}}{2},
  • sin⁡30∘=12.\sin 30^\circ = \frac{1}{2}.

Substituting these values yields:

sin⁡15∘=(22)(32)−(22)(12)=64−24=14(6−2).\sin 15^\circ = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{1}{4}(\sqrt{6} - \sqrt{2}).

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;